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9.2 – Space
9.2.1 – The earth has a gravitational field that exerts a force on
objects both on it and around it.
9.2.1.1 – Define weight as the force on an object due to a gravitational field.
On Earth, objects have a weight force equivalent to their mass time the acceleration due
to gravity.
W = mg
Where:
 W = weight (N)
 m = mass (kg)
 g = acceleration due to gravity (ms-2)
On Earth, the value for g is taken to be 9.8 ms-2. (There are subtle variations possible
which will be discussed later). On other planets etc. g becomes a different value and can
be calculated using the equation:
mp
“g = G
”
r  2p
Where:
 g = acceleration due to gravity (ms-2)
 G = gravitational constant (6.67 x 10-11 Nm2kg-2)
 r2p = radius of the planet (m)
 mp = mass of planet (kg)
Calculations
1) If Dean’s has a mass of 95.0 kg. Calculate his weight on Earth and on the moon
where the g value is approximately 1/6 that of Earth.
W = mg
= 95 x 9.8
= 931 N on Earth
W = mg
= 95 x (9.8 x 1/6)
= 1.6 N on the Moon
2) Lee has a weight of 15500 N on an unknown planet while his weight on Earth is
647 N. Calculate his mass and the g on this unknown planet.
m = W/g
= 647/9.8
= 66.0 kg is Lee’s mass
g = W/m
= 15500/66
= 234 ms-2 is the gravity on the unknown planet.
a = 34.0 ms-2
Fr
Fg
Scales
Fnet = Fr - Fg
Ma = Fr – mg
70 x 34 = Fr – 70 x 9.8
Fr = 70 x (34 + 9.8)
= 3066 N is the apparent weight
= 312 kgf is the apparent mass
In orbit FR = 0 because of a freefall situation where there is zero g.
Landing on the moon, decelerating at 5.0 ms-2:
Stationary on the moon:
Fnet = Fr - Fg
Ma = Fr – mg
Fr = m x (a + g)
= 60 x (5 + 1.6)
= 396 N is the apparent weight
= 40.4 kgf is the apparent mass
Fnet = Fg - Fr
Ma = mg - Fr
Fr = mg – Ma
= 60 x 1.6 – 60 x 0
= 96 N is the apparent weight
= 9.8 kgf is the apparent mass
9.2.1.2 – Explain that a change in gravitational potential energy is related to work
done.
On Earth:
GPE = mgh
Where: GPE = Gravitational Potential Energy (J)
m = mass (kg)
g = acceleration due to gravity
h = height (m)
1
70.0kg
20.0m
2
35.0m
3
70.0m
57.0m
4
0.0m
5
Situation 1: GPE = mgh
= 70 x 9.8 x 70
= 48020 J
= 48000 J (to 3 sig fig)
= 4.8 x 104 J
Total Energy = GPE + KE
Situation 2: GPE = mgh
= 70 x 9.8 x 50
= 34300 J
ET = GPE + KE
48020 = 34300 + KE
KE = 13720 J
KE = ½ x mv2
13720 = ½ x 70 x v2
v2 = 13720/35
v = √392
= 19.8 ms-1
Situation 3: GPE = mph
= 70 x 9.8 x 35
= 24010 J
ET = GPE + KE
48020 = 24010 + KE
KE = 24010
KE = ½ x mv2
24010 = ½ x 70 x v2
v = √24010/35
= 26.2 ms-1
On earth if we wish to lift an object up higher, we must do work on it.
W = Fs
Where: W = work (J)
F = force (N)
s = distance (m)
Here F must be equal to or greater than Fg (weight) = mg & s = height lifted = h
Therefore: W = mgh or GPE = mgh
In this case we chose the ground as our zero level where Ep (potential energy) = 0 J. When
objects are raised to say point x the value of Ep at this point is greater than zero.
In Space: On the larger scale we need to look at the situation differently. Using the of
universal gravitation (9.2.3), the force of attraction between a planet and an object will drop
to zero only at infinity (∞). For this reason the ∞ point is chosen to represent zero potential
energy (at ∞, Ep = 0).
Due to this choice, since work must be done on an object to move it from point x to ∞ so
that it gains Ep. This means that all other points have negative Ep values.
∞
x
Figure 1.5
9.2.1.3 – Define gravitational potential energy as the work done to move an object
from a very large distance away to a point in a gravitational field.
Ep  G
m1 m2
r
Where: Ep = Gravitational Potential Energy (J)
G = Gravitational Constant (6.67 x 10-11)
m1 = Mass of significant object (kg)
m2 = Mass of other object (kg)
r = distance between objects (m)
Problem:
mearth = 5.97 x 1024 kg
mmoon = 7.35 x 1022 kg
msun = 1.99 x 1030 kg
rearth – moon = 3.84 x 108 m
rearth – sun = 1.50 x 1011 m
(a) Calculate the gravitational potential energy of:
i. The moon within the earth’s gravitational field
ii. The earth within the suns gravitational field
iii. Kate the astronaut’s gravitational potential energy from the earth if
she is halfway to the moon if she weighs 100 kg.
i.
Ep  G
me  mm
r
Ep  6.67  10^ 11
(5.97  10^ 24)  (7.35  10^ 22)
3.84  10^8
= -7.62 x 1028 J
ii.
Ep  G
me  ms
r
Ep  6.67  10^ 11
(5.97  10^ 24)  (1.99  10^30)
1.50  10^11
= -5.28 x 1033 J
iii.
Ep  G
me  mk
r
Ep  6.67  10^ 11
= -2.07 x 108 J
(5.97  10^ 24)  100
(3.84  10^8)  0.5
9.2.2 – Many factors have to be taken into account to achieve a
successful rocket launch, maintain a stable orbit and return to
Earth.
9.2.2.1 – Describe the trajectory of an object undergoing projectile motion within
the Earth’s gravitation field in terms of horizontal and vertical components.
An object thrown into the air near to the Earth’s surface will follow a parabolic path. Its
horizontal component’s velocity will remain constant (neglecting air resistance), while
the vertical component will undergo a constant negative acceleration of 9.8ms − 2. The
overall velocity is the sum of these components. These two components are vectors and
can be summed by adding them head to tail, or as they are vertical and horizontal, using
Pythagoras theorem.
9.2.2.2 – Describe Galileo’s analysis of projectile motion.
Galileo was the first to propose that all projectiles took a parabolic path. He also
proposed that two objects (a stone and a feather) would reach the ground at the same time
if they were dropped from the same height at the same time provided there was no air
resistance. He also said that the horizontal and vertical components of projectile motion
occurred both independently and simultaneously.
An alternate interpretation:
Galileo’s analysis of projectile motion is that the motion of a body can be split into two
components, horizontal and vertical. Where the horizontal motion will have no
acceleration acting on it, and the vertical motion will have acceleration due to gravity
acting on it.
9.2.2.3 – Explain the concept of escape velocity in terms of the:
 Gravitational constant
 Mass and radius of the planet
By considering the kinetic and gravitational potential energy of a projectile, a
mathematical formula can be derived.
vescape =
2Gm( planet )
r ( planet )
Where: vescape = Escape velocity (ms-1)
G = Gravitational constant (6.67 x 10-11)
mplanet = Mass of planet (kg)
rplanet = Radius of planet (m)
Example: Using previously supplied data find the escape velocity for:
a) Earth
2Gm( planet )
vescape =
r ( planet )
(2  6.67  10^ 11)  5.97  10^ 24
6.38  10^6
= 11172.6 ms-1
= 11200 ms-1 (3 sig fig)
b) Mercury
2Gm( planet )
vescape =
r ( planet )
=
(2  6.67  10^ 11)  3.3  10^ 23
2440000
= 4247.6 ms-1
= 4250 ms-1 (3 sig fig)
c) Venus
2Gm( planet )
vescape =
r ( planet )
=
(2  6.67  10^ 11)  4.9  10^ 24
6052000
-1
= 10392.7 ms
= 10400 ms-1 (3 sig fig)
=
d) Io
vescape =
2Gm( planet )
r ( planet )
(2  6.67  10^ 11)  8.9  10^22
1815000
= 2557.6 ms-1
= 2560 ms-1 (3 sig fig)
e) Jupiter
2Gm( planet )
vescape =
r ( planet )
=
(2  6.67  10^ 11)  (5.97  10^ 24  318)
(6.38  10^6  11)
-1
= 60072.0 ms
= 60100 ms-1 (3 sig fig)
=
Since ‘G’ is a constant, although it determines the escape velocity it cannot change it.
The planets mass and radius can, however, alter the required velocity needed to escape
the planet’s gravitational field. As m increases, the escape velocity increases and as r
increases the escape velocity will decrease. Also as density increases, escape velocity
also increases.
9.2.2.4 – Outline Newton’s concept of escape velocity.
Isaac Newton wrote that it should be possible to launch a projectile fast enough so that it
achieves an orbit around the Earth. Through this principle it was found that escape
velocity is the initial velocity required by a projectile to rise vertically and just escape the
gravitational field of a planet. If this specified escape velocity is exceeded slightly then
the object will follow an elliptical orbit around the Earth. If the specified velocity is
exceeded further still, then the object will follow a parabolic or hyperbolic path away
from the Earth. This is the manner in which space probes depart the Earth and head off
into space.
9.2.2.5 – Identify why the term ‘g forces’ is used to explain the forces acting on an
astronaut during launch.
Your body is a mass lying somewhere in a gravitational field, and therefore experiences a
true weight. The sensation of weight that you feel derives from your apparent weight,
which is equal to the sum of the contact forces resisting you true weight. This includes
the normal reaction force of the floor and the thrust of a rocket engine. The term ‘g force’
is used to express a person’s apparent weight as a multiple of his/her normal true weight.
Hence:
g force =
apparent _ weight
normal _ true _ weight
(Figure 2.18)
During the launch of an astronaut into space the astronaut’s body is exerting a downward
weight force on the floor, and the floor meets this with an upward reaction force equal to
mg. In addition, the floor is exerting an upward accelerating force equal to ma.
Therefore:
mg  ma
9 .8 m
ga
=
9.8
g force =
Where: g = acceleration due to gravity at altitude (ms-2) (+ve away from earth, -ve
towards earth)
m = mass of astronaut (kg)
9.2.2.6 – Discuss the effect of Earth’s orbital motion and its rotational motion on the
launch of a rocket.
The fuel required to launch a rocket into orbit can be minimised if it is launched to the
east and as close to the equator as possible. This is because the Earth rotates from west to
east, so a rocket launched to the east will get a boost in speed from this rotation, and since
the surface speed is fastest at the equator this is where the largest boost will be obtained.
Transfer orbits between the Earth and other planets in the solar system can be achieved
by using the Earth’s orbital velocity around the Sun. If the rocket is launched against the
motion of the Earth’s orbit it will have an overall velocity of the Earth’s orbital velocity
minus the escape velocity, and it will consequently transfer to an orbital path nearer to the
sun than that of the Earth. If the rocket is launched with the Earth’s orbit it will be
travelling at the Earth’s orbital velocity plus the escape velocity and will hence transfer to
an orbit further away from the sun than that of the Earth
Because the Earth rotates, when a rocket is launched from Earth the horizontal velocity of
the rocket will be the same as the rotational speed of the Earth plus the orbital motion of
earth (depends where on earth you are). This will affect the path of the rocket in outer
space. It will also contribute to the rockets speed. The same applies for the Earth’s orbital
motion. So where on earth the rocket is launched from and what time of day and year will
affect where the rocket will initial be heading. So to fully utilise this and minimise fuel
usage the location on Earth of the rocket and what time of day and year it is is chosen to
get the rocket to head in a particular way.
Remember, the axis of Earth’s rotation on its own axis is from pole to pole. Hence
launching from the equator will result in the most velocity boost from Earth’s rotation.
Earth rotates on its own axis one revolution per 24 hours, and around the sun once every
year.
9.2.2.S1 - Solve problems and analyse information to calculate the actual velocity of
a projectile from its horizontal and vertical components using :
v2x = u2x
v = u + at
v2y = u2y + 2ay∆y
∆x = uxt
∆y = uyt + ½ ayt2
s = ut + ½ at2
Where: x = horizontal
y = vertical
u = initial velocity (ms-1)
v = final velocity (ms-1)
a = acceleration (ms-2)
t = time (s)
∆x = horizontal displacement
∆y = vertical displacement
s = displacement
There is no acceleration in the horizontal component, i.e. ux or vx is constant. Therefore
∆x = uxt.
Vertically there is constant acceleration due to gravity, hence the equations of motion.
Example: Kate drives a car at 30.0 ms-1 (horizontally), off the edge of an 895m cliff. Find
the following:
a) Time taken to reach the ground
b) Final vertical velocity
c) The range or distance from base of the cliff, and;
d) The final velocity
a) ∆y = 895m
ux = 30.0 ms-1
ay = 9.8 ms-2
uy = 0.0 ms-1
∆y = uyt + ½ ayt2
895 = 0.0 + 4.9t2
895
t=
4 .9
t = 13.5 s
b) vy = uy + ayt
= 0 + 9.8
= 132 ms-1
c) ∆x = uxt
= 30 x 13.5
= 405 m
tanθ = 132/30
Θ = 77.1°
d) 30m
132m
Examples: Lee standing on top of a bell tower throws an orange at 8.50 ms-1. It takes the
orange 3.05s to strike the ground. Find:
a) Height of the tower
b) Range
c) Its actual final velocity
Answers:
a) ∆y = uy + ½ at2
= 0.0 + 4.9 x 3.052
= 45.6 m
b) ∆x = uxt
= 8.5 x 3.05
= 25.9 m
c) vy = uy + ayt
=0 + 9.8 x 3.05
= 29.9 ms-1
v = 8.5^ 2  29.9^ 2
= 31.08 ms-1
A ball is thrown upwards from a 63.0 m cliff so that it reaches a maximum height of
139 m (from base). Calculate:
a) Initial velocity
b) Total time
a) v2y = u2y + 2ay∆y
0.0 = u2y – 19.6 x 76
u = 1489.60
= 38.6 ms-1
b) ∆y = uyt + ½ ayt2
76 = 38.6t + 4.9t2
 b  b ^ 2  4ac
t
2a
 38.6  38.6^ 2  4  4.9  76
t
2  4.9
t =3.94s
9.2.2.7 – Analyse the changing acceleration of a rocket during launch in terms of
the:
 Law of Conservation of Momentum
 Forces experienced by astronauts
A rocket engine is a little different to most other engines in that it must carry/contain both
the fuel and the oxygen supply (assuming a typical combustion reaction). Modern rockets
can use either solid or liquid propellants. The forward motion of the rocket can be
understood in terms of either Newton’s third law of motion (for every action there is an
equal and opposite reaction [Fab = Fba]); and the Law of Conservation of Momentum
(  Mbefore =  Mafter).
mrur + meue = mrvr + meve
0 = mrvr + meve
mrvr = -meve
  Mrocket = ∆Mengine
Since ∆M or ∆p = Ft 
Frocket = Fengine
ie. Newton’s 3rd Law
We assume that the rocket provides a relatively constant thrust and therefore the reaction
force propelling the rocket upwards should also be relatively constant. However the mass
of the rocket is continually decreasing due to the propulsion of exhaust gases; this means
that as mass decreases the acceleration increases using Newton’s Second Law (F = ma).
To more economically place a satellite into space and to minimize g-forces experienced
by astronauts, multi stage rockets are used. This creates a series of changes as shown in
figure 2.2 pg 31 – In stage 1 the rocket accelerates at an increasing rate as the mass of the
propellant decreases in the first booster (peaks at around 4g; when stage 1 breaks away,
the g-forces drop dramatically to 0; at this point the second stage rocket ignites and a
similar pattern of gradual increase in acceleration to about 1.8g occurs before again it
drop to 0 as the second booster is jettisoned; similar for stage three.
Forces experienced by Astronauts
As we discussed earlier in g-forces, an astronaut feels an apparent weight equivalent to
ma + mg where a is the acceleration away from the Earth’s surface. Since g-force =
ga
apparent _ weight
then g-force =
.
9.8
real _ weight
Referring to figure 2.2 on page 31, the maximum g-force experienced by Apollo
Astronauts, using the Saturn V rockets, was around 4g (at 4g’s most people begin to
loose some of their colour and peripheral vision). Soon after this most people will black
out (individual differences). In the first manned space flight by the US, astronaut Alan
Shepard had to tolerate a peak g-force of 6.3g. Rocket design has improved significantly
and the current space shuttle launch does not exceed 3g’s.
To minimize some of the effects of these g-forces on humans careful design has been
implemented. These include:
 That a transverse (lying down) position coped better with g-forces since the blood
was not as easily forced away from the brain.
 An ‘eyeballs-in’ application in g loads is easier to tolerate than an ‘eyeballs-out’,
ie. On lift off the astronauts should be facing up and also reenter backwards
facing up so that the g-forces are always directed upwards.
 Supporting the body in as many places as possible, most using a contoured couch
built of molded fiberglass specific to that astronaut.
Calculations actually show that even larger forces can be involved on reentry. Alan
Shepard experienced a maximum reentry force of 11.6g’s.
9.2.2.8 – Analyse the forces involved in uniform circular motion for a range of
objects, including satellites orbiting the Earth.
v (orbital speed/velocity)
Fc
r
Centripetal Force (Fc) =
mv^ 2
r
Where: Fc = Centripetal force (N) and is always directed to the center of the circle.
m = Mass (kg)
v = velocity (ms-1)
For an object on the end of a string swung around our head in a horizontal circle, the
force causing Fc is in fact the tension in the string (see prac on circular motion). If the
string suddenly breaks, the object on the end ‘fly’s off’ at a tangent.
For celestial objects (planets and stars) or for natural and artificial satellites the Fc is
Gm1m 2
mv^ 2 Gm1m 2

