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Fields Test Example Questions Guide
CAUTION: This guide covers 35 problems that will not appear on your Fields Test. Also under
stand that the guide makes statements that are usually true without adding qualifying remarks or
apologies. Advanced or special cases are not treated. In some cases, one or two steps are omitted.
Complete the solutions are you read through this guide.
STRATEGY: The best approach is to recognize the problem and solve it. If you are unsure, find
physical reasons to eliminate as many of the five options as possible. Make an educated guess as you
pick from among the remaining options. Mark the problem so that you can return to it later at which
time you can attempt a more thorough analysis. As you work through the guide, pay special attention
to the examples for which one or more of the options are found not to be viable.
DO NOT MEMORIZE: Memorizing values and facts from this guide is not recommended. Do
think about the concepts and values as you read this guide. Think and read just carefully enough to
be left with a hazy recollection of the experience. THINK, think, think while you take the test.
The sample questions appear boxed. Search for
Ex.
to find the next problem.
MECHANICS AND RELATIVITY:
Newton’s Laws: Use when asked about forces and acceleration at a single position and time.
2
acircle   v r rˆ  dv dt tˆ   2 r rˆ   r tˆ
Always account for the centripetal part. The tangential part applies if the speed is changing.
Static friction acts with a magnitude from 0 to s N to keep the contacting surfaces from slipping
relative to one another. One the surfaces are slipping, the friction force on each surface has
magnitude equal to k N directed to oppose the relative motion.
Factoid: k  s.
Work-Energy: Work is W = F  dr suggesting that work-energy by used when a change is
associated with a change in position.
Mechanical energy: Ei = K1 i + K2 i + … + UA i + UB i + …..
K = ½ mv2 + ½ Icm2 = ½ McmVcm2 + ½ mvrel 2 + ... = ....
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
U = ½ k (stretch)2 +
{mgh; - GMm/r} +
choose near-earth or Universal form
Ei + Wnon-conserve = Ef
{q V; kQq/r}
choose appropriate form
Wnc-frintion = - k N (distance that surfaces slip relative to one another)
A conservative is any force for which the work integral can be evaluated as the difference in the
values of a scalar function at the integration endpoints. (Can do if the work integral around any
closed path is zero; curl of the force is zero).
Momentum and Collisions: Impulse is

t2
t1
F dt suggesting that impulse-momentum methods be
tried for a change that occurs as time changes – even for the time change from before to after.
m
k
vk ,initial   Fnet ,external   mk vk , final
tf
Internal forces do not change the total momentum.
ti
(1) Collisions:
Rule 1: Conserve momentum vectorially
Rule 2: Kinetic energy is unchanged for an elastic collision. Some kinetic energy is converted to
other forms if the collision is partially or totally inelastic. As the total momentum is conserved, only
the kinetic energy associated with motion relative to the center of mass is available to be lost
(converted).
2
Ktotal   ½mi vi2  ½ M V 2   ½mi vrelative
, i  KCM  K relative ; KCM 
i
2
Ptotal
2M
i
M: total mass; V: speed of the center of mass (CM); vrelative, i : speed of mass i relative to CM.
In a totally inelastic collision, the particles stick together so Krelative is lost. KCM remains as required
by conservation of momentum.
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
Ex. 1) The figure to the left shows two
m, 2 vo
particles with masses and velocities as
indicated. The objects are moving on a flat,
frictionless surface. When they collide, the
300
objects stick together. Their speed after the
collection is most nearly:
300
(A) 0.67vo
(B) 0.87vo
(C)
vo
(D) 1.15 vo
2 m, vo
(E) 1.73vo
Rule 1 is to conserve momentum vectorially so a coordinate system is adopted.
m, 2 vo
y
300
x
300
2 m, vo
The momentum of the particles before the collision are p1  m (2v0 )[
p2  (2 m) v0 [
3
3
ˆ  ½ ˆj ] and
2i
ˆ  ½ ˆj ] . The total mass is 3 m and the total momentum is Ptot  4 mv0 [
2i
Physics Handout Series – fields.tank
3
ˆ] .
2i
page 3
Fields Test Example Questions Guide
v final 
Ptotal
4 m v0 [ 3 2 iˆ ] 2 v0 ˆ


i
M total
3m
3
(inelastic; stick together)
Was kinetic energy converted to other forms as a result of the collision?
The collision was totally inelastic. Was all the kinetic energy lost?
(2) Motion in a uniform gravitational field:
Ex. 2) Ball 1 is dropped form a height h and ball 2 is dropped form a height ½ h. Which of the
following gives the ratio of the speed of ball 1 to that of ball 2 just before they impact? (Assume that
air resistance is negligible.)
(B) 2 ½ (C) 1
(A) 2
(D) 2-½ (E) ½
First, the problem is a constant acceleration problem. Write down the master equations.
x(t) = xo + vo t + ½ a t 2;
v(t) = vo + a t ;
v 2 = vo2 + 2 a (x – xo);
vave = ½(v +vo)
The balls fall from rest for a distance H. Using v 2 = vo2 + 2 a (x – xo), v 2 = 2 g H or v  2 g H .
2 g H1
v1
h


v2
2 g H2
½h
Sanity Check. Three answers can be discarded immediately. Which three? Why?
If the answer were (A), ball 1 would have twice the average speed of ball 2. What would be the ratio
of the fall times in that case given that the first ball falls twice as far?
(3) Law of Universal gravitation:
Ex. 3) Two planets of mass m and M respectively have center-to-center spacing R. At what distance
from the planet of mass M do the gravitational forces of the planets cancel each other.
 


m R (B) 1  m R
(A) M
M
(C)
m R
M


m R
(D) 1  M
Physics Handout Series – fields.tank
(E)
R
m
1 M
page 4
Fields Test Example Questions Guide
One need only consider points along the line joining the planets. Why? What must be true about the
directions of two vectors that sum to zero?
Prepare a sketch: A large well-drawn sketch provides the greatest benefit. Assume M > m.
R-d
m
d
M
R
Thought 1: Gravitational forces are attractive. Therefore, the ‘zero – force’ point must be between
the planets and along the line joining them. Compare/contrast this situation with that in Ex. 10.
Thought 2: The force is an inverse square (with distance) law force.
Gm
GM
( R  d )2 m
Rd




 m
2
2
2
M
(R  d )
d
d
M
d
Think about the result. At the point where their influences balance, the distance from the smaller
planet is smaller than that from the large planet. (Include sanity checks as you proceed through your
solution.)
Rd  m
Since
m
M
M

d  R  d 1 m
M
  d  1 Rm
M
is less than one, d > ½ R. The point is farther from the more massive planet. Reread
the question. The distance d of the point from the more massive M planet is requested. The unknown
label d is assigned to the value requested. Always review the question to ensure that you have
answered the question that was asked.
(4) Statics and Archimedes Principle
In equilibrium, the sum of the forces acting on a rigid body is zero, and the sum of the torques (about
any axis  free to choose it where you wish) acting on the body is zero.
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
A body is buoyed up by the weight of the fluid that it displaces. FB = fluid Vdisplaced g.
Ex. 4) A metal block is suspended in an empty tank from a scale that indicates a weight of W. The
tank is then filled with water until the block is covered. If the density of the metal is three times the
density of water, what apparent weight of the block does the scale now read?
(A) ½ W
(B) 2/3 W (C) W (D) 3/2 W (E) 3 W
Examine your options. The new reading should be less than the original reading. Answer (B) could
be chosen without much more, but one should solve the problem before choosing an answer.
Prepare drawings
T = Wdry
T = Wwet
wVg
mVg
Wdry = m V g = 3 w V g
mVg
Wwet =m V g - w V g = (m - w) V g
* The ratio of the density of a material to that of water is called the specific gravity of the material.
Wwet = Wdry ( - w/m) = Wdry ( - /)
(5) Simple Harmonic Motion
Equation of Motion: x(t) = A cos[ t + ] = C cos[ t ] + D sin[ t ]
Take derivatives to find expressions for v(t) and a(t).
v(t) = -  A sin[ t + ];
a(t) = - 2 A cos[ t + ]
Energy Conservation: ½ m v2 + ½ k x2 = ½ m (vmax)2 = ½ k A2 = Etotal mechanical
=
k
m
 g
 
I
 stiffness
inertial property
Physics Handout Series – fields.tank

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Fields Test Example Questions Guide
A mass hanging from a linear spring executes simple harmonic motion about equilibrium.
Potential energies are associated with systems of interacting entities. Potential energies are
associated with pairs of things (at least) while a kinetic energy can be owned by a single entity. The
potential energy can be shifted by an additive constant. For SHM, it is often set to zero for a particle
positioned at equilibrium.
Ex. 5) A 1-kilogram particle is attached to a spring and exhibits one-dimensional simple harmonic
motion. The particle’s distance (!/* displacement) from the equilibrium position is given by the
expression: y(t) = A sin[ t + /2], where A = 1 meter and  = 0.5 rad/s. If the potential energy of the
particle (system) at its equilibrium position is null (or zero), which of the following gives the total
energy of the particle (system)?
(A) 2 J (B) 1 J (C) ½ J (D) 1/8 J (E) 0 J
v(t) = -  A sin[ t + ];
a(t) = - 2 A cos[ t + ]
½ m v2 + ½ k x2 = ½ m (vmax)2 = ½ k A2 = Etotal mechanical
The total energy is kinetic at equilibrium as the potential is set to zero at that point.
½ m (vmax)2 = ½ k A2 = Etotal mechanical
The relation v(t) = -  A sin[ t + ] shows that vmax =  A = 0.5 m/s. Using m = 1 kg,
Etotal = ½ m (vmax)2 = ½ (1 kg) (0.5 m/s)2 = 1/8 J.
Write down the equation that is to be used. Substitute numerical values with units for each symbol
writing the values in the same geometric pattern that was used for the symbols. Adopt procedures
that reduce the chance of making a careless error.
(6) Equilibrium with torques
In equilibrium, the sum of the forces acting on a rigid body is zero, and the sum of the torques about
any axis (  free to choose axis where you wish) acting on the body is zero.
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
  r  F ;    r F   r F  r F sin  
The displacement from the axis to the point of application of the force is r , CCW is positive, and 
is the angle that takes you from the direction of r to the direction of F .
Vector cross product: Extend the fingers of your right hand in along the direction of r . Orient your
hand so that the fingers can curl toward the direction of F . The thumb of the right hand will be in
the direction of r  F .
500
F
axis
r

r
line of action of F


Ex. 6) The figure shows a uniform rod of mass 4 kilograms that is pivoted at one end and supported
by a string at the other end. If the rod is a rest (in equilibrium) the tension in the string is most nearly
(A) 20 N
(B) 26 N
(C) 31 N
(D) 40 N
(E) 62 N
Use the figure The rod does not have an assigned length. Assign a length of , 1 m or even 4 m. If
the value is not given, the answer must be independent of its value.
Choose an axis: The pivot point is a natural choice for the axis. Picture the situation. If the tension
were changed, about what point would the rod rotate?
The tension force tends to cause a CCW (+)
½


rotation so the associated torque about the
500
0
40
rT 
40
pivot is T = + T  sin(400). To compute
torque, the weight of the boom can be
assumed to be applied at its center of mass.
W
W
T
0


The weight has a lever arm of ½  and tends
to cause a CW rotation. W = - W (½  ).
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
The sum of the torques must be zero so:
T
½ (4 kg ) (9.8 N kg )
½W
½mg


sin(400 ) sin(400 )
sin(400 )

½ (4 kg ) (9.8 N kg )
.707

20 N
 28.3 N
.707
We expect an answer larger than 28 N so response (C) is chosen. The actual value is about 30.5 N.
(7) Relativity
Relative speed v
Lengths are contracted along the direction of relative motion, but lengths running transverse to the
motion are unchanged.

1  (v c ) 2 ;
 

t  
A clock in motion relative to the observer runs slowly.
t
1  (v c) 2
If a time t = 10 s elapses as measured by a moving clock, and it appears to be running slowly as
viewed by the primed observer, then the corresponding interval t is longer than 10 s.
Consider a primed observer moving at v iˆ relative to an unprimed observer with the initial
conditions that their respective axes are parallel and the origins coincided at t = t  = 0.
Adopt the notations:   v c and   1  (v c) 2  1   2 .
x   [ x   ct ]
c t    [c t   x ]
y  y
z  z
 
t    1 t
Ex. 7) A stick of length L lies in the x-y plane as shown. An observer moving at 0.8 c in the x
direction measures the length of the stick. Which of the following gives the components of the
length as measured by the moving observer?
y
Lx
Ly
(A)
L cos
0.60 L sin
(B)
0.6 L cos
0.60 L sin
(C)
0.6 L cos
L sin
(D)
0.64 L cos
0.64 L sin
(E)
0.78 L cos
0.78 L sin
L

O
x
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
The component along the direction of motion should be contracted by a factor of  = 0.6 while the
transverse length should be unchanged.
Lx = 0.6 (L cos) ;
Ly = Ly = L sin
Note that only answer (C) has an unchanged transverse component.
ELECTROMAGNETISM AND CIRCUITS:
(8) Basic Circuits and Circuit Elements
The three basic passive elements are the capacitor, resistor and inductor. Each is characterized by its
VC =( 1/C) Q ;
voltage rule.
Series Combos:
Parallel Combos:
1
Cseries

1
1

C1 C2
Cparallel = C1 + C2
VR = I R ; VL = L dI/dt (Note 1/C in the VC relation.)
Rseires= R1 + R2
1
Rparallel

Lseries= L1 + L2
1 1

R1 R2
1
Lparallel

1 1

L1 L2
The voltage across a capacitor is a continuous function of time. Why?
The current through an inductor is a continuous function of time. Why?
Kirchhoff’s voltage rule: The sum of the voltages (potential changes) around a closed path is zero.
Kirchhoff’s current rule: The algebraic sum of the currents into a node is zero.
Elements are in series if the wiring of the circuit requires that the current through the elements is the
same. There can be no branch points between them.
Elements are in parallel if the wiring of the circuit requires that the voltage across the elements is the
same. One lead from each goes to a common point and the other two leads from the elements join at
another common point (connection path to join of high conductivity).
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
Ex. 8) A capacitor of capacitance 125 microfarads is initially charged such that the voltage across
the plates is found to be 100 volts. If the capacitor is the connected in series to (with) a pure inductor
of inductance 0.2 Henry, what is the maximum value of the current that is observed in the inductor?
(A) 1.0 A (B) 2.5 A
(C) 4.5 A (D) 5.0 A (E) 10.0 A
Prepare a sketch:
When the switch is closed, the two elements will be
connected in series and parallel. The capacitor initially
has a charge Q = C VC = (125 F)(100 V) = 12500 F.
C
100 V
Q
KVR:
/C + L dI/dt = 0. Using I = - dQ/dt,
L
125 F
d 2Q 1
d 2x

Q  0 compare
 2 x  0
2
2
dt
LC
dt
0.2 H
A simple harmonic oscillation at  = (LC)- ½.
Q(t) = Qi cos[t + ]; I(t) = -  Qi sin[t + ]
The current through an inductor is a continuous function of time. It is zero just before the switch is
closed so it is zero just after it is closed. I(t) = -  Qi sin[] = 0   = 0 and Qi is just Q(t = 0) or
12500 F. The frequency  is (LC)- ½ = (125 x 0.2 x 10-6 x 102)- ½ = (25 x 10-4 )- ½ = 20 rad/s.
I(t) = -  Qi sin[t + ]  Imax =  Qi = (12500 F) (20 rad/s) = 250000 A.
The potential drops as one moves down in both
Switch closed at t = 0.
elements. Note the pattern in which the static
charges accumulate on the inductor to cause the
dI
+Q
C
+++
dt

d 2Q
---
dI
/dt . Starting at the lower left and proceeding CW,
the sum of the potential changes is Q/C – (L dI/dt) =
100 V
-Q
dt 2
0.2 H
0.When the current is in the direction chosen as
positive for dI/dt, the charge Q decreases so I = dQ
/dt.
Better: Recall that an LC circuit is an
oscillator.
d 2Q 1

Q0
dt 2 LC
Physics Handout Series – fields.tank
page 11
Fields Test Example Questions Guide
Ex. 9) If the V is the potential difference between points I
R
R
and II in the diagram above and all three resistance have the
same resistance R, what is the total current between I and II?
I
(A) V/3R
II
R
(B) 3 VR (C) 2V/3R
(D) 3VR/2
(E) 3V/2R

What is not said? The lower conductor continues out of the field of view. The components illustrated
are embedded as a unit in some larger circuit.
Solution: Start with the voltage versus current relation: VR = I R. The voltage across a resistor is
equal to the current through it times its resistance. The first conclusion is that I = V/R for an
individual resistor with each symbol representing its value for that resistor.
1st Conclusion: Responses (B) and (D) have incorrect dimensions and so are eliminated.
2nd Conclusion: There is a current V/R in the lowest resistor. A smaller current exists in the upper
branch of two resistors. The currents are to be summed.  V/R < I < 2 V/R. They make it rather easy
as only response (E) corresponds to a current greater than V/R.
Related Techniques: When you encounter a network of passive elements, you should make series
and parallel combinations in sequence to reduce the network. The upper branch resistors are
combined in series (What is required for two elements to be in series?) to yield an effective
resistance of 2R. That 2R and the R in the lower branch are then combined in parallel to yield a net
effective resistance of 2/3 R. If one is asked about the current, voltages, … for individual elements,
one should start with the fully collapsed network and find all the values. Next back out one step (in
Physics Handout Series – fields.tank
page 12
Fields Test Example Questions Guide
our case to R and 2R in parallel. Analyze to find all the values. Next, back out one more step and
analyze …. .
(10) Coulomb’s Law and Electrostatics
A physics major should know Coulomb’s law, the Biot-Savart law and Maxwell’s equations.
equation
integral form

Gauss’s Law
V

Faraday’s Law
A
A

Charge Conservation
(Continuity)
r

4  0 rp  rs
p
rp  rS
all rs
 rs 
rp  rs
2
J (rs ) d 3rs   rp  rS 

 E   Bt
B  nˆ dA  0
A
  J    Et   nˆ dA
0
0 0
dQinside
dt
J  nˆ dA  
 (rs ) d 3rs
all rs

B(r )  0
4
0


3



all r
/
 B  0
 B  o J  o o Et
 V t dV
Fq  q  Eothers  vq  Bothers 
Lorentz Force Law
E (rp ) 
V
B d 
V
 E  
 V 0 dV
 