supplied by Fg (Fg =
). Therefore, Fc = Fg,
where m1 is the
d ^2
r
d ^2
larger mass and m2 is the smaller mass undergoing circular motion. Also r = d.
m2v ^ 2 Gm1m2
Gm1
2r


and finally v ^ 2 
. Furthermore, v 
.
d
d
d
T
9.2.2.9 – Compare qualitatively low Earth and geo-stationary orbits.
Low Earth orbits are above the Earth’s atmosphere but below the Van Allen radiation
belts.
Most of these are polar (north to south) elliptical orbits. They are a lower orbit than
geostationary orbits and thus have a faster orbital velocity. Some examples of uses for
low Earth orbits are to survey the surface of the Earth for minerals, weather and imaging
satellites and as an orbit for the space shuttle.
Geostationary orbits have an orbital period equal to the rotational period of the Earth.
They orbit above the equator. Because of this the satellites footprint remains over the
same place on the surface of the Earth for its whole period. These satellites are used as
weather and communications satellites. This means that for satellite TV, the receiver
(satellite on your house) can just point in the same place above the sky and the satellite
will always be there.
9.2.2.10 - Define the term orbital velocity and the quantitative and qualitative
relationship between orbital velocity, the gravitational constant, mass of the central
body, mass of the satellite and the radius of the orbit using Kepler’s Law of Periods.
r ^3
GM

T ^ 2 4 ^ 2
Where: r = Radius (m)
T = Period (s)
G = Universal Gravitational Constant
M = Mass of central object (kg)
Calculate the period an orbital velocity for the following satellites:
a)
Altitude 250 km
b)
Altitude 400 km
c)
Altitude 40000 km
Earth’s average radius = 6.38 x 106 m
Earth’s average mass = 5.97 x 1024 kg
G = 6.67 x 10-11
Answers:
(6.38  10^6  250000)^3 ((6.67  10^ 11)  (5.97  10^ 24))

a)
T ^2
4 ^ 2
((6.38  10^6  250000)^3)4 ^ 2  ((6.67  10^ 11)  (5.97  10^ 24))T ^ 2
((6.38  10^6  250000)^3)4 ^ 2
T ^2 
((6.67  10^ 11)  (5.97  10^ 24))
T  28893500.25
T  5375.27s
2r
T
2  (6.38  10^6  250000)
v
5375.27
v  7750 ms-1 = 27900 km/h
v
b)
(6.38  10^6  400000)^3 ((6.67  10^ 11)  (5.97  10^ 24))

T ^2
4 ^ 2
((6.38 x10^6  400000)^3)4 ^ 2  ((6.67  10^ 11)  (5.97  10^ 24))T ^ 2
((6.38  10^6  400000)^3)4 ^ 2
T ^2 
((6.67  10^ 11)  (5.97  10^ 24))
T  30899300.88
T  5558.71s
2r
T
2  (6.38  10^6  400000)
v
5558.71
v  7660 ms-1 = 27600 km/h
v
c)
(6.38  10^6  40000000)^3 ((6.67  10^ 11)  (5.97  10^ 24))

T ^2
4 ^ 2
((6.38 x10^6  40000000)^3)4 ^ 2  ((6.67  10^ 11)  (5.97  10^ 24))T ^ 2
((6.38  10^6  40000000)^3)4 ^ 2
T ^2 
((6.67  10^ 11)  (5.97  10^ 24))
T  9891264254
T  99454.84s
2r
T
2  (6.38  10^6  40000000)
v
99454.84
-1
ms
=
10 550km/h
v  2930
v
9.2.2.11 – Account for Orbital decay of Satellites in Low Earth Orbit.
When a satellite is in a low Earth orbit it will collide with small particles in the
atmosphere and will hence experience drag. This causes the satellites orbit to decay (or
lose altitude). As the altitude is reduced the atmosphere becomes denser and the process
of orbital decay is accelerated as drag is increased.
9.2.2.12 - Discuss issues associated with safe reentry into the Earth’s atmosphere
and landing on the Earth’s surface.
As a spacecraft enters the Earth’s atmosphere an immense amount of heat is produced.
This is because the craft is traveling at an extremely high velocity, and is colliding with
particles in the atmosphere. If this heat is not dissipated the craft may burn up. This can
be combated by using heat shields spread over large surfaces (so that the heat is spread
evenly). It can also be minimised by increasing the time taken for re-entry.
As the spacecraft enters the atmosphere and heat builds up around the craft atoms in the
air become ionised and surround the craft, resulting in a phenomenon known as ionisation
blackout. These ions deflect any radio signals sent to or from the craft. This can be
dangerous if there is contact needed between the craft and the Earth at this phase of the
flight.
The deceleration during re-entry can cause high g forces to be experienced by the
astronaut, which can cause blood to rush out of the brain, causing loss of consciousness
or even death. This loss of blood can be minimised by reclining the astronaut and
applying these forces perpendicular to the astronaut’s long axis. The g forces can be
minimised by increasing the re-entry time, slowing the rate of descent. The human body
can stand g forces of up to about 4g without comfort, with 10g being tolerable for short
periods of time along the long axis of the body.
The spacecraft must touch down softly on the surface of the Earth. This can be achieved
by using parachutes to slow the craft down and splashing into the ocean, or by landing on
an airstrip.
9.2.2.13 - Identify that there is an optimum angle for safe re-entry for a manned
spacecraft into the Earth’s atmosphere and the consequences of failing to achieve
this angle.
The angle of re-entry must be between 5°-7°. If the angle is too shallow the spacecraft
will bounce off the atmosphere and back into orbit. If it is too steep the craft will burn up
due to excessive drag.
9.2.3 – The Solar System is held together by Gravity.
9.2.3.1 – Describe a gravitational field in the region surrounding a massive object in
terms of its effects on other masses in it.
Any field is a region of influence cause by such things as charge, magnetism or mass.
These fields are vector quantities (the have strength/magnitude and direction).
A gravitational field is a little different to others in that it is always directed towards the
large mass which causes the gravitational field.
9.2.3.2 – Define Newton’s Law of Universal Gravitation.
F G
m1m2
d^2
9.2.3.3 – Discuss the importance of Newton’s Law of Universal Gravitation in
understanding and calculating the motion of satellites.