0
E  d    Bt  nˆ dA
A
Gauss’s Law - Magnetism
Ampere (Maxwell's 4th)
Qinside
E  nˆ dA 
differential form
  J   t
Fq  q  Eothers  vq  Bothers 
 (rs ) d 3rs
qs
rˆsp 
rˆsp
2
4  0 rsp
4  0 rsp2
Coulomb
0 I d s   rp  rS 
 qs vs   rp  rS 
 0
3
3
4
4
rp  rS
rp  rS
Biot-Savart
Ex. 10) A charge of +Q is placed on the x-axis at x = - 1
meters, and a charge of -2Q is placed at x = + 1 meters,
y
as shown to the left. At what position on the x-axis will a
test charge of +q experience zero net force?
(A)  (3  8) m
-2Q
Q
1
m
1
m
x
(D)
1
/3 m
(B) – 1/3 m
(E)
(C)
0m
(3  8) m

Physics Handout Series – fields.tank
page 13
Fields Test Example Questions Guide
One should think about a problem before launching into a solution. The net force on the test charge
due to the two existing charges is to be zero. Two vectors can sum to zero only if they are antiparallel. A little reflection makes it clear that the solution points must lie on the (extended) line
joining the two charges. That is: The solution point is somewhere on the x axis.
y
y
q
I
-2Q
Q
1
m
1
m
II
Q
III
-2Q
x
x
In order to sum to zero, the forces must have equal magnitude and opposite direction. For points on
the x-axis, the test charge must lie to the right of both charges or to the left of both charge to meet
the opposite direction requirement. To meet the equal magnitude requirement, the test charge must
be closer to the smaller magnitude charge. That places the solution point in region I, x < - 1 m. The
particular value of x is found using the inverse square law nature of the Coulomb force.
Look for equal magnitudes for points in region three. The distance from Q is | x – (-1)| and the
distance from -2Q is | x – 1|.
kQ
k (2 Q)

 2( x  1) 2  ( x  1) 2 or x2 + 6 x + 1 = 0
2
2
( x  1)
( x  1)
This equation has a root (3  8) m in region I and a root (3  8) m in region II. Our previous
arguments have identified region I (x < - 1 m) as only region in which both the equal magnitude and
opposite direction requirements can be met. Recall that quadratic equations often return one
physical result and one not-so-physical result.
Review your options. Which of the proposed answers corresponds to a point to the left of both
charges? Test them, simplest value first, to see if the equal magnitude condition is met.
(11) Gauss’s Law, Ampere’s Law and E&M Knowledge Points:
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
Gauss:

V
E  nˆ dA  ( 0 )1   dV  Qenc 0
V

V
E  nˆ dA  EA  Qenc 0
Qenc
directed away
 o A
E
A : The enclosing surface through the point where the field is to be determined is to be made of
portions perpendicular to the field and parallel to it. A : net area that is perpendicular to the field.
A cyl = 2  r L ;
A plane = 2 A ;
A sph = 4  r2
For applications to conductors, make a sketch. (For equilibrium,) The net charge is on the surface(s).
The field at points in the conducting material is zero. A small patch on the surface is essentially
planar, and the field only pokes outside. Qenc = local A; A = A; E 
 local
 away.
o
Ampere:

A
B  d   I enc    J  nˆ dA   I enc

A
B  d   Bi
A
i

i
 B
o I enc
  0
i
  I enc
i
directed to circulate (RHR)
The encircling path through the point where the field is to be determined is to be made of portions
perpendicular to the field and parallel to it.
constant non-zero magnitude.
long st. wire
: the net parallel length along which the field has a
=2r;
solenoid
= ;
toroid
=2r
Knowledge points:
(a) Field and potential of a point charge: V (rsp ) 
q
4 0 rsp
; E (rsp ) 
q rsp
4 r
3
0 sp

q rˆsp
4 0 rsp2
rsp : displacement from source point to field point
(b) The field of a uniform shell of charge is zero inside and the same as a point charge with qtotal
located at the center outside.
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
(c) The field of a spherical ball of charge increases linearly with r inside and is the same as a point
charge with qtotal located at the center outside.
(d) The field of a long uniform line of charge is the charge per length divided by 2 times the
distance from the wire. The field is directed away from a uniform line of positive charge.
(e) much, much more …
Ex. 11) A uniform insulating sphere of radius r has a total charge Q and a uniform charge density.
The electric field at a point R/3 from the center of the sphere is given by which of the following?
(A)
Q
12   0 R
2
(B)
Q
8  0 R
(C)
2
Q
6 0 R
(D)
2
Q
4 0 R
2
(E)
3Q
4  0 R2
By knowledge: The field grows linearly inside a uniformly charge sphere so the field has 1/3 of the
 3  4  Q R
strength at r = R or 1
2
. The direction is radially away by symmetry.
0
By Gauss: E 
Qenc
directed away.
 o A
For the point R/3 from the center, the Gaussian surface is a
sphere surface of radius R/3 concentric with the sphere. (The point at which the field is to be found
must lie on the Gaussian surface. The surface has the symmetry of the problem.)
Qenc = (1/3)3 Q; A = (1/3)2 (4 R2)
(12) Faraday’s Law

A
B
 nˆ dA
A t
E  d  
Faraday’s Flux Rule
 = Emf = -
dB
dt
 B   B  nˆ dA  B A
A
Ex. 12) Four meters of wire form a square that is placed perpendicular to a uniform magnetic field of
strength 0.1 Tesla. The wire is reduced in length by 4.0 centimeters per second while still
maintaining its square shape. Which one of the following gives the initial induced emf across the
ends of the wire?
(A) 1 mV
(B) 2 mV
(C) 4 mV
(D) 8 mV
Physics Handout Series – fields.tank
(E) 16 mV
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Fields Test Example Questions Guide
The magnetic field is not varying in time, so we will use the flux rule form: Emf = -
dB
.
dt
The easy way: The side s is the perimeter divided by 4. The side at time zero is s(0) = 1 m. At one
second, s(1 sec) = 0.99 m. Using  B  B A , B(0 s) = (0.1 T) (1 m2) and B(1 s) = (0.1 T) (.992 m2)
Emf = -
dB

dt
 B (1 s)   B (0 s)
(0.1T )(1 m2 )  (0.1T )((.99)(.99) m2 )


 (0.1T )[1  0.98]V = 2 mV
1s
1s
As a function of time.
2
 B  B A = B side2 = B (perimeter/4)2 = (0.1 ) (4.0 – 0.04 t ) /16 Tm2
d/
dt
B =(0.1 ) ( [2 (4.0 – 0.04 t) (- 0.04)]/16) Tm2s-1
At time zero, the magnitude of the emf is |(0.1) ( [8 (- 0.04)]/16)| Tm2s-1 = 2 mV.

OPTIC, WAVES, THERMO AND STAT:
(13) Properties of an E&M plane wave and the Poynting Vector
EM waves are transverse which means that the electric and magnetic fields are perpendicular to the
direction of propagation of the wave. The electric and magnetic fields are mutually perpendicular.
The wave propagates in the direction of the Poynting vector, S ( E  H  01  E  B  ) . The E and
B fields oscillate in phase with one another and B/E = v, the wave speed (= c in vacuum). The
Poynting vector has units W/m2 and represents the power flow of the wave.
Energy Densities: E = ½ o E2 ; B = ½ 2/o
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
Ex. 13) A linearly polarized electromagnetic plane
z
wave carries energy in the positive z direction. At
y
some positions r and time t, the magnetic field points
along the positive x-axis, as shown in the figure to the
right.
At that position and time , the electric field points
x
B
along the
(A)
positive y-axis
(B)
negative y-axis
(C)
positive z-axis
(D)
negative z-axis
(E)
negative x-axis
EM waves are transverse which means that the electric and magnetic fields are perpendicular to the
direction of propagation of the wave. The electric and magnetic fields are mutually perpendicular.
As the wave propagates in the z direction and at that point and time has the magnetic field in the x
direction, the electric field must be in the  y direction. The Poynting vector S represents the
intensity of the wave times its propagation direction. Find the direction of S .
S  E  H  01 ( E  B)
S  01 ( E  B) and ˆj  iˆ  kˆ and ( ˆj )  iˆ  kˆ
The electric field is in the negative y direction at that point and time.
(14) Diffraction and Interference Standing waves
As a first guess, diffraction limits angular resolution to min = /d where d is the width of the aperture.
(If the aperture is circular, use min = 1.22 /D where D is the diameter of the aperture.)
(Ex. 14) An observer looks through a slit of width 5 x 10-4 meters at two lanterns a distance of 1
kilometer from the slit. The lantern emits light of wavelength 5 x 10-7 meters. The minimum
separation of the lanterns at which the observer can resolve the lantern lights is most nearly
(A)
0.01 m
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
(B)
0.1 m
(C)
1m
(D)
10 m
(E)
100 m
In this case,  = x 10-7 meters and d = 5 x 10-4 meters so min = 10-3 radian or one milli-radian. One
should prepare a sketch, but one can guess that the answer is 10-3 of one kilometer or one meter.
Polazation, polarizers: The interaction of E-M waves with matter is described by the Lorentz force
law, Fq  q  Eothers  vq  Bothers  . For a wave, |B|  c-1 E so Fmagnetic  (vq/c) Felectric. The electric
interaction dominates so the polarization of the wave is associated with the direction of the electric
field. That field is transverse to the direction of propagation and there are two independent transverse
directions so there are two independent polarizations for a given propagation directions. These
polarizations might be chosen to be vertical and horizontal linear polarizations or left- and righthanded circular polarizations. Circular polarizations carry angular momentum along the direction of
propagation. Linear polarizations carry zero net angular momentum. Natural (unpolarized light) can
be assumed to be an equal mix of the two independent polarizations.
A linear polarizer absorbs the component of the polarization perpendicular to its pass axis. It does
this by adding a wave polarized perpendicular to the pass axis that is  out of phase with the
incident wave and of equal amplitude.
The incident wave has an E field that makes an

Incident
angle  w.r.t. the pass axis. The charges in the
components
polarizer move perpendicular to the pass axis and
added by polarizer
absorb energy effectively adding a field
perpendicular to the pass axis that is  out of
transmitted
phase. Superposing, the perpendicular part is
pass axis

eliminated on only the part along the axis passes.
The wave immediately after a polarizer has a polarization parallel to the pass axis and an average
electric field strength of Eincident cos. As the intensity of a wave is proportional to the square of its
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
amplitude, Itransmitted = Iincident cos2. (The Law of Malus.) Natural light is an equal mix of the two
polarizations. If natural light is incident on a polarizer, light of one-half the intensity that is 100%
polarized along the pass axis is transmitted.