9.2.3.4 – Identify that a slingshot effect can be provided by planets for space probes.
9.2.4 – Current and emerging understanding about time and
space has been dependent upon earlier models of the
transmission of light.
9.2.4.1 – Outline the features of the aether model for the transmission of light
9.2.4.2 – Describe and evaluate the Michelson-Morley attempt to measure the
relative velocity of the Earth through the aether
9.2.4.3 – Discuss the role of the Michelson-Morley experiments in making
determinations about competing theories
9.2.4.S1 – Gather and process information to interpret the results of the MichelsonMorley experiment
The analogy below can be used to model the concept of aether that Michelson and
Morley were trying to prove. In this analogy the aether is represented by the water
current; boat A is traveling in the same direction/plane as the aether whilst boat B is at
90º.
B
Current
Boat A and B both travelA2km one way and 2km back. They also both travel at 5km/h in
still water.
Boat A
Upstream trip
Speed = 5 – 3
= 2 km/h
Time = distance/speed
= 2/2
= 1 hr
Downstream trip
Speed = 5 + 3
= 8 km/h
Time = distance/speed
= 2/8
= 0.25 hrs
Total trip = 1 hr + 0.25 hrs
= 1.25 hrs
Boat B
For boat B to travel straight across, it has to aim upstream.
3 km/h
4 km/h
5 km/h
By Pythagora’s theorem, the measured speed straight across the river is 4 km/h, therefore
the total round trip:
Conventional Mirror
Time = distance/speed
V (aether wind)
2a/b
Original Light
Ray
M
1a/b
Conventional
= 4/4
= 1 hr
In the above simplified diagram, light from the light source traveled horizontally to the
special mirror (M) (half silver mirror) which splits the light into two separate ways.
Half passes straight through (1a) where it strikes a conventional mirror, reflects back to
M (1b) and in turn is reflected back to the observer (1c).
The other half of the original ray is reflect by M upwards (2a); reflects from this
conventional mirror (2b) and passes straight through M to reach the observer (2c).
If, in fact, there is an aether wind as shown (V) then 2c should reach the observer slightly
more quickly than 1c (see previous boat analogy). This means that 1c and 2c will be out
of phase with each other (to some extent).
In 1887 when the experiment was conducted, this piece of apparatus was capable of
picking up minute differences in the time or path length. Small differences in path length
may have been attributable to the pieces of equipment eg. Phase changes on reflection,
thicknesses of mirrors etc.
However, when the apparatus is rotated so that one of the rays was no longer exactly in
the same direction as the suspected aether wind, there should have been a subsequent
change in the interference pattern (since the path difference would also have changed). In
fact there were no changes even when the apparatus was rotated through a variety of
angles up to 360º. This was therefore a nil result – it failed to detect the aether.
The experiment was repeated numerous times at different times of the day and year and
some groups repeated a similar experiment with even more sensitive apparatus. No
evidence could ever be found.
The scientific community was not quick to abandon the aether model and may tried to
adapt or modify the theory. One such suggestion was that a large object like a planet
could drag the aether along with it; another was that objects contract in the direction of
the aether wind. None of these modifications, however, survived careful scrutiny.
It was not until Einstein in 1905 provided an alternative which did not require the aether
construct. This alternative explanation for light and the inability to detect aether
eventually led to this aether wind theory being rejected and Einstein’s special relativity
became the new accepted theory.
9.2.4.4 – Outline the nature of internal frames of reference.
300 hundred years before Einstein, Galileo proposed the ‘principle’ of relativity – ‘all
steady motion is relative and cannot be detected without reference to an outside point’.
The principle of relativity applies only for non-accelerated motion (at rest or moving with
constant velocity). This is referred to as an inertial frame of reference. Situations
involving acceleration are called non-inertial frames of reference.
This principle states that within an inertial frame of reference you cannot perform any
mechanical experiment or observation that would reveal to you whether you were moving
with uniform velocity or standing still.
9.2.4.5 – Discuss the principle of relativity.
(See earlier dot point [9.2.4.4] on Galileo and the principle of relativity)
The belief in aether posed a real problem for the principle of relativity, since the aether
was supposed to be stationary and light had a fixed velocity relative to the aether. This
would mean that if a scientist tried to measure the speed of light from the back of a train
carriage to the front and the train was traveling into the aether it should be a slower value
than moving away. This means a ‘simple’ optical experiment would violate the principle
of relativity.
9.2.4.6 – Describe the significance of Einstein’s assumption of the constancy of the
speed of light.
At around the turn of the 20th century, Einstein puzzled over this apparent violation of the
principle of relativity posed by the aether model. He presented many of his explanations
as thought experiments (‘Gaedaken’). He used the analogy of a train traveling at the
speed of light, holding up a mirror in front and posing the question would he see his own
reflection. If the aether model was right, there would never be a reflection since the light
would not be able to catch up to the mirror (this would violated the principle of relativity
since the absence of a reflection would be a way of detecting motion). On the other hand,
if the principle of relativity were not to be violated the reflection must be seen normally
BUT this would mean an observer on an embankment would see light traveling at twice
its normal speed (using conventional physics).
Einstein believed firmly in the principle of relativity which meant that the aether could
not exist. To solve the two dilemmas, Einstein decided that both the train rider and the
person on the embankment must both observe the light traveling at its normal speed (3 x
108 ms-1) but if this is to occur since speed = distance/time, than the distance and time
witnessed b both observers must be different or relative.
9.2.4.7 – Identify that if c is constant then space and time become relative.
(See previous dot point [9.2.4.6])
Einstein’s ideas were published in a 1905 paper ‘On the Electrodynamics of Moving
Bodies’. The main ideas were:
 The laws of physics are the same in all frames of reference (ie. The principle of
relativity always holds)
 The speed of light in empty space always has the same value c independent of the
motion of the observer (ie. Everyone always observes the same speed of light
regardless of their motion)
 He stated that the luminiferous aether is superfluous
This allows us to generate another series of equations for the length an time based on the
relative velocity based on the frames of reference.
9.2.4.8 – Discuss the concept that length standards are defined in terms of time in
contrast to the original meter standard.
Originally defined in 1973
9.2.4.9 – Explain qualitatively and quantitatively the consequence of special
relativity in relation to:
 The relativity of simultaneity
 The equivalence between mass and energy
 Length contraction
 Time dilation
 Mass dilation
Relativity of Simultaneity
Einstein contended that if an observer sees two events to be simultaneous then any other
observer in relative motion to the first would not judge them to be simultaneous. Einstein
offered the following modified thought experiment/explanation.
The operator of a lamp sits in the middle of a train carriage. The doors at either end are
light operated and another observer on the embankment (stationary) is exactly in line
with the lamp operator when they switch the lamp on to open the doors.
The operator of the lamp in the moving train will see both doors open simultaneously
because light travels at the same speed forward and backward (c).
The observer on the embankment, however, sees something different. Light from the
lamp travels at a constant speed forward and backward but the light moving forward
travels a longer distance to reach the front door because the door has moved forward
whilst the train is moving. Similarly, the light will reach the back door earlier since the
back door ahs moved forwards ie. The stationary observer sees the back door open before
the front door.
Both observers are correct from their different frames of reference – this is a direct
consequence of the constancy of the speed of light.
The Equivalence Between Mass and Energy
As a consequence of mass dilation, then there is also equivalence between mass and
energy. As force is applied to create acceleration, this acceleration leads to higher
velocities that eventually create in creases in mass. As mass increases towards infinity,
then an infinite force would be needed to cause the acceleration – this is why no object
can be accelerated beyond the speed of light.
If a force is applied on an object then work is done or energy has been given to an object
(in this case kinetic energy [Ek]). However, near the speed of light the object does not
acquire velocity but rather mass (or inertia). The amount of energy or work the has been
applied has been used to increase the objects mass. This leads to the equation:
E = mc2
Where: E = energy (J)
m = mass (kg)
c = speed of light (3 x 108 ms-1)
Example
Calculate the energy equivalent of the typical uranium atom with a mass of 238 amu (1
amu = 1.661 x 10-27 kg).
E = mc2
= (238 x [1.661 x 10-27]) x (3 x 108)2
= 3.558 x 10-8 J
Length Contraction
Again as a consequence of the constancy of c, length/distance is also relative and will be
perceived differently by two observers in relative motion to each other. A derivation of
the equation shown below can be found in the textbook (pg 77-80) but it is complex.
lv = l0 1


v^2
c^2
Where: l0 = Normal length as seen by someone within the moving frame (or if the object
was stationary) (m)
lv = Length perceived by the other observer where the object is moving relative to
them (m)
v2 = Velocity of the moving object/frame
c2 = Speed of light (3 x 108 ms-1)
Example
When stationary, the carriages on a very fast train are 20.00m long. How long would
these appear to a person standing on a station platform if the train is traveling at:
a) 330 ms-1
b) 50% the speed of light
c) 95% the speed of light
v^2
a)
lv = l0 1
c^2

lv = 20.00 x 1

330^2
(3 10^8)^2
= 20.00 m

v^2
lv = l0 1
c^2
b)

lv = 20.00 x 1

0.5  [(3 10^8)^2]
(3 10^8)^2
= 17.32 m

v^2
lv = l0 1
c^2
c)

lv = 20.00 x 1

0.95  [(3 10^8)^2]
(3 10^8)^2
= 6.24 m
 by another train and the observer in this train estimates Lee’s width to be
Lee is passed
15 cm. How fast was this train going?
lv = l0 1


v^2
c^2
0.15 = 0.4 x 1
v^2
(3 10^8)^2
v^2
1 - (0.375)2 =
(3 10^8)^2

v2 = -0.140625 x (9 x 1016)

v = 1.0265625 10^17
= 2.828 x 108 ms-1

Time Dilation
Similarly, tie is also relative, and the equation which relates to this is:
tv =
to
v^2
1
c^2
Where: tv = Time as measured by a frame moving relative to the object (s)
t0 = Time measured within the frame of reference (as if the frame were stationary)

Example
A train driver sneezes just as the train passes a station platform. A person on the train
measures the time interval for the sneeze to be 1.000 seconds. What time will an observer
standing on the platform measure this sneeze to be if the train was traveling at:
a) 50% of c
b) 2.95 x 108 ms-1
a)
tv =
tv =

to
v^2
1
c^2
1.000
1 (0.5)^2
= 1.155 s

b)

tv =
to
v^2
1
c^2
tv =
1.000
1 (2.95 10^8)^2  (3 10^8)^2
= 1.155 s
Mass 
Dilation
Example
Calculate the mass of an electron at the following speeds if its rest mass is 9.109 x 10-31
kg:
a)
5.65 x 104 ms-1
b)
80% the speed of light
c)
99.9% the speed of light
a)
mv =
mv =

mo
v^2
1
c^2
9.109 10^31
(5.65 10^4)^2
1
(3 10^8)^2
= 9.109 x 10-31 kg

b)
mv =
mv =


c)
9.109 10^31
0.8  ((3 10^8)^2)
1
(3 10^8)^2
= 1.518 x 10-30 kg
(This is approximately 1.7 times the rest value)
mv =

mo
v^2
1
c^2
mo
v^2
1
c^2
mv =

9.109 10^31
0.999  ((3 10^8)^2)
1
(3 10^8)^2
= 2.037 x 10-29 kg
(This is approximately 22 times the rest value)
9.3 – Motors and Generators
9.3.1 – Motors use the Effect of Forces on Current-carrying
Conductors in Magnetic Fields
A current-carrying wire in a magnetic field experiences a force. This is the motor effect.
In electric motors, electrical energy is transformed into mechanical energy.
9.3.1.1 – Discuss the Effect on the Magnitude of the Force on a Current-carrying
Conductor variations in:
 The strength of the magnetic field in which it is located
 The magnitude of the current in the conductor
 The length of the conductor in the external magnetic field
 The angle between the direction of the external magnetic field and the
direction of the length of the conductor.
9.3.1.S3 – Solve problems and analyse information about the force on current
carrying conductors in magnetic field using:
F = BIlsinØ
I
B
Figure 1.1 – Right Hand Palm Rule
The direction of the force acting on a current carrying wire is perpendicular to both the
direction of the current and the direction
of
the magnetic field. If the fingers of the
right hand point in the direction of the
magnetic field and the thumb points in
the
direction of
current, then
in the
force.
Where:
the conventional
the palm points
direction of the
F = force on the wire in Newtons (N)
B = magnetic field in Tesla (T)
I = current in Ampere (A)
l = length of the wire in B in meters (m)
Ø = angle between the wire and B
(directions determined by right hand rule)
It becomes obvious then that the force acting on the wire will increase if:
 Magnetic field strength is increased (using permanent or electromagnets)
 The current is increased by increasing voltage across the wire
 Increase in the length of the wire within B by using multiple coils.
 Increasing the angle between l and B up to 90° (If the angle is 0°, then there is no
F)
We can use the above equation to solve problems.
9.3.1.2 – Describe qualitatively and quantitatively the force between long parallel
current carrying conductors:
F
I1I2
K
l
d
9.3.1.S1 – Solve problems using the above.

Where:
F
I1I2
K
l
d
F = force of current carrying wires (N) between two parallel lines
l = Common length of the two wires (m)
K = magneticforce constant
I1 = current in wire 1 or 2x10-7 NA-2
I2 = current in wire 2 (A)
d = distance between two wires parallel only (m)
As can be seen from the above equation, the following factors will increase the
magnitude of F between the two wires: Increase current in I1 or I2, decrease distance
between wires and increase the length. As can be seen below, ‘Like’ current attract and
‘unlike’
current repel.
9.3.1.3 – Define torque as the turn moment of force using:
τ= Fd
F = force applied ‘at the edge’(N)
d = distance from turning point (m)
Where:
τ= torque in Newton meters (Nm)
This principle was used in the junior school when dealing with the turning effect or
moment of a lever. In the case of a first order lever, eg. A see-saw, where the two
moments or turning effects are balanced.
9.3.1.4 – Identify that the motor effect is due to the force acting on a current
carrying conductor in a magnetic field.
9.3.1.5 – Describe the forces experienced by a current carrying loop in a magnetic
field and describe the net result of the forces.
Initially
B
C
b
l
A
D
+ve
-ve
Axle/Spindle
Forces on AB only: Due to the motor effect F = BIlsin Ø.
In this case F = BIl and F is into the page
Forces on CD only: Similarly F = BIl where F is out of the page
Forces on BC: F = 0 since Ø = 0, sin Ø = 0
Forces on AD: Similarly F = 0
B
After Half Turn (180°)
C
B
B
b
l
D
A
-ve
+ve
Axle/Spindle
Forces on AB: F = BIl and is into page
Forces on CD: F = BIl and is out of the page
Forces on CB and AD: F = 0
This
coil as
simply
flop
rotating.
DC
device to
direction
coil. The
is a split
means that the
pictured will
oscillated or flip
rather than
To convert this
apparatus into a
motor we need a
alter the relative
of current in the
simplest of these
ring commutator.
As the coil and split ring commutator rotate, the direction of the current relative to the
coil reverses each half turn (180°). This will the result in the coil rotating rather than
oscillating.
There is still a problem of changing magnetic force as the coil undergoes each rotation.
The force on each of section AB and CD is a maximum when each section of the wire is
perpendicular to the magnetic wire or when the plane of the coil itself is parallel to the
magnetic field (B). If the coil part way (90°) from the diagram so that it is perpendicular
to the magnetic filed then there is no force acting on the coil. Part way in between there is
some force but not the maximum. This therefore creates variable torque.
To overcome this issue of variable torque, many motors and generators use a radial
magnetic field.
This means that a uniform magnitude magnetic field (but not direction) is produced so
that the force on the coil at almost any angle is also of constant magnitude (ie. Torque is
constant).
9.3.1.S4 – Solve problems and analyse information about simple motors using:
τ= nBIAcosθ
Where:
τ= torque (Nm)
B = magnetic field (T)
n = number of coils
I = current (A)
A = area of coil (m2)
Θ= angle between the plane of the coil and B
B
C
B
b
l
A
D
+ve
-ve
Axle/Spindle
Net Forces: FAB = BIlsin Θ
= BIl (into)
FCD = BIl (out of)
Net Torque = Fd
τ = τAB + τCD
b
b
= BIl   BIl 
2
2
= BIlb and b x l = Area of the coil
τ= BIA for 1 coil only
 
In a typical motor there is not just one coil but many, each making its own contribution.
The number of coils is shown by the symbol n.