Io natural
½ Io cos2 along last pass axis
½ Io vertical
½ Io cos2cos2 along last pass axis
One half the intensity of the incident intensity of natural light passes through the first polarizer. All
the light that passes is polarized along the pass axis direction. Incident polarized light obeys the Law
of Malus and the transmitted light is 100% along the pass axis of the last polarizer.
(15) Lens Problems
(Ex. 15) Two thin converging lenses A and B, each having a focal length of 6 centimeters. are
placed 10 centimeters apart, as shown in the figure above. If an object is placed 10 centimeters to
the left of the lens A, the final image is
(A)
30 cm to the right of lens B
(B)
30
(C)
30
(D)
30
/11 cm to the right of lens B
/10 cm to the right of lens B
/11 cm to the left of lens B
Physics Handout Series – fields.tank
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
Fields Test Example Questions Guide
(E)
30
/10 cm to the left of lens B
A positive lens converges the light that passes through it. Light diverges for a real object. Real
objects have positive object distances. Light converges to a real image with a positive image
distance.
The focal length of the lenses in this problem is f = 6 cm. The object is 10 cm before the left lens
and so it has an object distance + 10 cm. Using the lens equation,
1
1
1


dObj dImg f

1
1
1
3
1
5
1
2


 



10 dImg 6
30 dImg 30
dImg 30
Lens A would form an image 15 cm to its right if there were no second lens. That would be 5 cm to
the right of lens B. Lens B is converging so the rays should converge closer to lens B than 5 cm.
(Lens B is converging so we are focusing on responses (B) and (C).) The appearance of 5 cm and 6
cm makes the 30/11 cm value look attractive.
The light arriving at lens B is converging. That case is called a virtual object and corresponds to a
negative object distance. (Light diverges from a real object and converges to form real image. These
cases correspond to positive object and image distances. Converging light incident on a lens and
diverging light leaving the lens corresponds to virtual objects and virtual images  negative
distances for the lens formula.) Applying the formula to lens B,
1
1
1
1
1
1
6
1
5
1
11










dObj dImg f
5 dImg 6
30 dImg 30
dImg 30
The image distance is positive so the light converges to a point on the side of the lens opposite to
that from which the light was incident.
Ray Trace of the action of the first lens only.
NOT drawn to scale!!!!!
15 cm
Real image: the rays actually converge to and pass through a real image point.
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
Rays are converging as they reach the second optic  lens two has a virtual object  negative
object distance.
add construction ray through lens center toward lens 1 image; undeflected
15 cm
action of the second lens
in parallel; out through focal point
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
Virtual image: The rays appear
to diverge from an image point
that the rays did not converge to
and pass through. Rays are
diverging immediately after the
last optic.
(16) Basic Wave Properties
Basic Wave Plug*:
v=f
The wave-speed is the frequency of the wave times its wavelength.
THE BASIC WAVE PLUG (BWP) is the most important wave property relation.
Wave Speed: v =
Tension
stiffness

mass / length
inertial density
Traveling Wave:
y(x,t) = A sin[ k x - t + ]
Standing Wave:
y(x,t) = A cos[t + ] sin[ k x + ]
k  = 2;  T = 2
node to node spacing: ½ node to anti-node spacing: ¼ 
Wave types: longitudinal and transverse
There are two transverse directions so transverse waves have polarizations
Waves reflect with a sign change when they reflect from a region with lower wave speed.
Waves reflect with a sign change when they reflect from a fixed end.
(Ex. 16) Two identical sinusoidal waves travel in opposite directions in a wire 15 meters long and
produce a standing wave in the wire. The traveling waves have a speed of 12 meters per second and
the standing wave has 6 nodes, including those at the two ends. Which of the following gives the
wavelength and frequency of the standing wave?
Wavelength
Frequency
(A)
3 m
2 Hz
(B)
3m
4 Hz
(C)
6m
2 Hz
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Fields Test Example Questions Guide
(D)
6m
3 Hz
(E)
12 m
2 Hz
Prepare a sketch! The four interior nodes divide the wire into five equal length segments, node-tonode spacings. Thus, 5 (/2) = 15 m or  = 30 m/5 = 6 m. From the BWP, f = v/ = 12/6 s -1 = 2 Hz.


(17) Kinetic Theory of Gases
Gas molecules have an average translational kinetic energy of 3/2 k T (three-halves times
Boltzmann’s constant times the absolute temperature). The pressure arises due to the many
molecular collisions per second with the walls so it increases for higher speeds v (v increases if
temperature increases) and for higher density. The molecules have a cross sectional area for
interaction  and so sweep out a volume  v t in time t. The mean time between collisions is found
as the time for a molecule to sweep out the average volume per molecule.
free
 v tcollision
= V/N so tfree = V/Nv
The mean free path free is the distance traveled to sweep out V/N.
 free = V/N.
Ex. 17) Cubical tanks X and Y have the same volumes and
share a common wall. There is 1 gram of helium in tank X
and 2 grams of helium in tank Y, and both samples are held
at the same temperature. Which of the following is the same
for both samples?
(A) the number of molecular collisions per second on the
common wall
(B) the average speed of the molecules
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
(C) the pressure exerted by the helium
(D) the density of the helium
(E) the mean free path of the molecules
The number of collisions per second, pressure and density for sample Y are double the X values. The
mean free path for sample Y is half that for sample X because there are twice as many things to hit in
the same volume. The average speed depends on the temperature and molecular mass. It is identical
for samples X and Y.
(18) Ideal Gas Law
isothermal: process at constant temperature
isobaric: process at constant pressure
iso-entropic: process at constant entropy
adiabatic: process in which there is no heat transfer; insulated from the environment
isochoric, isometric, iso-volumetric -------- process at constant volume
The Ideal Gas Law:
PV=nRT
PV=NkT
number of moles: n
ideal gas constant R = 8.31 (J/ mol-K)
number of molecules: N
Boltzmann’s constant: k = 1.38 x 10-23 J/K
LaChatlier’s Principle: A stable system acts to counter any change in its parameters. If a the
volume of a sample of gas is reduced, its pressure increases to fight additional decreases in volume.
(Ex. 18) If one mole of an ideal gas doubles its volume as it undergoes an isothermal expansion, it
pressure is
(A)
quadrupled
(B)
doubled
(C)
unchanged
(D)
halved
(E)
quartered
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
The process is isothermal (same temperature). P1 V1 = n R T = P2 V2
V
P2  P1  1   ½ P1
 V2 
Use ratios whenever possible.
(19) Thermodynamics and efficiency
Entropy Change: S 
Qrev
T
Use absolute temperatures.
Ideal Heat Engine Efficiency: e 
Thigh  Tlow
Thigh
0 0C = 273 K
Ex 19.) A power plant takes in steam at 527 0C to power turbines and then exhausts the steam at
127 0C. In any given time, it consumes 100 megawatts of heat energy from the steam. The maximum
output power of the plane is
(A) 10 MW
(B) 20 MW
(C) 50 MW
(D) 75 MW
(E) 100 MW
The problem statement is a little fuzzy with regard to its use of the terms power and energy!
Using 0 0C = 273 K, Thigh = 800 K and Tlow = 400 K. The ideal thermal efficiency for a reversible
engine operating between these temperatures is 50% so the maximum possible mechanical/electrical
power out is 50 MW.
QUANTUM MECHANICS AND ATOMIC PHYSICS:
Particles have associated wave properties. As a crude characterization, particles propagate like
waves and interact as (point) particles. It may be helpful to assume that each particle has a guide
wave that feels out the space to generate the probability distribution for the particle’s location.
Waves such as optical radiation also display particle-like character. When they interact, energy
transferred into and out of the wave in quanta (photons) each with energy hv (and momentum hv/c.)
  h p  k;
E  hv  ;
h  6.626 1034 Js;
1.0546 1034 Js
The information that can be known about a system is encoded in a wavefunction (x,t) that satisfies
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Fields Test Example Questions Guide
the Schrödinger equation. The probability to find the particle between x and x + dx is
(x,t)(x,t)dx (Born). Additional information can be extracted from the wavefunction using the
operator for the item of interest. The operator for x is x. The operator for px is - i x.
Dynamic quantities are functions of position, momentum and time:
Q(x,y,z, px,py,pz,t) Q(x,y,z, –i x, –i y, –i z, t); K  
2
2m
2 ;
V (r )  V (r )
Shrödinger’s equation identifies two operators for energy, the Hamiltonian and i t. In other
words, the Hamiltonian represents the total energy (kinetic + potential) and is the time-development
operator for wavefunction.
 2 2