Similarly, as discussed earlier the force and therefore torque will vary as the angle
between the coil and the magnetic field changes (it is a maximum as shown parallel to the
field and a minimum when perpendicular).
All of these factors combine for the equation:
τ = nBIAcos Θ
To overcome this problem of variable torque a radial magnetic field is used as mentioned
earlier.
9.3.1.6 – Describe the main features of a DC motor and the role of each feature.
9.3.1.7 – Identify that the required magnetic fields in a DC motor can be produced
either by current carrying coils or permanent magnets.
Feature
Magnets
Role








Coil/Coils
Current (power source)


Split Ring Commutator

Creates B
Permanent or Electromagnet
Radial Magnet
Carries I
Coil is insulated (usually copper)
Usually a large number of coil
Different planes
Eg. Transformer, battery etc.
V
I
R
Relative to the coil, it reverses the
direction of the current each half
turn
Brushes

Axle/Spindle


Armature/Stator


Completes the circuit I from
power to coil
Made from wire, often graphite
Provides rotation for coils (low
friction)
Stationary – motor body and
magnets
Rotating frame made of
ferromagnetic material.
9.3.2 – The relative motion between a conductor and a
magnetic field is used to generate an electrical voltage.
9.3.2.1 – Outline Michael Faraday’s discovery of the generation of an electrical
current by a moving magnet.
After Oersted proved that a magnetic field could be produced from an electrical current
people were sure that the reverse could also be done. This was referred to as
electromagnetic induction which involves the conversion of mechanical energy to
electrical energy. This discovery was landed to Michael Faraday who was successful in
noticing that a changing magnetic field was necessary to produce electricity.
Faraday explained electromagnetic induction as whenever there is relative motion
between a conductor and a magnetic field such that field lines are ‘cut’ and emf (voltage)
is induced and current will flow if provided.
9.3.2.2 – Define magnetic field strength B as magnetic flux density.
9.3.2.3 – Describe the concept of magnetic flux in terms of magnetic flux density and
surface area.
The number of magnetic lines of force emerging through an imaginary surface in a
magnetic field is called the magnetic flux of the field. Magnetic Flux density is the
magnetic flux per unit of area and is the unit of measure for magnetic field
strength/intensity. The larger the flux density the more intense the field in that region.
Therefore the magnetic flux density (φ) is equal to the area (A) times the magnetic field
strength (B). Mathematically:
φ = Bacosθ
9.3.2.4 – Describe generated potential difference as the rate of change of magnetic
flux through a current.
Experiments show that the size of the induced emf (potential difference) depends upon:
 The value of B
 The speed at which the conductor ‘cuts’ the flux lines; and
 The number of conductors.
This is stated in Faraday’s Law which says that the induced emf is proportional to the rate
of change of flux through a circuit.
9.3.2.5 – Account for Lenz’s Law in terms of
conservation of energy and relate it to the
production of back emf in motors.
9.3.2.6 – Explain that, in electric motors,
back emf opposes supply emf.
Lenz’s Law states that the direction of the
induced emf is such that the current it produces
creates a magnetic field opposing the change
that produced this emf.
With reference to the conservation of energy
whereby energy cannot be created nor
destroyed the induced current must the
change that gives rise to it ie. Back emf.
A back emf will be produced in the coil to oppose the motion of the coil as given by
Lenz’s Law. This back emf helps limit the current in the coil and hence its speed. As the
applied current increases, the motor speeds up. This produces increasing emf that is in
opposition to the applied voltage.
9.3.2.7 – Explain the production of eddy currents in terms of Lenz’s Law.
Eddy currents are caused by a moving (or changing) magnetic field intersects a
conductor, or vice-versa. The relative motion causes a circulating flow of electrons, or
current, within the conductor. These circulating eddies of current create electromagnets
with magnetic fields that oppose the effect of the applied magnetic field as per Lenz’s
Law.
Eddy currents can be used for various
applications including induction cooking
and electromagnetic braking (refer to
9.3.2.S3 & 9.3.2.S4) the latter of which
can be seen below.
9.3.3 – Generators are used to provide large scale power
production.
A generator is a device for producing electrical energy from mechanical energy (the
reverse of an electric motor).
9.3.3.1 – Describe the main components of a generator.
9.3.3.2 – Compare the structure and function of a generator to an electric motor.
The prime mover is the mechanical device that is used to drive the generator. It could be
a steam turbine driven by steam produced by burning coal, oil or natural gas or one
driven by moving water ie. A water turnbine. The other main components of a generator
include the armature, a field structure which can be from an electro or permanent magnet,
slip rings and brushes to take the induced current away, a spindle, a stator, and, if it is a
DC generator only, a commutator.
A generator can function as a motor and, similarly, a motor can function as a generator.
9.3.3.3 – Describe the differences between AC and DC generators.
The essential difference between an AC and a DC generator is the nature of the
connection between the rotor coils and the external circuit. In an AC generator, the
brushes run on slip rings which maintain a constant connection between the rotating coil
and the external circuit as can be seen in the above diagram. This means that as the
induced emf changes polarity with every half-turn of the coil, the voltage in the external
circuit varies like a sine wave and the current alternates direction. In a DC generator, the
brushes run on a split-ring commutator which reverses the connection between the coil
and the external circuit for every half-turn of the coil refer to syllabus point 9.3.1.5. This
means that as the induced emf changes polarity with every half-turn of the coil, the
voltage in the external circuit fluctuates between zero and a maximum while the current
flows in one constant direction.
9.3.3.4 - Discuss the energy losses that occur as energy is fed through transmission
lines from the generator to the consumer.
Power stations are usually located large distances from major population areas (in NSW
most are located near Newcastle and in the Hunter Valley; a ready coal supply and still
close to Sydney and water supplies).
This presents problems associated with loss of energy as the current flows through the
transmission lines due to Ohmic heating.
R = ρL / A
ρ = RA / L
= Ωm
Where:
 p is resistivity
 R is resistance
 L is length
 A is the cross sectional area
And V = IR and P (power) = VI combining P = I²R
This is the power loss due to heating of the wire/conductor.
Example
Q1. A power station generates electricity at 120 KW. It sends this power to a town 10KM
away through transmission lines with a total resistance of 0.40Ω. If the power is
transmitted at 240V, calculate:
a) I
b) The voltage drop across the transmission lines
c) The voltage available to the town
d) Power loss
a) I = P / V
I = 120000 / 240
= 500 A
b) V = IR
= 500 x 0.4
= 200 V
c) Vtown = 240 – 200
= 40 V
d) P = I2R
= 5002 x 0.4
= 100000 W
= 100 KW
Obviously, this forms problems as most of the power is lost before reaching the town.
To overcome this transformers can be used to step up the voltage (V) before transmission
so that a lower current (I) flows through the transmission lines and thus reducing
dramatically the heating losses. Eg. If V is made 100 times larger, I becomes 100 times
smaller and P becomes 10000 times smaller.
In NSW the Bayswater Power Station in the upper hunter has 4 x 660 mW generators
each with an output voltage of 23 kV. It is then stepped up with transformers at a
transmission sub-station to 330 kV. Near the end of transmission lines it stepped down to
66 kV then to 22 and 11 kV and finally the pole transformers step down from higher
values our 240 V domestic standard.
9.3.3.5 – Assess the effects of the development of AC generators on society and the
environment.
AC Generators
 What is it?
 Why AC?
o Most widely used,
o transformed/change in V to reduce I to minimise heat losses (P = I2R)
 Societal impacts?
o Large scale electrical production,
o lifestyle ease with appliances,
o city development/urbanization/jobs/employment,
o changes in lifestyle eg shift work,
o overdependence on AC,
o unskilled labour
 Environmental impacts?
o Coal burning/CO2/greenhouse,
o global warming, rising sea levels, changing climate eg droughts etc.,
o light/noise pollution,
o issues associated with coal mining,
o air/water pollution,
o laying of cables,
o hydroelectricity
9.3.4 – Transformers
9.3.4.1 – Describe the purpose of transformers in electrical circuits.
Transformers are devices that change (increase or decrease) an AC voltage. They may be
designed to be step up transformers (eg in CRT sense) and step down transformers (eg
computers, cordless phones etc.)
They consist of two coils of insulated copper wire – the primary and secondary wrapped around a soft iron core. See figure 8.18 pg 144. They are designed so that when
an AC flows in the primary coil (due to an external AC voltage source) constantly
changing magnetic flux is created which induces an AC voltage at the terminals of the
secondary coil.
The factor determining any changes in voltage is the ratio between the number of
turns/coils in the primary relative to the secondary.
9.3.4.2 – Compare step-up and step-down transformers.
Step–up transformer – Here the voltage in the secondary coil is greater than that in the
primary. VS > VP (nS > nP where n is the number of coils)
Primary – 12V
Secondary – 24V
VP
Vs
This is determined by the number of turns in each of the coils.
Step-down transformer – As above however vice versa.
9.3.4.3 – Identify the relationship between the relationship between the number of
turns in the primary and secondary coils and the ratio of primary to secondary
voltage.
9.3.4.S2 – Solve problem and analyse information about transformers using:
Vp np

Vs ns
Where:
 VP = Voltage in primary coils (V) “Input”
 VS = Voltage in secondary coils (V) “Output”
 nP = Number of turns in primary coil
 nS = Number of turns in secondary coil
This equation applies to an ideal transformer assuming that all the energy/power in the
primary coil is transferred to the secondary coil.
Therefore in a step-up transformer not only is Vp < VS but nP < nS.
Example
Q1. The transformer in an electric piano converts 240V AC into 12V AC. If there are 30
turns in the piano coil, how many in the primary?
Vp np

Vs ns
240 np

12 30
12nP = 7200
nP = 600
Q2. A neon sign needs 12kV to operate. If the primary coil has 100 turns and mains
electricity is supplied, how many turns in the secondary coil?
Vp np

Vs ns
240
100

12000 ns
240nS = 1200000
nS = 5000
9.3.4.4 - Explain why voltage transformations are related to the conservation of
energy.
Using the law of conservation of energy we know that in theory the total energy input
must equal the total energy output. The equations and derivation below show how this
conservation of energy can be applied to modify the equation from the previous syllabus
point.
P = VI
W
P=
t
W = Pt
Where:
 W = Work
 P = Power
 W = VIt
or E = VIt
Energy in primary (p) and secondary (s)
EP = ES
VPIP = VSIS
Vp Is np


Vs Ip ns
Example
Q1. A power station generates electric power at 120 kW. It sends this power to town 10
km away through transmission lines that have a total resistance of 0.40Ω. If the power is
transmitted at 5.00 x 105V, calculate:
a) the current in the transmission lines
P = IR
120000
I=
500000
= 0.24 A
b) the voltage loss
V = IR

= 0.24 x 0.4
= 0.096 V
c) the voltage available to the town
= 500000 – 0.096
= 5 x 105 V
d) the power lost due to heating
P = I2R
= 0.242 x 0.4
= 0.023 W
9.3.4.S3 – Gather, analyse and use available evidence to discuss how difficulties of heating
caused by eddy currents in transformers may be overcome.