Hˆ  (r , t )  
  V (r )  (r , t )  i
 (r , t )
t
 2m

 (r , t )  u(r ) T (t );
(Use separation.)
 2 2

  2m   V (r )  un (r )  Enun (r );


En Tn (t )  i

Tn (t )
t
The values En are the energy eigenvalues, the total energies, a set of real values that are
characteristic of the problem and the associated temporal dependence is: Tn(t) = e in t  e

A general solution has the form,  (r , t ) 
i En t
.
an un (r ) ein t .
all states n
Eigenvalue equation:
px (x,t) = –i x(x,t) = (momentum eigenvalue) (x,t)
H n(x,t) = En  n (x,t)
Operators for physical observables are Hermitian (or self-adjoint) an, as such, they have real
eigenvalues and expectation values.
Wavefunction Behavior:  2 (r , t )  
2m
2
 En  V (r ) (r , t )
In a classically allowed region, the
kinetic energy K, E – V, is greater than zero so the net curvature of  has a sign opposite to that of .
The function value oscillates back and forth across zero. In the case that V > E or K < 0 which can
not happen classically, the region is classically forbidden, and the curvature and function have the
same sign  behavior like growing and decaying exponentials.
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Fields Test Example Questions Guide
All the functions oscillate in the classically allowed region and decay in the forbidden regions. The
decay is more rapid for larger energy deficits. Each higher state has an additional node.
Free Particle States for piecewise constant potentials are of the form ei ( kx t ) where
k
 2m 
E  V ( x) . At any abrupt change in V(x), the wave is partially transmitted and partially
reflected. The wave function ca tunnel through classically forbidden regions (V > E  k = i and the
wave is like ex).
Fails to show the reflected wave!
Energy eigenstates are stationary  the associated probability density * is time-independent.
 n (r , t )  un (r ) ein t  un (r ) e
i En t
;
 n (r , t ) n (r , t )  un (r )
2
 22 
Energy Level Spacing: Energy levels in an infinite well with width a have the form n 2 
so
2 
 2ma 
 22 
the n to n + 1 spacing is (2 n  1) 
; the spread grows with n and is larger for smaller a
2 
 2ma 
(tighter confinement). The finite-well states penetrate into the forbidden region and so are less
tightly confined leading to lower energies as compared to the infinite well with the same inside
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Fields Test Example Questions Guide
width. For the hydrogen atom problem, the electron’s range approaches infinite as the energy
approaches zero from below. The level spacing approaches zero in this limit. For positive energies,
the particle is not confined and the allowed energies run continuously (zero spacing).
Commutation: If the operators for two quantities do not commute, then those quantities have a
minimum uncertainty product. [x,px] = i   x px  ½ .
If operators commute, they can have simultaneous eigenvalues. For the hydrogen problem, H, L2 and
Lz (plus the operator for electron spin) commute. The hydrogen atom states are labeled by nm
representing the eigenvalues of energy ( - 13.6 eV/n2), orbital angular momentum squared ([+1]2)
and z component of orbital angular momentum (m) (plus spin – up or down).
Generalized Uncertainty Principle. Experimental measurements are performed on systems in
order to extract average values and the corresponding uncertainties. In quantum systems, constraints
can arise between two uncertainties due to the nature of the corresponding operators. Given two
operators A and B, and knowledge of the commutator [A,B] = iC, then there exists a hard lower limit
on the product of uncertainties A B >= ½ <C>. Example: [x,px] = i   x px  ½ .
If two operators commute, then there is no lower-level uncertainty constraint on knowledge
of the eigenvalues, and the expectation values of these operators serve as a very useful label to
identify and characterize a system. For the bare Coulomb potential (simple H-atom problem), the
operators H, L2, and Lz commute, therefore we identify individual states by the expectation values
(energy, a.m. magnitude, a.m. z-comp) or their corresponding labels n,,m (quantum numbers).
Expectation value: Q  everywherethat  *(r , t ) Qˆ  (r , t ) d 3 r
 is non  zero
Mixed State:  (r , t ) 

an un (r ) ein t more than one an  0. Here, an is the amplitude to be
all states n
found in state n and | an |2 is the probability for the system to be found in state n. As energy
eigenstates have been used above, E

2
an En . A precise measurement of the energy will
all states n
yield one of the energy eigenvalues. For a set of identically prepared systems, the eigenvalue En will
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Fields Test Example Questions Guide
be found with probability | an |2. Making a precise measurement and finding the value En places the
system in a state that has that energy. If a measurement on the state:
 hydrogen  a100  100  a210  200  a211 210  a211 211  a211 211  a310  310
returns the energy, - 3.40 eV, then the atom must be in a state of the form
 after  b210  200  b211 210  b211 211  b211 211
immediately after the measurement. It will time develop as directed by the Hamiltonian and is not
necessarily a state consistent with the measured eigenvalue after some time has passed. If the
operator for the eigenvalue commutes with Hamiltonian, the expectation value for the quantity will
be time independent.

i
Q
  Hˆ , Qˆ   
t

i
  Hˆ , Qˆ  
d Q
dt
d Q
dt
time development of expectation values
for operators Q without explicit time dependence
Example: p  i  
d p
dt

i
The time rate of change is momentum is the ‘force’.
  Hˆ , p    V (r )
Ehrenfest’s Theorem
Every Hermitian operator has a complete set of eigenstates (Q m = qm m) which can be used as an
basis set for representing a general solution.  (r , t ) 

cm  m (r , t ) . It follows that:
eigenstates m
Q 

cm
2
qm . A precise measurement of Q will return an eigenvalue and place the system
eigenstates m
in a state with that eigenvalue immediately after the measurement.

(20) Ionization and the Periodic Chart
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The inner electrons screen the nuclear charge seen by the outer electrons. The apparent unscreened charge increases as
you move across the table filling an electron shell. It is a maximum for group VII and a minimum for the closed shell
group VIII’s. The effective unscreened charge increases as you move to down the table to the higher Z atoms. Group I
has one electron outside a closed shell and are the easiest to ionize (easiest at top after H; lowest screened charge). In
group VII, the effective charge gets higher as you move down the table  most electron hungry (highest electron
affinity). Electron affinity increases as you move down and across from I to VII. VIII is the closed shell nobles.
Outer valence shell has two S plus 6 P states.
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Electron shell filling sequence. Except for (H, He), the first electron in
an S shell leads to an alkali metal (strong electron donor). Two s
electrons yield a metal. The nth valence shell consists of the two ns and
six np states. Beginning with period n = 4, inner shell (d and f electrons
of lower n) fill after the nS, but before the nP. Shells fill first with spins
parallel (say up) and add spins anti-parallel only after the sub-shell is
half filled. Elements with approximately half-filled d and f sub-shells
exhibit magnetic properties associated with the unpaired spins.


Ex 20.) For which of the following elements is the ionization energy of a neutral atom the lowest?
(Z) is the atomic number.
(A) Oxygen (Z = 8)
(B) Fluorine (Z = 9)
(C) Neon
(Z = 10)
(D) Sodium (Z = 11)
(E) Magnesium (Z = 12)
The first period of elements includes hydrogen and helium with one and two electrons. In the next
period, there are eight electrons in the valance shell. Oxygen and fluorine need electrons to close the
shell and so are hard to ionize. Neon is the very stable closed shell configuration. Sodium has just
one electron outside a closed shell and so is easy to ionize. For Magnesium, the nucleus plus inner
electron shells combined have a net charge of plus 2 and so hold the outer electrons more tightly
bound than does the sodium nucleus plus inner shell electrons (net charge +1).
Sodium is a light alkali metal and so is an electron donor. In compounds such as NaCl, it is strongly
ionic with the sodium being positive which means its electron has transferred to the chlorine. After
hydrogen, the Group I atoms are willing electron donors and easy to ionize. Group 7 is the group of
the strongest electron acceptors. Top Left  strong donor; Bottom Right (grp.7)  strong acceptor.
(21) Hydrogen Atom spectrum levels, …
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En  
   E
13.6 eV
n2
E  h  h c
upper
 1
1 
 Elower  13.6 eV  2  2 
 nlower nlower 
Ex 21.) The Paschen series for hydrogen corresponds to transitions that end in the state of quantum
number n = 3. The shortest wavelength line in the Paschen series is closest to which of the
following? (The ionization energy of hydrogen is 13.6 eV and hc = 1,200 eV nm)
(A) 125 nm
(B) 250 nm
(C) 400 nm
(D) 800 nm
(E) 1800 nm
The shortest wavelength would correspond to the transition with the largest E from the highest
bound level down to the n = 3 level.
    13.6 eV  31  1   13.69eV ;
E  h c
2
2

9 hc
9 1200 eV nm

 794 nm
13.6 eV
13.6 eV
(22) Blackbody radiation
Power per area = Stephan Boltzmann constant times T to the fourth power =  T 4
Wien’s Displacement Law: peak T = constant = 2.898 x 106 nm  K
Planck’s Law: I ( ) dv 
2hv 3
1
dv
hv
2
c
e kT  1
I ( ) d  
2c 2 h

5
1
e
hc
 kT
1
d
Wien: I() is a function of T so the peak occurs
for a set value of T  peakT = constant.
Stephan:


0
I ( ) dv  ? change of variable
3
3
2hv3
1
 kT  2hu 1  kT 
dv

 h  c 2 eu  1  h  du
c 2 ehv kT  1


0
I ( ) dv  T 4
2k 4
h3c 2


0
u du
 T 4 
u
e 1
Ex 22.) Which of the following describes the effect of doubling the absolute temperature of a
blackbody on its power output per square meter and on the wavelength where the radiation
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distribution is a maximum?
(A) The output power is increased by a factor of 16 and the maximum of the distribution shifts to
twice its original wavelength.
(B) The output power is increased by a factor of 16 and the maximum of the distribution shifts to
half its original wavelength.
C) The output power is increased by a factor of 8 and the maximum of the distribution shifts to twice
its original wavelength.
(D) The output power is increased by a factor of 16 and the maximum of the distribution shifts to
half its original wavelength.
(E) The output power is increased by a factor of 2 and the maximum of the distribution shifts to four
its original wavelength.
Higher temperature suggests higher energy and hence shorter wavelength. Answers (B) and (D) are
viable! The power increase factors of 24 or 21 for (B) and (D). The fourth power should be familiar.
E    hv  hc 
1 eV  1240 nm
hc = 197 eV·nm (atomic physics) = 197 MeV·fm (nuclear physics)
color
 (nm)
v (1012 Hz)
E(eV)
IR
> 700
< 428
< 1.77
red
625-700
405-480
1.77-2.0
orange
585-620
484-513

yellow
585-570
513-526

green
570-505
526-594

blue
500-440
600-682

violet
440-400
682-750

UV
< 400
> 750
> 3.10
ranges are approximate and arbitrary
no universal agreement on the ranges
Ex 23.) Let H denote an Hermitian operator and suppose that H |= a |, where | is an
eigenvector of H. Which of the following is true of the eigenvalue a? (The symbols Re and Im
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Fields Test Example Questions Guide
denote the real and imaginary parts, respectively.)
(A) Re(a) = Im(a)
(B) Re(a) = - Im(a)
C) Re(a) = 0
(D) Im(a) = 0
(E) H = a
The eigenvalues of a Hermitian operator are real. The eigenvectors associated with distinct eigenvalues are
orthogonal.
Ex 24.) The quantum numbers used to label the radial wave function solutions to the Shroedinger
equation for hydrogen atom are the principal quantum number n and the angular momentum
quantum number . If the principal quantum number is n = 2, which of the following gives the
possible values for the angular momentum quantum number ?
(A) 1, 0
(B) 
(C) 2, 1, 0
(D) 
(E) 3/2, 1/2
The states of the hydrogen atom are labeled by the quantum numbers n,  and m plus the intrinsic
spin. The allowed values are n = 1, 2, 3, ….;  = 0, 1, … , (n -1); m = -, - +1, … , 0, 1, …, ).
That is: m assumes the 2+ 1 values spaced by 1 running from - to .
For n = 2,  can be 0 or 1 and m can be -1, 0, +1. Each combination can be combined with spin up or
down. The value  is the orbital angular momentum which assumes integer values. The total angular
momentum is the orbital plus spin, and it assumes half-integer values for an electron in hydrogen.
Ex 25.) The three operators (Lx, Ly, Lz) for the components of the angular momentum commute with
the Hamiltonian of a particular particle. Therefore, the angular momentum of the particle is
(A) equal to zero
(B) equal to the energy in magnitude
(C) always equal to Lz
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(D) a unit vector
(E) a constant of the motion
The Hamiltonian is the energy and time-development operator for a quantum system. It results that
d
Q
dt
 i
[ H , Q] for time-independent operators. Quantities with time-independent operators
that commute with the Hamiltonian are constants of the motion. A quantum state can be labeled with
good quantum numbers corresponding to it eigen-energy quantum number and quantum numbers
each additional member of the largest set of mutually commuting operators that includes the
Hamiltonian. For example, the members of the set (Lx, Ly, Lz) do not commute with one another.
They do commute with L2 = Lx2 + Ly2 + Lz2. Considering that it is a given that (Lx, Ly, Lz) commute
with the Hamiltonian, then L2 will also commute with the Hamiltonian. In such a case, there would
be good quantum numbers corresponding to H, L2 and one of the components of angular momentum.
Lz has the simplest representation in spherical coordinates so it is usually chosen. For the hydrogen
atom problem, the quantum numbers n,  and m are associated with H, L2 and Lz.
Ex 26.) A particle of energy E is in an eigenstate of the square well potential shown above, with
wave function (x). Which of the following is a correct expression for the expectation value of x2
for this particle?
a
(A)  x 2  ( x ) dx
a
(B)

(C)

(D)



x 2  ( x) dx
 *( x) x ( x) dx


a
a
 *( x) x 2  ( x) dx
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(E)


 *( x) x 2  ( x) dx

The correct procedure to compute an expectation value is to slap the operator for the quantity
between *(x) and (x) and then to integrate of the full range of x for which x) is non-zero. This
square well has a finite depth so the particle penetrates into the classically forbidden region |x| > a. In
this case the integrals must run over the full range (-,).

SPECIAL TOPICS:
(27) The Pauli Matrices:
x
0 1 
= 
;
1 0 
y
0 i 
= 
;
i 0 
z
1 0 
= 

0 1
Ex 27.) Given the Pauli spin matrices shown above and the operators defined by
-
=
x
(A)
-i
y,
=
x
+i
y
and
which of the following is NOT correct.
1   0 
    (B)
0  0
+
+
0 
1 
 2   (C)

1 
0
+
1 
0
 2   (D)

0 
1 
-
z
1  1 
0  0 (E)
   
0
0
  

1 
1 
z
Method: Realize that you have no idea what the problem is about, but also realize that it just
requires some matrix multiplication. Find the explicit forms for
+
and
-,
and brute force compute
each matrix product until you find the one that is NOT correct.
Background: ** Skip to the next problem if you are in a hurry.  (28) Counting Statistics
Suppose that a set of 2 x 2 matrices is sought such that any 2 x 2 matrix can be represented as a sum
of the members of that set. Clearly the set must have at least four members as 2 x 2 matrices have
four independent elements. One choice that could be made is:
 1 0   0 1   0 0   0 0  

,
,
,

  0 0   0 0  1 0   0 1  
This set of matrices is not very exciting. Spice it up; add the requirement that each member of the set
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Fields Test Example Questions Guide
be Hermitian (equal to the complex conjugate of its transpose). The simplest matrix that has
imaginary elements that meets this requirement is:
0 i 
i 0 


An independent off-diagonal matrix needs to have elements that are equal rather than the negative of
one another. Equal off-diagonal elements and Hermitian restricts the elements to be real.
0 1 
1 0 


Following the equal and negative scheme for the on diagonal matrices, the remaining members are:
1 0 
1 0 
0 1  and 0 1




This particular set of 2 X 2 matrices is the set of Pauli matrices. They have assigned labels:
1 0 
= 
;
0 1 
x
=
1
0 1 
=
;
1 0 
y
=
2
0 i 
=
;
i 0 
z
=
3
1 0 
= 

0 1
The first matrix is the identity, and the final three are the Pauli matrices. The set of the four matrices
is a basis for the collection of all 2 X 2 matrices.
 a b   a  d  1 0  b  c  0 1  c  b  0 i   a  d  1 0 
 c d    2  0 1   2  1 0   2 i   i 0    2  0 1




 
 
 
 


An arbitrary 2 x 2 matrix can be represented as a linear combination of the four matrices, but not as
linear combination of any set with fewer than four members. (A basis must be complete in the sense
that all elements (matrices) of interest can be represented as linear combinations of the members of
the set and economical in the sense that every member is necessary. If even one member of the set is
removed then there will be at least one 2 x 2 matrix that cannot be represented as a linear
combination of the remaining members.)
Why are the Pauli matrices introduced?
The Pauli matrices are introduced to act on two-row column matrices. Matrices of the forms:
 a  1 
0 
 b  ,  0  and  1  . The column vectors
   
 
1 
0 
 0  and  1  are a basis set for all the
 
 
a 
 b  . The column
 
1 
0 
vectors   and   are to be called UP () and DOWN ().
0 
1 
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If you are not familiar with modern physics, skip this paragraph.
In quantum mechanics, the Pauli matrices add the flexibility that allows a wavefunction to represent
the spin character of an electron. A two row column vector is appended to a function of position and
time.
a 
   (r , t )  
b 
1 
The column vector   represents spin up while
0 
0 
 1  represents spin down. The spin part is to be
 
 a   a   a * b *  a 
normalized independently so:     
 b   a a * bb*  1 .
b  b 
 
†
Special properties of the Pauli matrices:
1
1
=
2
2
=
3
3
=
The determinants are each of the three Pauli matrices is – 1. The determinant of the identity is, of
course, equal to one.
| 1| = | 2| = | 3|= 1
Actions of the Pauli Set:
x
The operator
x
y
z
 0  0 1  0 1
 1  = 1 0  1    0
   
  
1 
0  =
 
0 i  1  0
 i 0   0  i  1 ;

   
y
0 
1  =
 
0 i  0
 1
 i 0  1   i 0

 
 
lowers the UP state to i times the DOWN and raises the DOWN state to –i times UP.
z
The operator
x
lowers the UP state to DOWN and raises the DOWN state to UP.
y
The operator
 1   0 1  1   0
 0  = 1 0   0    1  ;
   
  
1 
0  =
 
1 0  1 
1 
0 1 0   (1) 0  ;