Eddy current
o 3D
o Solids
o Conductors
o Cause – induction (Lenz’s Law)
o Changing B
Core
o Why have iron core?
Reducing Eddy currents
o Lamination (reduces currents to one plane)
o Ferrite Material (intensifies B but not very conductive)
Cooling
9.3.4.5 – Explain the role of transformers in electricity sub-stations.
9.3.4.S4 – Gather and analyse information to discuss the need for transformers in
the transfer of electrical energy from a power station to its point of use.
There is usually a substation in each regional center and in each municipality of a city.
Here the voltage is stepped down to values of 11 kV and 22 kV using transformers.
Power substations can perform three tasks including:
 Step down the voltage using transformers
 Split the distribution voltage to go in different directions
 Enable, using circuit breakers and switches, the disconnection of the substation
from the transmission grid to be switched on and off.
Transformers are used for the transfer of energy from the power station to the point of use
as it assists in overcoming the problems faced with power losses due to heating and
transmitting electrical energy at low voltages over long distances. As mentioned above,
by increasing the voltage transmitted the current decreases hence reducing the power loss
as Ploss = I2R. Using transformers enables electricity to be supplied over large distances
without wasting too much energy. If transformers were not used power station would
have to be built in towns and cities or the users of electricity would have to live near to
the source leading to increased societal pollution and increased health issues.
Power Station
Approx. 20000 V
Step-up
Substation
High V Transmission
Wires to Grid
Approx. 250000 V
Step-down
Home/Household
240 V
Step-down
Town/Suburb
Approx. 11000 V
Step-down
District Area
Distribution
Approx. 132000 V
9.3.4.6 - Discuss why some electrical appliances in the home that are connected to
the mains domestic power supply use a transformer.
Many appliances inside the home still require a transformer since they work more
efficiently and/or a designed to operate at a voltage different to the 240 V AC
domestically supplied (this 240 V is the RMS value, it actually fluctuates between ±339
V).
CRT TV’s have a picture tube that requires a voltage of about 1500 V to operate. It
therefore contains a step-up transformer.
Most electronic circuits (involving microprocessors, transistors etc.) use voltages
between 3 and 12 V DC. Such devices have a step-down transformer and also incorporate
a rectifying unit to convert AC into DC. For example, laptops, cordless phones etc.
9.3.4.7 - Discuss the impact of the development of transformers on society.
This syllabus point fits closely to 9.3.3.5 on AC supply.



What is a transformer?
o Step-up/step-down
o AC
o Primary/secondary coil
o Faraday/Induction
o Eddy currents
o Annotated diagram
o Pp = Ps
Vp np

o
Vs ns
Pros
o Reduce heating loss in transmission
o P = I2R
o Power to remote location
o Don’t need as many power generators near to town centers (less pollution)
o Cost/Economics on society
o Appliances need high or low V and even DC can be used
Cons
o Extra danger with high V
o Extra danger with AC – affect heart rhythm
o Still heating losses in wires
o Heating losses in metal towers due to induction
o Aesthetic spoiling of environment
o Loss of unskilled labor/reliance on TV and multimedia/reliance on
electricity
9.3.5 – AC Motors
9.3.5.1 - Describe the main features of an AC electric motor.
There are two types of electric motors – the universal motor (which is essentially similar
in construction to a DC motor and can run on either AC or DC) and the induction motor.
Both are described in more detail from page 160-165 in our text.
The Induction Motor – consists of a stator which is a series of wire coils wound on soft
iron cores that surround the rotor. They provide usually three pairs of opposite radial
fields. They are connected to a central external power supply in such a way that the
magnetic field polarity B rotates at constant speed in one direction (uses either the three
phases of power supply or manipulates the one phases using capacitors).
The rotor consists of coils wound on a laminated iron armature mounted on an axle.
These rotor coils are not connected to an external power supply and there is no need for
commutators or brushes. Eddy currents are induced in the rotor coils by the rotating
magnetic field or the stator. The eddy currents in turn produce magnetic fields which
interact with the rotating stator fields to produce a torque on the rotor in the same
direction as the rotation of the stator field.
The rotor coils are often simplified to single copper bars (which can carry a large current)
embedded in the surface of the armature connected at the ends by a ring or disc of copper
which allows current to flow in a loop between opposite bars. This physically resembles a
‘squirrel cage’ (like the exercise unit for mice).
As mentioned earlier, because there is no need for an external current to be supplied to
the rotor, there is no need for slip-rings, brushes or commutators. This simplifies the
motor, reducing maintenance, wear and tear, etc.
The motor works only on AC and rotates at a constant speed equivalent to the
speed/frequency of the AC supply (50 Hz or 3000 rpm). If faster or slower speeds are
needed, gear or pulleys are used.
The simplicity and reliability of the induction motor greatly outweighs the limitation of
speed and it is estimated that AC induction motors make up 95% of all motors.
(see figures 9.3, 9.5, 9.6, 9.7, 9.8 and 9.9 in text).
9.3.5.S2 - Gather, process and analyse information to identify some of the energy
transfers and transformations involving the conversion of electrical energy into
more useful forms in the home and industry.
Electricity is such a widely used form of energy since it can be so easily converted into
other energy forms.
Electrical energy is transferred from the primary to the secondary coil in a transformer.
Similarly it is transferred when it passes into any electrical conductor such as a lead etc.
It is then transformed (changed) into another useful form of energy. The electrical
devices which do this do so with variable ease and efficiency.
In Home
Electrical energy is transformed into
radiant energy.
 Light in light globe
 Heat in toaster and kettle
 Microwaves in microwave ovens
 Radio/microwaves in
In Industry
Electrical energy is transformed into
radiant energy.
 Light in laser circuit printing
 Heat in induction ovens
 Microwaves in wood curing
 Radio/microwaves in
communications
Electrical energy is transformed into
mechanical energy.
 Rotation in a food blender (motor)
 Vibration in a stereo system
 Electric motor in lawn mower
Electrical energy is transformed into
chemical energy.
 Recharging batteries
communication
 X-rays in medical
Electrical energy is transformed into
mechanical energy.
 Rotation in industrial motors
 Vibration in TV speaker
 Kinetic energy and GPE in fun park
rides
Electrical energy is transformed into
chemical energy.
 Electro plating/electrolysis
9.4 – From Ideas to Implementation
9.4.1 – Cathode Rays to TV
9.4.1.1 – Explain why the apparent inconsistent behaviour of cathode rays caused
debate as to whether they were charged particles or electromagnetic waves.
In 1855 a German physicist Geyssler refined a vacuum pump capable of lowering the
pressure inside a glass tube to 0.01% (a near vacuum). His friend Plucker placed
electrodes at either end and applied a high voltage and produced a green glow in the tube.
This green glow appeared to be coming from the glass at the anode (opposite the
cathode). Debate occurred as to what caused these fluorescents and in fact to the
properties of the cathode rays as they became known.
Further experiments by a variety of scientists but in particular William Crookes (1875)
are summarised below:
[Figure 10.2 pg 171]
As mention there was debate as to whether cathode rays were electromagnetic waves or
streams of charged particles. The properties of cathode rays were:
 [Dot points on pg 180]
9.4.1.2 – Explain that cathode ray tubes allowed the manipulation of a stream of
charged particles.
Most of the experiments mentioned above and in our practicals (9.4.1.S1 and 9.4.1.S2)
involved the manipulation of the charged particles (cathode rays).
They could be affected by:
 Pressure of the gas inside the discharge tube
 They could be blocked by pieces of metal etc. (eg. Maltese Cross) or restricted to
a narrow beam by a slit in the metal
 Can be view more readily using fluorescent screens
 Deflected by a magnetic field (B) (Cathode rays/electrons travel in the opposite
direction to conventional current)
Cathode Ray 1
X
X
X
X
X
X
X
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
Cathode Ray 2
x x x
x x x
x x x
x x x
x x x
x x x
x x x

Deflected by an electric field
- ve
Cathode Ray 1
E
+ ve
It is also possible to initiate chemical reactions with cathode rays eg. Darkening of silver
salts.
9.4.1.3 – Identify that moving charged particles in a magnetic field experience a
force.
As we can see from the previous syllabus point, electrons and other charged particles eg.
Protons, have a force exerted on them by a magnetic field (B). This is a maximum when
the velocity (v) and B are perpendicular to each other and is zero when they are parallel.
We can express this mathematically as:
F = qvBsinθ
Where: F = Force (N) direction from right hand palm rule
q = Charge (Coulombs [C])
v = Velocity of the charged particle (ms-1)
B = Magnetic field strength (Tesla [T])
θ = Angle between B and v
Example: An electron of charge -1.6 x 10-19 C is projected into a magnetic field of
strength 2.0 x 10-2 T into the page while its velocity is 2.5 x 104 ms-1 due east. Draw a
diagram to show the path of the electron and calculate the force acting on the electron. If
this force is exerted for 1.00 seconds, calculate the final velocity of the electron as it
leaves the field at mass 9.109 x 10-21 kg.
Electron @ -1.6 x 10-19 C
v = 2.5 x 104 ms-1
X
X
X
X
X
X
X
X
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x B = 2.0 x 10-2 T
F = qvB
= (-1.6 x 10-19) x (2.5 x 104) x (2.0 x 10-2)
= -8.0 x 10-17 N South
F = ma
8.0 x 10-17 = (9.109 x 10-21) x a
a = (8.0 x 10-17)/(9.109 x 10-21)
= 8.78 x 1013 ms-2
v = u + at
= 0 + 8.78 x 1013
= 8.78 x 1013 ms-1
Similarly, the magnetic field would have to be huge in dimensions if the particle were
trapped for one second horizontally as at first assumed.
9.4.1.4 – Identify that charged plates produce an electric field.
d
E
E = V/d
Where: E = Electric field strength (NC-1)
V = Potential Difference/Voltage (V)
d = Distance between plates (m)
Also from earlier work,
Where: q = Charge (C)
F = qE
V
And,
W = Fd and therefore, W = qEd
Where: W = Work done on or by the charge (J)
F = Force (N)
d = Distance F is applied in the same direction as the force (m)
When work is done on a charge, it will acquire energy – in this case KE (EK = ½ mv2).
Combining these equations then, for a charge entering an electric field due to a parallel
plate capacitor, it experiences a force and gains energy given by W = qV = ½ mv2.
Example: Find E between two parallel plates separated by 5.0 mm if the V applied is 48
Volts.
E
V
d
E
48
0.005

E = 9600 NC-1

How much work is done moving a charge of 3.6 micro Coulombs through a potential
difference of 15 Volts.
W = qV
W = 3.6 x 10-6 x 15
W = 0.000054 J
Two parallel plates are 5.0 mm apart with a V of 200 Volts across them. A small object
of mass 1.8 x 10-12 kg and charge of 12 micro Coulombs is released from rest near the
positive plate (ignore gravitational effects). Calculate the velocity gained as it strikes the
negative plate.
Example: Two parallel plates 10 cm apart with a potential difference of 20.0 V between
the plates. Calculate the electric field, if a charge of +2.0 x 10-3 C is placed in the field
calculate the force acting on it, if this charge is released from rest near the positive plate,
calculate it’s acquired maximum velocity and if this same charge is travelling
horizontally at 750.0 ms-1 near the positive plate and the plates are 25.0 cm in length,
calculate the path (final velocity) of the charge and determine whether it passes through
the plates.
E

V
d
qV = ½ mv2
E

20
0.1
0.002 x 20 = ½ x 1.8 x 10-12v2
4.44444444444 10^10
E = 200 NC-1
v=
F = qE
v = 2.11 x 105 ms-1 towards negative plate

F = 2.0 x 10-3 x 200
v2 = u2 +2as
(2.11 x 105)2 = 0 + 2 x 0.1 x a
F = 0.4 N
a = 2.23 x 1011 ms-2
v = u + at
2.11 x 105 = 0 + 2.23 x 1011t
t = 9.47 x 10-7 s
Range = uxt
= 750 x 9.47 x 10-7
= 7.1 x 10-4 m
Therefore the charge would descend within 7 mm of entering the electric field.
9.4.1.5 – Describe quantitatively the force acting on a charge moving through a
magnetic field.
F = qvBsinØ
[See 9.4.1.3]
Example: A charge of 5.25 mC moving with a velocity of 300 ms-1 northeast enters a
uniform magnetic field of strength 0.310 T directed vertically downwards into the page.
Calculate the magnetic force on the charge.
F = qvBsinØ
F = 5.25 x 10-3 x 300 x 0.310 x 1
F = 4.88 x 10-1 N Northwest
Example: An electron travelling at an unknown speed enters a magnetic field of strength
0.666 T east and experiences a force of 3.80 x 10-7 N initially south. Calculate the
magnitude and direction of the electron velocity.
F = qvBsinØ
3.80 x 10-7 = 1.602 x 10-19 x v x 0.666
v = 3.57 x 1012 ms-1 Out of the page*
*Physical impossibility as it is faster than c
9.4.1.6 - Discuss qualitatively the electric field strength due to a point charge,
positive and negative charges and oppositely charged parallel plates.
[Pg 173 in text]
9.4.1.7 – Describe quantitatively the electric field due to oppositely charged parallel
plates.
See earlier notes.
E = V/d or V = Ed
9.4.1.8 – Outline Thomson’s experiment to measure the charge:mass ratio of an
electron.
In the 1890’s, J. J. Thomson carried out a series of brilliant experiments to investigate the
properties of cathode rays and in particular to design an intricate experiment to measure
the charge to mass ratio of these cathode rays (or electrons as they become known as).
As discussed earlier, Thomson was able to cause cathode rays to be deflected by electric
fields produced by parallel plate capacitors. He then went on to examine the effect of B
and E in combination and alone – in doing so was able to derive an expression for q:m or
q
.
m

He did this is two stages:
1. By combining and adjusting magnetic field and electric field he was able to
balance B and E so that the cathode rays passed through without being deflected.
This meant that
FE = FB.
Therefore FE =
qV
. . . . . .
qE or
and
Cathode Ray e
d
. . . . . . E
FB = qvBsinØ
B
. . . . . .
Pass through undeflected
+

2. Thomson then removed the E while maintaining the same B
This means if B was strong enough the path of the cathode ray/electron would be circular
and that the force due to B (FB = qvBsinØ) is supplying the centripetal force (Fc = mv2/r).
By combining equations from both experiments he was able to determine a mathematical
value for q/m (charge to mass ratio).
From experiment 1:
FE = qE
FB = qvBsinØ (Ø = 90º)
But FE = FB
Therefore, qE = qvB
V = E/B
From experiment 2:
FB = qvB
FC = mv2/r
But FB = FC
Therefore, qvB = mv2/r
q/m = v/Br
But from experiment 1:
V = E/B
Therefore, q/m = E/B2r
E can be calculated/measured (E = V/B). Similarly B can be calculated/measured. r can
be measured from the curvature from experiment 2. Thus, Thomson could calculate the
charge to mass ratio (q/m or q:m). Thomson found exactly the same ratio regardless of
the material the cathode was made from and also that this value was incredibly large
(1.76 x 1011 C/kg-1). This value was around 1800 times greater than the ratio obtained by
using the lightest know element and producing irons by electrolysis (H+). This means that
it is likely to be smaller (less mass) than the lightest known element and since it was
common to all cathode materials it must be a subatomic particle.
Thomson is therefore accredited for discovering the electron.
*Outside course*
It was not until 1909 that an American physicist Robert Milicon, performed an
experiment to actually calculate the charge on an electron (qe). Here he balanced the
electrical force (FE = qE) on an oil drop with gravity (Fg = mg).
Equating these two equations mg = qE and q = mg/E.
E and g could or were known and m could be calculated using diameter, density and fluid
physics equations. This allowed Milicon to calculate the charge on one oil drop. He
repeated this for thousands of oil drops and found each charge was a multiple of 1.6 x 1019
C – this was accepted as the charge on an electron.
9.4.1.9 – Outline the role of – the electrodes in the electron gun, the deflection plates
or coils, the fluorescent screen – in the cathode ray tube of conventional TV displays
and oscilloscopes.
[Excel textbook]
9.4.2 – Black Body Radiation/Photoelectric
9.4.2.1 – Describe Hertz’s observation of the effect of a radio wave on a receiver and
the photoelectric effect he produced but filed to investigate.
9.4.2.2 – Outline qualitatively Hertz’s experiments in measuring the speed of radio
waves and how they relate to light waves.
Early history – Maxwell’s theory of EM waves
Based on observations that a changing magnetic field produced/induces and electric field
around a magnet and that a current carrying wire induces a magnetic field, Maxwell
concluded an interrelationship electric and magnetic fields. The following sequence
explains how an EM wave propagates:
 A time varying electric field in one region produces a time and space varying
magnetic field at all points around it.
 This varying B similarly produces a varying E in it’s neighbourhood
 Thus, an EM disturbance starting at one location (eg. Vibrating charges in a hot
gas or in an antenna) creates a disturbance which travels out through mutual
generation of E and B
 The E and B propagate through space in the form of an EM wave (ie. Self
propagating)
Hertz’s Experiments with Radio Waves
Two of the more important predictions of Maxwell and his equations were:
 EM waves could exist with many different frequencies
 All EM waves would propagate through space at the speed of light
In 1886, Heinrich Hertz conducted a series of experiments verifying these predictions.
Hertz reasoned he should be able to produce EM waves of different f values by creating
rapidly oscillating electric field with an induction coil. He used the induction coil to
produce sparks between the spherical electrodes of the transmitter (see figure 11.6/7/8).
He observed that when a small length of wire was bent in a loop with a small gap
between the ends (and held in the right plane), a spark would jump across this detector
loop even though this loop was not connected to any source of current. Hertz concluded
that electromagnetic waves had travelled from the transmitter to the detector.
He then showed that these EM waves could be reflected from a metal mirror and
refracted as they passed through a prism made from pitch ie. They had wave properties
not light.
He was also able to show that these ‘new’ EM waves could be polarised. If the detector
loop was perpendicular to the transmitter gap, the radio waves produced no spark;
however if it was placed parallel to the spherical electrodes of the transmitter the spark
was at a maximum. See figures 11.7 (perpendicular, no spark) and figure 11.8 (parallel,
maximum spark).
Hertz reasoned that if the EM waves travelled from the transmitter to the detector that
there should be a small delay between the transmitted spark and the detector spark. He
measured this speed in 1888 using a predetermined frequency from an oscillating circuit
and determined it’s wavelength by measuring it’s interference effects (refer to laser
refracting experiment done later). These measurements allowed Hertz to calculate the
velocity of this wave (v = ƒλ) and he achieved an answer identical to the speed of light.
Hertz’s experiments and procedures were the first involving EM waves of a know
frequency. His waves are now classified as radio waves. Hertz never transmitted his radio
waves over more than a few hundred meters. It was not until the later work of Marconi
(late 1890s to early 1900s), that radio waves became used in communication. He found
that longer wave lengths penetrated further than short wave lengths and that tall aerials
were more effective for producing the highly penetrating radio waves (long wave length)
than short aerials. (1895 sent a radio message about 3 km and by 1901 across the Atlantic
from England to Canada).
Different radio frequencies can be generated easily and precisely by oscillating electric
currents (AC) and the length/s of the aerial.
Hertz also noticed the photoelectric phenomena but did not carry out any further research
on it. He noted that if the detector loop was darkened/enclosed then the maximum spark
length became decidedly smaller (ie. Illuminating the spark gap with UV light gives
stronger sparks).
9.4.2.3 – Identify Planck’s hypothesis that radiation emitted and absorbed by the
walls of a black body cavity is quantised.
When an object such as a light filament is heated it glows with different colours (black,
red, yellow, blue/white), as it gets hotter. To explain the emission of radiation from such
objects explaining how the landa/frequency, vary with temperature, we use objects called
black bodies.
[]
The modification was seen by Planck as a small correction to a ‘classical’
thermodynamics but it finished up being one of the most significant steps in the
development of a new branch of physics – quantum theory. Essentially he had developed
a mathematical ‘trick’ to explain the results of black body radiation and he failed to
accept the quantisation of radiation until much later in his career.
Classical physics in broad terms is described as the physics up until the end of the 19th
century. It relied on Newton’s mechanics and included Maxwell’s theories of
electromagnetism (This is still largely of use today for typical situations encountered on
earth). Classical physics predicted that the emission of EM radiation is continuous were
as quantum physics are stated that the energy occurred in discreet packets or quanta.
9.4.2.4 – Identify Einstein’s contribution to quantum theory and its relation to black
body radiation.
As mentioned earlier, Hertz was the first to notice this photoelectric effect when studying
the spark in the detector loop.
In 1899, J. J. Thompson set up some apparatus similar to that shown in figure 11.14. He
found that UV light caused electrons to be emitted from the metal surface of the cathode
(in this case zinc). The new feature of this experiment was that the electrons were ejected
from the metal by the radiation rather than the strong electric field used in cathode ray
tubes. It was shown that these ejected particle (electrons) were identical to cathode ray
particles.
In 1902, Von Leonard studied how the energy of emitted photoelectrons varied with the
intensity of light used. He observed that:
 Doubling the light intensity double the number of electrons emitted (greater
current)
 There was no change in the maximum KE when the light intensity was increased
(measured by the voltage required to stop the electrons reaching the other
electrodes [Vstop])
 The maximum KE of the electrons depended on the frequency of the light
illuminating the metal.
Classical physics was unable to explain these results. It predicted that energy from the
light could be absorbed slowly over time until the electrons gain sufficient energy to
leave the metal surface. The existence of a threshold frequency (f0) could not be
explained. The table below summarises the classical predictions and the actual
experimental results (which Einstein’s model was able to explain).
Einstein successfully explained the photoelectric effect using Planck’s theory in which
the particles of light (photons) carried energy that could be transferred to matter. He
proposed that:
1. Light exists as photons, each with an energy (E = hf)
2. Light intensity depends on the number of photons
3. Photons with the highest energy correspond to light with the highest frequency
4. *To produce the photoelectric effect (freeing an electron from the surface of the
metal) the energy in the photon must be equal to or greater than the energy
holding the electron to the surface. The energy require to release an electron from
the surface is called the work function (value is different for different metals)
5. *If the energy of the photon is greater than the work function, then this difference
provides kinetic energy to the photoelectron (ie. A high frequency photon is more
likely to pass on a higher difference in energy and therefore the photoelectrons
released will have a higher maximum kinetic energy. This KEmax can be measured
by changing the voltage across the plates until the flow of photoelectrons just
ceases [this is known as the stopping voltage or Vstop])
Example:
Q1. A beam of monochromatic light (lambda = 5.0 x 10-7 m) hits a perfect black body
and imparts 0.10 mW of power to it. Calculate:
a. f
b. Energy per photon
c. Number of photons striking per second
a.
c = fλ
3.8 x 108 = f x 5.0 x 10-7
f = 6.0 x 1014 Hz
b.
E = hf
E = 6.63 x 10-34 x 6.00 x 1014
= 3.98 x 10-19 J
c.
P = 0.10 mW
= 1 x 10-4 W
P = W/t
W = Pt
= 1.0 x 10-4 x 1
= 1.0 x 10-4 J
1.0 10^4
3.98 10^19
= 2.52 x 1014
Therefore, No. Photons =
Q2.
a.

f = 7.0 x 1015 Hz
KEmax = 9.0 x 10-19 J
W = KEmax
W = qVstop
9.0 x 10-19 = -1.6 x 10-19 x V
V = -5.62 Volts (to repel photoelectrons)
b.
E = hf = WF + KEmax
hf = WF + KEmax
6.626 x 10-34 x 7.0 x 1015 = WF + 9.0 x 10-19
WF = 3.74 x 10-18 J
c.
hf0 = WF
(threshold frequency = f0)
3.74 10^18
f0 =
6.626 10^34
= 5.64 x 1015 Hz