 
 
z
1 
0  =
 
1 0   0 
 0
0 1 1   (1) 1 

 
 
on the UP state to 1 times the UP and, on the DOWN state, returns – 1 times the
DOWN state. The operator
z
measures the up or down character of the state.
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The combination of operators
+
The operator
+
+
=
x
+i
y annihilates
 1   0 2  1   0 
 0  =  0 0   0    0 ;
   
  
+
UP and raises DOWN to 2 times UP.
 0  0 2 0
1
 1  =  0 0  1   2  0
 
 
  
is called the raising operator.
Exercise: Find the action of
-
=
x
+i
y on
the states UP and DOWN. Propose a name for
-.
** The Pauli matrices are multiplied by ½  when they are to be used in anger. They become the
operators for the components of the electron’s spin angular momentum.
z
=½ 
z
1 0 
= ½  
;
0 1
The eigenvalues of
z
2
=(½ )2 [
2
x
+
y
2
2
z ]
+
are  ½ , and the eigenvalue of
2
1 0 
= 3/4 2 

0 1 
is 3/4 2 = s(s + 1)2
(28) Counting Statistics
We will assume that the quick and dirty answers are enough to fake our way through the problem.
When a counting experiment yields N counts, the uncertainty in that number is N ½ (or a fractional
uncertainty of N -½  large count numbers are relatively more precise.). The value N ½ is the
expected standard deviation. If the counting experiment is repeated many times, the expected results
are assumed to follow a standard distribution.
68% of the time the result is within  
95% of the time the result is within  2 
99.7% of the time the result is within  3 
The average count number is N , and we assume
the standard deviation  is N ½.

** Know: 68% with  and 95% within 

p ( n) 
Physics Handout Series – fields.tank
( n N ) ]2
1

e½[
 2
page 40
Fields Test Example Questions Guide
Ex. 28) An experimenter measures the counting rate from a radioactive source as 10,150 counts in
100 minutes. Without changing any conditions, the experimenter counts for one minute. There is a
probability of about 15% that the number of counts recorded will be less than
(A) 50
(B) 70
(C) 90
(D) 100
(E) 110
The first line established the average count rate as 101.5 per minute. We expect 101.5 counts in one
minute on average. The standard deviation of the count number in one minute is assumed to be the
square root of that value or 10.075. There is a 68% probability that the counts falls between 101.5 –
10.075 and 101.5 + 10.075 or a 32% chance that the number of counts is outside this range. The
counts can be high or low, so there is a 16% chance that the number of counts in a minute is less
than 101.5 – 10.075 = 91.42.  about 15% likelihood to be less than 90.
We expect 101.5 counts means that if the one minute count were repeated a large number of times
that various integer results would be recorded each time. The results 101 and 102 would occur about
equally, and larger and smaller values would occur at a lower frequency. The values around N  
occurring about 0.6times as often as those around N .
p ( n) 
( n N ) ]2
1
 , e-½  0.6.
e½[
 2
(29) Graphs and Basic Lab
Ex.29) A motion sensor is used to measure the
position x versus time t for a cart traveling down
a ramp. A spreadsheet is then used to make a
linear fit to the plot of x vs. t, as shown in the
graph to the left. The equation for the best fit line

2
appears on the graph. Which of the following
gives the acceleration of the cart?
x(t) = 0.3190 t + 0.0013

Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
This one is a gift! The acceleration is the second derivative of x(t) with respect to t or 0.6380.
Adding dimensions, x(t) = 0.3190 m/s 2 t 2 + 0.0013 m yielding a = 0.6380 m/s 2. Finally, everyone
knows that we could not have gotten data that was accurate to even two significant figures.
Whenever possible, work from the definition. You are always correct to do so.
(30) Lagrangian and Hamiltonian Mechanics:
We restrict our treatment to the simple cases for which the constraints are good, the potentials are
time independent and for which there is no dissipation.
The Lagrangian L is T – U, the kinetic energy minus the potential energy. The Lagrangian is a
natural function of the coordinates qi, the coordinate velocities qi and time t.
The Lagrange equations of motion are:
d  L

dt  qi
  L
  
  qi



The signs and details can be recovered by considering: L( x, x, t )  ½ m x 2  V ( x) .
d  L   L 
V
 Fx
      m x 
dt  x   x 
x
(recovered Newton’s 2nd Law)
It is crucial to note that the time derivative is a total derivative while the others are partial
derivatives.
The momentum conjugate (associated with) a coordinate is found as: pi 
L
. Using the example
qi
L( x, x, t )  ½ m x 2  V ( x) , px  m x . Another familiar result is p  m r 2  which appears for the
planar orbit problem for which L(r , , r , )  ½ m[r 2  r 2  2 ]  V (r ) .
Conservation or constants of the motion:
pi 
L
qi
d  L

dt  qi
  L
  
  qi

 L
d
   pi   
 dt
 qi
Physics Handout Series – fields.tank



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Fields Test Example Questions Guide
The momentum pi is a constant of the motion if the Lagrangian does not depend on the coordinate
conjugate to that momentum. Given L(r , , r , )  ½ m[r 2  r 2  2 ]  V (r ) , there is no dependence of
the angular coordinate  so the conjugate momentum p  m r 2  is a constant of the motion.