9.4.3
– Semiconductors and Transistors
9.4.3.1 – Identify that some electrons in solids are shared between atoms and move
freely.
Different materials vary greatly in their ability to conduct electricity (as a solid). Their
conductivity depends on the ease with witch the electrons are able to move within the
crystal lattice or arrangement of atoms/ions.
Covalently bonded substances (usually between non-metallic elements) involve the
sharing of electrons. These electrons are therefore held tightly and are therefore usually
unavailable for electrical conductivity/electron flow. Such substances (which do not
conduct electricity easily) are referred to as electrical insulators.
Ionic compounds for an intricate lattice held together by the attraction of positive and
negative ions (the metals loose electrons to form positive ions while the non-metals gain
electrons to form negative ions). Ionic compounds will conduct electricity when molten
or dissolved because the ionic lattice is broken and the ions are free to move. However, in
the solid state the ions are fixed and so no conductivity occurs.
Metals, on the other hand, consist of an orderly array (a lattice) of positive metal ions
surrounded by the de-localised valance or other electrons. These de-localised electrons
or sea or electrons are able to move randomly throughout the lattice and will display a
net movement when an electric field (E) or voltage is applied. This means that metals are
good conductors of electricity. [Refer to figure 12.2, pg 209]
9.4.3.2 – Describe the difference between conductors, insulators and semiconductors
in terms of band structures and relative electrical resistance.
When atoms of any type of substance (conductors, semiconductors and insulators) are
very close together as in a solid, there highest electron energy levels (valance levels)
form a band called the valance band. In a conductor, this valance band is only partly
filled and the valance band and conduction band above overlap.
In an insulator, electrons have been shared to completely fill the valance band and so
there is a very large energy gap between this valance band and the conduction band
(energy level needed to ‘escape’ the atom and therefore conduct).
In the case of semiconductors, there is an energy gap between the valance band and
conduction band but it is much smaller than the case with the insulator. [figure
12.5,Band
table
Conduction
12.1].
Conduction Band
Valance Band
Valance Band
Energy Gap
Valance Band
With metals, increasing temperature increases resistance (decreases conductivity).
Increasing temperature increases lattice vibration which ‘upsets’/restricts the flow of
electrons.
In the case of semiconductors, however, an increase in temperature results in a decrease
of electrical resistance (increase in conductivity). The increase in temperature allows
some electrons to move across the energy gap from valance to conductive and once there
are free to move under the influence of E. At low temperatures eg. Absolute zero (273ºC) a semiconductor would act like an insulator.
9.4.3.3 – Identify absences of electron in a nearly full band as holes, and recognise
that both electrons and holes help to carry current.
The semiconductor material widely used today is silicon (historically germanium was the
first used). Silicon is in group 4 of the periodic table and has 4 outer (or valance)
electrons. To achieve chemical stability the silicon covalent bonds with neighbouring
silicon atoms to produce a network covalent solid (an intricate lattice).
.. .. ..
: Si : Si : Si :
.. .. ..
.. .. ..
: Si : Si : Si :
.. .. ..
9.4.3.5 Early diodes and transistors were made from germanium because suitable industrial
techniques were developed to purify germanium to the ultra pure levels required to act as
a semiconductor during WWII.
The major problem with Germanium, however, is that it becomes too good of a conductor
when it gets too hot – allows too much current to pass through which can damage the
electronic equipment. (The resistance to electric current flow that makes a semiconductor
useful also generates heat).
Silicon was the other element with semi conducting properties predicted to be ideal as a
semiconductor. Silicon in fact is quite common (most sand is made of Silicon Dioxide
[SiO2]). It is however much more difficult to purify and extract. However, once Silicon is
purified to the ultra pure level needed, it is affected much less by higher temperatures
while maintaining it’s performance level. The first transistors were made in 1957 by
Gordon Teal (Texas Instruments). From the 1960’s onwards Silicon has become the
material of choice.
9.4.3.6 – Describe how ‘doping’ a semiconductor can change its electrical properties.
9.4.3.7 – Identify differences in p and n-type semiconductors in terms of the relative
number of negative charge carriers and positive holes.
As mentioned earlier, pure semiconductors contain the precise number of electrons to fill
their valance band. They can only conduct if electrons are introduced into the conduction
band or are removed from the valance band to produce holes (intrinsic semiconductors).
To enhance the conductivity of a semiconductor, a tiny amount of an impurity atom can
be introduced into the crystal lattice to alloy with the material. This is called ‘doping’.
There are two types of extrinsic semiconductors – n type (negative) and p type (positive).
n-type
These are formed when a group 5 impurity atoms such as Phosphorus and Arsenic is
substituted into the crystal lattice. Group 5 elements have 5 valance electrons (compared
to Silicon which has 4); 4 of the 5 Arsenic electrons will pair up with corresponding
Silicon atoms in the lattice but this leaves 1 ‘extra’ electron left over. This extra electron
is easily promoted to the conduction band. This increases the number of negative charged
carriers in the conduction band and such semiconductors that produce excess of negative
charge carriers are called n-type semiconductors. (figure 12.8 pg 215)
p-type
Here an atom from group 3 such as Boron or Gallium is substituted into the crystal
lattice. This atom has 3 valance electrons which means that there is one electron short in
the lattice thus creating a hole.
9.4.3.8 – Describe the differences between solid state and thermionic devices and
discuss why solid state devices replaced thermionic devices.
In electrical devices and circuits there is often a need to control the direction of current
flow by either converting AC to DC (rectifying); switch current flow on or off or amplify
a current. Prior to the invention of semiconductors and the associated solid state devices,
thermionic devices accomplished the task.
Thermionic devices use heated filaments and terminals/electrodes set in vacuum tubes
(eg. The valves in old radio sets).
The filament in the vacuum valve is heated by an electric current, causing it to liberate
electrons and act as a cathode (negative electrode). These electrons are then accelerated
by a high potential difference towards an anode. The simplest of such a valve is the diode
shown below. (Figure 12.11 pg 216)
The cathode may be heated directly (with current flowing through the cathode) or
indirectly (with a separate filament). With the negative terminal of the high voltage
source attached to the cathode, electrons can flow through the diode creating an electric
current. If this battery or high voltage source were connected in reverse, no current would
flow. Such ‘unidirectional conduction’ makes a diode suitable as an ‘electronic switch’
and for converting AC into DC current (rectification).
Later scientists (around 1900) added a positively charged plate or grid within the vacuum
tube which enabled the apparatus to amplify the input signal (known as a triode).
[Put Ron’s notes here]
Electrons can only flow in one direction using such combinations when an electric field is
applied – they can only move from n to p (with an n semiconductor, there is an excess of
electrons which can flow across the junction into the p-type to fill the positive holes
available when the field is applied; electrons cannot flow from the p to the n). This
ensures current flows in one direction only so that a simple p/n combination could act as
a diode to ensure unidirectional current.
Transistors usually are used to amplify current/readings and use two junctions, usually
either npn or pnp (see pg 220 for further explanation).
9.4.4 – Superconductivity
9.4.4.1 – Outline the methods used by the Braggs to determine crystal structure.
9.4.4.2 – Identify that metals possess a crystal lattice structure.
Interference/Diffraction Introduction
We know from earlier prelim studies that one of the wave properties is that they undergo
diffraction and show interference patterns.
For example, when water waves interfere from 2 source points (coherent), the waves
spread out, interfering with each other to produce a regular interference pattern consisting
of regions of maximum amplitude (constructive interference) and ‘zero’ amplitude
(destructive interference).
For constructive interference, the waves from
each source must arrive in phase; for
destructive interference they must be exactly
out of phase (half lambda, one and a half
lambda etc.). It can be shown that that light
waves undergo similar interference,
producing similar patterns.
Experiment/Exercise
Lambda (nm)
612
555
492
428
f (Hz) x 1014
4.9
5.4
6.1
7.0
Colour
Red
Orange
Yellow
Blue
Frequency vs. Back Voltage
Back Voltage/Vstop (V)
Back Voltage/Vstop (V)
0.349
0.552
0.777
1.141
[Insert Figure 13.3 Here]
It can be mathematically shown that:
n =
yd
L
Where: n = integer (which maxima 0, 1, 2, 3, 4….)
 = wavelength (m)

to n maxima (m)
y = distance from central maxima
d = distance between the two sources (m)
L = distances from sources to screen (m)

Example:
d = 3.333333333 x 10-6
y = 0.3675
L = 1.727

n 
yd
L
n 
3.333333 106  0.3675
1.727
= 7.09 x 10-7

= 709 nm
Light and/or other waves either bend around an obstacle or spread into a shadow zone
provided the gap or barrier is comparable to the wavelength of the wave.
X-ray Diffraction
X-rays were discovered by Röntgen in the 1890’s and were shown to be electromagnetic
waves with wavelengths much shorter then that of visible light (in the order of 10-10 m).
Within a short period of time they could be reliably produced with a specific frequency.
In 1912, the German physicist Max Von Laue proposed that the regular spacing of a
crystal such as sodium chloride, might form a natural 3D diffraction gradient for x-rays.
His colleagues, Friedrich and Knipping obtained a diffraction pattern on 3D film when
they bombarded a crystal of zinc sulphide.
9.4.4.3 – Describe conduction in metals as free movement of electrons unimpeded by
the lattice.
9.4.4.4 – Identify that resistance in metals is increased by the presence of impurities
and scattering of electrons by lattice vibrations.
9.4.4.5 – Describe the occurrence in superconductors below their critical
temperature of a population of electron pairs unaffected by electrical resistance.
The explanation for how superconductivity is occurs will be covered I the next syllabus
point on BCS theory (here pairs of electrons).
9.4.4.6 – Discuss the BCS theory
9.4.4.7 – Discuss the advantages of using superconductors and identify limitations to
their use.
9.4.4.S5 – Process information to discuss possible applications of superconductivity
and the effects of those applications on computers, generators and motors and
transmission of electricity through power grids.
The major problem (limitation) with wide scale use of superconductivity is obtaining and
maintaining the very low temperatures (tc) needed. Those superconductors which operate
at ‘higher temperatures’ eg. Metal ceramics; do not have suitable properties for
durability/longevity and is often brittle and will react with oxygen.
Further research however may be able to produce superconducting materials with more of
the desired metal-like characteristics, at temperatures which can be more economically
achieved.
There are numerous current applications and potentially limitless applications if these
limitations problems are overcome. Current can potentially be infinitely stored in circular
circuits (no joule heating effects [Ploss = I2R]); transport of electricity (DC) large
distances without loss of energy as well. Also hug super magnets can also be produced
with their own resulting applications.
Some current and potential applications are listed below:
 Power transmission – traditional transmission lines (Copper and/or aluminium)
loss an appreciable amount of energy due to the resistance of the wires and the
heating effect (at this stage set as high V AC to reduce this effect).
Superconducting wires could carry 5 or more time the current (DC). At this stage
however the ‘high temperature’ superconductors are far too brittle to be of use.
One current experimental technique is to use a ‘High Temperature’
Superconductor (HTS) (wound around a hollow helium liquid core).





Power generation – could use superconducting magnets requiring a much smaller
iron core (fraction of the size and mass of present generators). Potentially more
efficient – uses less fossil fuels to generate similar amounts of electricity.
Power storage – power stations cannot store electricity easily – most is used
immediately after production. This means that power stations/generators cannot
work at their optimum efficiency levels but are rather ‘turned up or down’ to meet
demand. Similarly other alternative energy sources for large-scale electricity
production are often inconsistent in their output eg. Solar, wind, tidal etc.
Superconducting Magnetic Energy Storage (SMES) is one possible solution. The
DC current produced by whichever method can be passed into the SMES’s
circular interior path and flow indefinitely without energy loss until it is needed.
Magnetically Levitated Trains (Maglev) – see assessment task. 2 types currently
used: EMS in Germany using conventional electromagnets to lift the train above
the track and EDS in Japan which uses superconducting magnets.
Electronics – computers are limited to some extent by generation of heat due to
resistance and the speed with which signals can be sent. Potentially
superconductors may solve some of these problems. Already in use is the
Josephson Junction a very fast switching device with current uses in sensitive
detectors used in geology.
Medical diagnosis – currently used to produce the intense magnetic fields used in
Magnetic Resonance Imaging (MRI) instruments. A superconducting alloy of tin
– niobium is used capable of producing magnetic fields greater to or equal to 4 T.
MRI machines effectively measures the concentration of hydrogen atoms and
from this a measure of the soft tissue of the patient can be computer generated.
9.8 – From Quanta to Quarks
9.8.1 – Early Models of the Atom
9.8.1.1 – Discuss the structure of the Rutherford model of the atom, the existence of
the nucleus and electron orbits.
Our current model of the atom has evolved over time. One of the earlier scientists to
propose/discuss atoms was John Dalton.
Later contributors around the turn of the 20th century included Thompson, Lenard and
Rutherford.
After his ‘discovery’ of the electron, J. J. Thompson went on to propose a model for the
atom (commonly know as his ‘plum pudding’ model). He visualised a spherical positive
mass with electrons embedded like plums in a pudding (positive and negative charges
cancel out). He even made an attempt to place these embedded electrons in shell type
structures to represent periodicity/electron configuration.
However, experimental evidence failed to support Thompson’s and later Lenard’s
models.
In 1907, Ernest Rutherford (Kiwi) moved to Manchester in England and returned to
investigations of the scattering of alpha particles, this time by very thin metal foils. He set
work for 2 scientists/undergraduate student (Geiger and Marsden). The approximate
apparatus and set up is shown below.
Lead Box
Scintillation
scope
Alpha source
Think metal sheet
with hole
Zinc of sulfide
screen
UV thin sheet of
metal (gold) foil
Geiger and Marsden found that the majority of alpha particles passed straight through the
thin foil as expected (red arrow) based on the then current theories. However, a
significant number were found to be deflected at much larger angles. It was Rutherford
that took these results and developed a new atomic theory to account for these
deflections.
He proposed that the atom must be made of almost empl
9.8.1.2 – Analyse the significant of the hydrogen spectrum in the development of
Bohr’s model if the atom.
The spectrum of hydrogen was well known and each spectral lines frequency and
wavelength had been accurately measured and recorded. In fact, several mathematicians
had derived several equations to match these measured wavelengths. Balmer derived the
n2
equation:
 = b( 2 2 )
n 2
Where: b is a constant
N is an integer > 2


This was modified by Rydberg to produce the more familiar equation:
1
1
1
 RH ( 2  2 )

nf
ni
Where:  is the wavelength of the spectral line
RH is the Rydberg constant
 level/shell the electron finishes at
nf is the electron energy
ni is the electron energy shell/level that the e originally has.

Bohr knew that the electrons in atoms must somehow produce the characteristic radiation
of the spectrum for each element. He was also familiar with Planck’s earlier description
of ‘atomic oscillators’ (see photoelectric effect). Bohr attempted to introduce the quantum
ideas of Planck to the atom and once he was mad aware or Balmer’s equation, was able to
explain the spectrum for hydrogen.
Exercise: Calculate and compare to the accepted values the first four lines in the Balmer
series (nf = 2).
n.d. RH = 1.097 x 107
1

1


 RH (
1
1
)
2 
nf
ni2
1.097 107 (
1 1
 )
22 32
  6.56 107

1



1.097 107 (
1 1
 )
22 4 2
  4.86 107
1


1.097 107 (
1 1
 )
22 52
  4.34 107

1


1.097 107 (
1 1
 )
22 62
  4.10 107

Calculate frequency for each wavelength and  E for each result.

n=6
n=5

Paschen Series (infra-red)
n=4
n=3
n=2
n=1
Brakett Series (infra-red)
Lyman series (ultraviolet)
Pfund series (infra-red)
Balmer series (visible light)
Principle Quantum Number
(n)
Energy E
(eV)
7
6
5
-0.28
-0.38
-0.54
4
-0.85
3
-1.51
2
-3.4
Pfund
series
Paschen
series
Balmer
series
Brakett
series
1
Lyman
series
-13.6
9.8.1.S1
9.8.1.S2
Aim: To observe and match the spectra of hydrogen (emission) and to explain the Balmer
series diagrammatically.
Method:
Enclosed in Box
Induction Coil
H2 Lamp
Spectroscope
6V DC







The apparatus was set up as per above.
Care taken with induction coil (x-rays, high voltage, damage if not connected to 6
volt DC).
Similar x-ray dangers with the hydrogen spectral tube.
If necessary the H2 tube was enclosed by a cardboard box with viewing slits to
eliminate background light.
The spectroscopes used were made from diffraction gradients which spread the
spectra when viewing and provide a scale to estimate the wavelength of spectral
lines observed.
The lines observed were recorded in a table and a diagram; and compared to the
accepted values for hydrogen from a spectral chart or the text on pg 423 figure
22.9.
Diagram/s representing the Balmer series were also drawn.
Results:
Observed
Wavelength
Red
Blue
Not observed
Not observed
4.10
Theoretical
Wavelength
Red
Blue
Violet
Violet
6.60 x 10-7 m
4.90 x 10-7 m
4.35
4.85
6.55 x 10-7 m
4.85 x 10-7 m
4.35 x 10-7 m
4.10 x 10-7 m
6.55
n=6
Principle Quantum Number
(n)
n=5
Energy E
(eV)
7
6
5
n=4
n=3
4
n=2
3
n=1
2
Balmer
series
1
Balmer series (visible light)
Eg. The 1st line is the Balmer series.
9.8.3 – Nucleus to Nuclear Energy
9.8.3.1 – Define the components of the nucleus (protons and neutrons) as nucleons
and contrast their properties.
All particles found within the nucleus (protons and neutrons) are known as nucleons.
The table below summarises and contrasts their properties:
Particle Symbol/Representation Charge
Mass
Mass
(approx)
(accurate)
-19
1
Proton
+1
or
+1.6
x
10
1.0
amu
1.673 x 10-27 kg
1p
[Reflected by E
and B]
Neutron


1
0
n
0
[No deflection in
E or B]
1.0 amu
1.675 x 10-27 kg
Location Penetration
Nucleus
(Nucleon)
Reasonably high
Nucleus
(Nucleon)
Very high
*Introductory Discovery of Radioactivity*
In the 1890s, Becquerel was studying the phosphorescence of minerals (exposed to
sunlight) then reemitted radiation in the dark later.
On a particularly cloudy day he threw a mineral sample (without being exposed to
sunlight) into a draw containing a photographic plate. The mineral produced radiation
that penetrated the black paper surrounding the photographic plate (which would
normally block any light and darkened the plate. The properties of these new rays seemed
similar to recently discovered x-rays of Roenteng.
Rutherford was able to show in 1898 that this particular mineral emitted two types of
radiation (alpha and beta); and the third gamma was discovered in 1900 by Villard.
As we investigated in the preliminary course, alpha, beta and gamma have their own
unique properties.
Name
Symbol Charge
Mass
Penetration
Ionising Ability
4
Alpha (  ) 2 He
+2
4 amu
Poor
Excellent (due to
(10-20cm air, thin
size and charge)
metal, paper)
+1
Reasonable but
Reasonable
  ) 01 e
0
(
Beta

annihilates
when
[Positron]
contact with normal
electron

 

0
-1
Reasonable
Reasonable
0
Beta (  ) 1 e
[Electrons
]

 

Gamma
0
0 (Wave) Very High
Very Poor (lack
or hf
photon
of mass)
()


9.8.3.2 – Discuss the importance of conservation laws to Chadwick’s discovery of the
neutron.
After the discovery of the nucleus it seemed logical to assume that the nucleus contained
protons and electrons (which would explain the emission of alpha and beta) however
there were major problems with this concept and in 1920 Rutherford proposed a neutral
party (neutron) with a mass comparable to a proton.
In 1919, Rutherford bombarded nitrogen gas with alpha particles (from Bi-214). A
positively charged particle more penetrative than an alpha was produced and later
identified as a proton. This was the first artificially induced transmutation.
In 1930, Bothe and Becker fired alpha particles at Beryllium and found that a highly
penetrating radiation was produced. The radiation seemed to be similar to gamma but
even more penetrative and had a much higher energy value (10 MeV) than any previously
observed gamma ray. Frederick Joliot and his wife Irene Curie allowed this radiation to
fall on a block of paraffin wax (a hydrocarbon rich in hydrogen atoms – protons.
Some leading physicists were adamant that they would but others including Bohr thought
otherwise (it wasn’t until 1936 that Bohr dropped his ideas of non-conservation of
energy). The diagram and equations below show/summarise the overall development.
Beryllium
Paraffin Wax
1
0
1
0
5
Conventional Mirror
n?
n?


 source
9
4
Be 24 He 
 126 C 01n ?

9.8.3.3 – Define the term ‘transmutation’.
9.8.3.4 – Describe nuclear transmutations due to natural radioactivity.

Transmutation is the process where a new element (daughter) is formed from a different
element (parent).
In the preliminary course wee covered much of this and wrote equations for alpha and
beta decay. We also need to incorporate neutrinos and antineutrinos associated with beta+
and beta- decay (see later syllabus points).
4
U 
 234
90Th 2 He (Alpha decay)
238
42

_
0
Bi 
 212
84 Po1e(electron)  U (antineutrino) (Beta- decay)
212
83
O
 157 N 01e( positron)   (neutrino) (Beat+ decay)
15
8

Uranium-238 undergoes both alpha and beta (assume beta-) decay until a stable isotope
of Pb-206 is reached. Calculate the number of alphas and betas.

He-4  each decay reduces mass by 4.
Total mass lost = 238 – 204
= 32
Therefore, No. of Alpha = 32/4
=8
Each alpha decay reduces charge by 2
(8 x -2  16)
But each beta- increases each charge by 1
92 – (8 x alpha) + (? X beta) = 82
92 – 16 + b = 82
b=6
Elements are naturally radioactive for usually one of three reasons:
 Too many neutrons compared to the number of protons (neutron turns into proton
and ejects an electron/beta-)
 Too many protons for the number of neutrons (proton turns into a neutron and
ejects positron/beta+)
 The element is too big/heavy – elements above atomic No. 83 (Bi) are likely to
undergo alpha decay
Gamma radiation often accompanies either alpha or beta as the nucleus de-excites.
Gamma is not a transmutation since a new element is not formed.
As well as natural radioactivity, artificial transmutation can also be induced (see
Rutherford’s first artificial transmutation in the previous section on the discovery of
neutrons).
27
13
10
5

9
4
28
Al 12H 
 14
Si 01n
B 24He 
 136 C11p
Be11p
 36 Li 24He

As can be seen from the artificial transmutations above, alpha and beta particles can and
were used to supply an understanding of physics atomic principles by probing other
atoms/nucleon. The 
resulting protons and neutrons sometimes produced could also be
used as a probing/diagnostic tool, particularly neutrons which have excellent penetration
(due to its neutral charge) but still with a significant KE/momentum.
9.8.3.5 - Describe Fermi’s initial experimental observation of nuclear fission.
After the discovery of the neutron, Fermi and his co-workers in Rome carried out many
experiments with the neutron bombardment of many elements. They noticed that for
some of the heavy elements there was often a delayed emission of a beta particle
resulting in an element with a higher atomic number.
Uranium (atomic no. 92) is the heaviest naturally occurring element and when Fermi’s
team fired neutrons at it they were able to make the first transuranic element. In a similar
fashion, Np-239 can be converted to Pu-240.
1
0
0
n238
 239
 239
92U 
92U
93 Np1e
1
0
0
n 239
 240
 240
93Np 
93 Np
94 Pu1e

This can continue until eventually Am-241.
Transuranic 
elements heavier than Americium cannot be made by neutron capture
(nowadays this neutron capture is achieved by placing a sample of suitable material in or
near a fission nuclear reactor where there is an abundance of neutrons); for elements with
atomic number greater than 95, particle accelerators must be used (see 9.8.4).
We now know that when Fermi and his associates bombarded Uranium with neutrons,
they must have produced nuclear fission. The production of transuranic elements would
also have occurred but the observations of radioactivity that Fermi assumed was solely
due to transuranic element formation was in fact partly due to the production of fission
products.
The first observations confirming nuclear fission were made by Otto Frisch using an
ionisation chamber to detect emitted particles. During the course of their neutron
bombardment experiments, Fermi’s co-workers (Amaldi and Pontecorvo) discovered that
the same experiment yielded different results when performed in different parts of the
room (the radioactivity was much greater after the substances had been irradiated with
neutrons on a wooden rather than marble table). Fermi decided to investigate this
anomaly by placing a paraffin block between the neutron source and the target - he
discovered that slow neutrons were much more likely to cause a reaction than fast ones
(the De Broglie wavelength of slow neutrons was longer and meant there was a greater
possibility of capture by the nucleus). The paraffin was effectively acting as a moderator
(see nuclear reactors lately). Fermi was awarded the noble prize in 1938 for the discovery
of transuranics (and associated use of slow neutrons) - it allowed he and his Jewish wife
to accept the prize in Sweden, but they ‘escaped’ to the USA.
Near the end of 1938, Lise Meitner, Otto Hahn and Fritz Strassmann studies the new
heavier elements produced by neutron bombardment and discovered Barium as one of the
products rather than the previously identified Radium. The group began to consider
Bohr’s ‘liquid drop’ model of the nucleus. They considered that if the nucleus was like a
liquid drop, it could become unstable and split in two, the parts forced apart but
electrostatic repulsion.
Using this model, the velocities of each main part should have had an energy of about
200MeV. Meitner calculated this energy using mass defects and E = mc2 and showed that
an energy of about 200MeV would be released. He had proved mathematically that
fission had occurred. In fact we can write an equation for this (several possibilities).
Slow
Neutron
Begin to
Nucleus
1
236
141
92
U
n


U


Ba 36
Kr 301n
 92  56separate
0
unstable
Split (2
fragments)
with high
235
1
236
147
87
1
velocity
  92U
 57 La 35Br 20 n
92U 0 n 
due to

electrostatic
Frisch returned to Copenhagen and informed Bohr of their theory just as he was to travel
repulsion
to the US. Frisch performed the first experiment that actually confirmed the fission
 of fission reached the US even before Frisch and Meitner had published
reaction. News
their papers and many others performed similar experiments.
235
92
9.8.3.6 - Discuss Pauli’s suggestion of the existence of neutrino and relate it to the
need to account for the energy distribution of electrons emitted in beta-decay.
Even though scientists confirmed that beta- decay was in fact an electron (and that
therefore a neutron had decayed into a proton, in the process ejecting and electron) they
knew that an electron could not possibly be confined to the nucleus (De Broglie
wavelength of the electron was much larger than the radius of the nucleus). Prior to WWI
Chadwick was able to show that beta particles were emitted with a wide spread of
energies (figure 24.10, pg 460 in text shows these). There was a dilemma as how the
same beta decay from the same nucleus (to produce the same new nucleus) with different
energy values.
To resolve the paradox, Wolfgang Pauli predicted there must be another sub-atomic
particle - if the laws of conservation were to remain valid, the expulsion of beta particles
must be accompanied by a very penetrating radiation of neutral particles which had not
yet been observed. He referred to the yet unknown particle as a neutron - it was later
renamed the neutrino to avoid confusion with Chadwicks neutron.
By late 1933 Fermi had completed his famous paper on beta decay. He included Pauli’s
suggestion of the neutrino (along with the electron) and the suggestion by Heisenberg
that the nucleus contained only proton and neutrons. He proposed that the number of
electrons and neutrinos was not constant and that they could be created or disappear just
like photons. In a complex series of quantum aspects outside of this course he could
produce the shape of the beta decay spectrum (it assumed the mass of the neutrino was
zero or very close to it.
By the 1950s, beta decay was believed to be associated with the weak nuclear force. This
joined gravitational force, electromagnetic force and the strong nuclear force as a
fundamental force of nature.