In Hamiltonian mechanics, the coordinates and the momenta are the natural variables. The
Hamiltonian is defined to be H =
 pi qi - L. Here pi is the momentum conjugate to qi as defined
i
based on the Lagrangian above. For the simple case, L( x, x, t )  ½ m x 2  V ( x) , we found px  m x .
 H(x, px) = px x  (½ m x 2  V ( x))  px x  ½ m x 2  V ( x ) . We find:
2
H ( x, px )  ½ m x 2  V ( x)  px 2 m  V ( x)
The Hamiltonian is to be expressed as a function of the coordinates and momenta, so the second
form is the one that must be used. For time-independent problems of the type usually encountered,
the Hamiltonian represents the total energy of the system.
The equations of motion are: qi 
qi 
H
pi
2
H
H
and pi  
. Using the example: H ( x, px )  px 2 m  V ( x) ,
pi
qi
 x
px
H
and pi  
qi
m
Constants of the Motion: The relation pi  
 mx 
V
 Fx
x
H
shows that pi is a constant of the motion of the
qi
Hamiltonian is independent of qi. (Recall that we are restricting our attention to Hamiltonians that do
not depend explicitly on time.)
Ex. 30) Two masses m1 and m2 on a
horizontal straight frictionless track are
connected by a spring of spring constant k, as
shown in the figure to the left. The spring is
 initially at its equilibrium length. If x1 and x2
give the displacements of the masses from their equilibrium positions, the Lagrangian L for the
system is given by which of the following? The dot denotes differentiation with respect to time.
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
The Lagrangian is to be the kinetic energy minus the potential energy. The kinetic energy is the sum
of the kinetic energies of the two particles, T (or K) = ½ m1 x12  ½ m2 x22 , and the potential energy is
½ times the spring constant times the stretch of the spring squared, U = ½ k (x2 – x1)2. Note that it
helps to roll the words around in your mind as you attempt to decipher a problem.
Conclusion: L( x1 , x2 , x1 , x2 )  ½ m1 x12  ½ m2 x22  ½ k ( x2  x1 ) 2
Note that four of the candidates identify the kinetic energy as: ½ m1 x12  ½ m2 x22 , and three identify
the potential as ½ k (x2 – x1)2. Only two have the negative sign for L = T – U. Possibility (E)
includes the awe-inspiring reduced mass; it has the wrong sign for the potential. In addition, (E)
depends only on the relative position and relative velocity. That is: the center of mass motion has
been suppressed. For example, if q = x2 – x1, then (E) is ½  q 2 + ½ k q2. Forgetting that it should be
½  q 2 - ½ k q2, there is the issue of the kinetic energy.
½ m1 x12  ½ m2 x22 = ½ (m1  m2 ) X 2 + ½  q 2 where X is the center of mass velocity.
Kinetic energy in terms of the center of mass and relative coordinates.
Let V be the velocity of the center of mass of a system of particles, vi be the velocity of particle i 
and ui be the velocity of particle i relative to the center of mass.
Ktotal  KCM  K relative or
½ m v
i
2
i
 ½ M V 2   ½ mi ui2
i
i
For a system with only two particles, the reduced mass can be used to get the special form:
2
½ m1 v12  ½ m2 v22  ½ M V 2  ½  | v2  v1 |2  ½ M V 2  ½  vrel
where M = m1 + m2,  
m1 m2
and vrel  v2  v1
m1  m2
(31) Nuclear Decays
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
Each nucleus has Z protons (aka atomic number) and N neutrons with a mass number of A
=Z+N. A particular nucleus is denoted
A
Z
X N ; such as 146C8 , however the Z and N are usually
omitted because they are redundant and are known from A and the symbol used for X.
particle
nucleus AX
photon 
alpha 
beta minus - or ebeta plus+ or e+
Electron-neutrino ve
electron-antineutrino
ve
charge
Z
0
2
-1
+1
0
nucleon #
A
0
4
0
0
0
lepton #
0
0
0
+1
-1
+1
0
0
-1
Alpha particle is a tightly bound system of 2 protons and 2 neutrons (A=4). The very high
binding energy (~28 MeV) of the alpha and its very high first internal excited state make it favorable
to be treated as a unit object. (The same could be said for the proton and neutron which in reality are
quark composites.)
Nuclei will spontaneously decay into another nucleus if the final system has lower rest-masss
energy than the initial. There are four decay modes; gamma decay, beta decay, alpha decay, and
fission. The conserved quantities are total energy, linear and angular momentum, along with
nucleon number and lepton number. The rest-mass energy difference between the initial and final
systems is called the Q-value and appears as kinetic energy and internal excitation energy of the
products.
In gamma decay, an excited nucleus decreases its level of excitation by emitting a gamma ray
(photon)The process may be denoted: A(X*)  AX + 0 or just AX +  The number of neutrons
and protons is unchanged. No leptons are involved so lepton number need not be considered.
In alpha decay, a nucleus emits an alpha particle (2 protons and 2 nucleons for a total of 4
nucleons). The final state nucleus, the daughter, has Z–2 protons, N-2 neutrons and A-4 nucleons.
This can be denoted
A
Z
X
A 4
Z 2
Y + 42  or just A-4Y + . No leptons are involved so lepton number
need not be considered.
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
There are three fundamental processes in beta decay. For convenience they can be considered to
be (with Q-values given).
p  e   ve  0.78 MeV
beta minus decay
n 
beta plus decay
p  n  e   ve   x.xx MeV
electron capture
p  e
 n  ve   x.xx MeV
Beta minus decay can occur to an isolated neutron, but because the Q-values for the remaining two
processes are negative, the reactions can only occur in regions where the proton is ‘disturbed’; that
is, in a nucleus. Electron capture occurs when a ‘proton grabs an electron out of its s-state orbit’.
Note that the reactions above conserve charge, nucleon number, and lepton number.
For nuclei undergoing beta decay, the reactions would be written,
beta minus decay
A
Z
X
A
Z 1
Y + 01 e or just
beta plus decay
A
Z
X
A
Z 1
Y + ee
electron capture
A
Z
X 
A
Z 1
A
Z 1
Y + e. + ve
Y + e
Note that the reactions above conserve charge, nucleon number, and lepton number. The existence
of three-bodies in the final state means that the available energy or Q value of the reaction is not
apportioned to the particles in a set ratio, but rather the particles of each type emerge with a
distribution of energies.
In fission, the nucleus breaks into two large pieces (fission fragments) and a few neutrons with Qvalues on the order of 200 MeV. A commonly used radioactive fission source is 252Cf. The nuclei
235
U and 239Pu do not themselves spontaneously fission, but fission can be induced by striking them
with a neutron. The conserved quantities are charge, total energy, linear and angular momentum,
along with nucleon number and lepton number
***************
WORDS NOT USED:
Alpha particles have a nuclear binding energy of 28.3 MeV or about 7.1 MeV per nucleon. The
relatively high binding energy of the alpha makes it favorable to be ejected as a unit. The binding
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
energy per nucleon increases form 7.1 MeV for the alpha (He4) to a maximum of 8.8 MeV for Fe56.
Nuclei lighter than iron-56 are candidates for fusion; those heavier are candidates for fission.
Heavier nuclei have a larger ratio of neutrons to protons than light nuclei so after a sequence of a
few alpha decays, the resulting daughter is neutron rich compared to the stable nuclei of that atomic
number. At this point the nucleus needs to reduce the number of neutrons and increase the number of
protons. Beta decay does it.
The net change in the nucleus is that one neutron has changed into a proton. The appearance of an
electron means that the lepton number has grown by one. STOP: Lepton number is conserved. The
leptons to consider are the electron and the electron neutrino e plus the positron e+ and the electron
anti-neutrino ve . The first two are particles and have lepton number +1 while the final two are antiparticles with lepton number -1. (The e- and e+ are also called the beta minus and beta plus.) After
conserving lepton number, the decay becomes:
XA  Z+1YA + e ve which for lepton number is: 0  0 + (+1)(-1)
Z
The existence of three-bodies in the final state mean that the available energy or Q value of the
reaction is not apportioned to the particles in a set ratio, but rather the particles of each type emerge
with a distribution of energies.
For electron capture by a nucleus:
XA + e-  Z-1YA + e
Z
For a beta plus decay:
XA  Z-1YA + ee
Z
The basic beta decay process is:
n  1p1 + e- + ve
0 1
The neutron has a rest energy of 939.573 MeV; the proton has 938.280 MeV; the electron has 0.511
MeV; the neutrino has zero rest mass and therefore rest energy. The energetics of the reaction are:
939.573 MeV  938.280 MeV + 0.511 MeV + Q
The Q value is 0.782 MeV so the electrons are emitted with a distribution of kinetic energies ranging
up to almost 0.782 MeV. Similarly, the neutrinos have an energy distribution ranging up to almost
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Fields Test Example Questions Guide
0.782 MeV. The lifetime of a free neutron is about 886 seconds. Neutrons inside a nucleus have a
binding energy of about 8 MeV that stabilizes them against decay.
********

Ex. 31) Thorium with atomic mass 228, decays by alpha emission to a daughter nucleus which also
decays by alpha decay to radon. Which of the following is true of the decay product, radon? (The
atomic number of thorium is 90.
Atomic Mass
220
220
224
224
228
(A)
(B)
(C)
(D)
(E)
Atomic Number
82
86
82
88
91
The process is a sequence of two alpha decays:
A
Z
X 
A 4
Z 2
Y + 42 
A 8
Z 4
Y + 42 + 42 
Atomic Mass: A = 228  Afinal – 4– 4
Atomic Number: Z = 90  Zfinal = 90 – 4 = 86
228
Th  224Ra +  220Rn + + 
Thorium  Radium + alpha Radon + alpha + alpha
(31) Spontaneous Elementary Particle Decays:
As with other decays, the electric charge, energy, linear and angular momentum must be conserved.
Because there is only one initial particle, the Q-value for the reaction must be positive. The excess
energy or Q appears as kinetic energy and excitation energy of the products. Anti-particles have the
same rest energy as their corresponding particles.
Charge is conserved.
mpc2 = 939.573 MeV *
mec2 =
0.511 MeV
*
mpc2 = 939.6 MeV *
mc2 = 106 MeV
*
mc2 = 1775 MeV
Physics Handout Series – fields.tank
mc2 = 0 *
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Fields Test Example Questions Guide
mc2 =
140 MeV charged; 135 MeV neutral
mc2 = 0 *
Be aware of the values marked * and that a pion is a little more massive than a muon.
Ex. 32) Which of the following decays is possible in a vacuum?
(A)
+  - + 
(-  + + e- + e+ 
(C)0  e- + p 
(D)p  n + e+ + e 
(E)n  p + e- +  e 
(A) Fails. Does not conserve charge. Q-value > 0 is OK.
(B) Fails for two reasons. Does not conserve charge. Does not conserve energy because Q<0, that
is, the products have more mass than the initial particle.
(C) Fails. Conserves charge, however the products have much more massive than the pion.
(D) Fails. Conserves charge, however, the neutron is slightly more massive than the proton.
(E) Possible. Conserves charge and the Q>0. An isolated neutron is unstable and decays as shown
with a lifetime of 886 seconds. Neutrons are stabilized in the nucleus by their binding energy which
is greater than the Q of decay (E).
Because the fundamental building blocks are the 6 quarks (u,d,s,c,b,t) and 6 leptons (x,x,x,x,x,x)
there are actually additional conservation rules. These are not as trivial to apply because one must
be told the quark content of the composite particles.
**** Not used below
There are three families of leptons (electron, muon, tau) each with its anti-particle and neutrino/antineutrino pair). The muonic family group is {-, +, the anti-particle, the  and the anti-neutrino   )
with muon-lepton numbers {1,-1,1,-1}. Lepton number is conserved separately for each family.
For the decay to proceed spontaneously, the total rest energy of the products must be less than that of
the initial state particles.
***
Physics Handout Series – fields.tank
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Fields Test Example Questions Guide
Particles and Intrinsic Spin

Intrinsic spin S of most objects (electron, proton, neutron, quarks, leptons) is described by the label
(quantum number) s = ½ . The length is the intrinsic spin vector is calculated as
s(s  1)  with
possible z-components of ms  , where ms can have the values + ½ or – ½ . These z-components
are often referred to as spin-up and spin-down, respectively.
Particles with half-integer spins are fermions, particle that obey the Pauli exclusion principle, only
one particle per each distinct quantum state. The pattern of electrons filling atomic states is an
example of state filling by fermions. The building blocks of matter are typically fermions. Particles
with integer-valued intrinsic spin are bosons which can, and in fact prefer, to multiply occupy states.
Examples of bosons include pions, photons, gluons and gravitons. Many bosons are associated with
forces or interactions. Those with zero rest mass give rise to forces with infinite range (1/r2) forces
such as the Coulomb and gravitational forces. Quantum states with many bosons in the same or
nearby states gives the classical behavior of electromagnetic waves, etc. Interactions exchanging
massive bosons decrease exponentially for large separations.
Mediator
pion
gluon
photon
graviton
Spin
0
1
1
2
Interacting Pair
nucleon-nucleon
quark-quark
charge-charge
mass-mass
Composite Particles: A group of particles that is bound together acts like a particle of the net spin
as long as it is probed only weakly compared to its binding energy. Pairs of quarks can bind to form
mesons (bosons, ½ and ½  integer value). Triplets of quarks for fermions such as protons and
neutrons (fermion, ½ and ½ and ½  ½ integer value). Effective bosons such as helium atoms (2
protons + 2 neutrons + 2 electrons  integer) can collect in large numbers in a single (ground) state,
the condensate, leading to the superfluid state of liquid helium below 2.2 K. Electrons bound in pairs
(bosons) form similar condensates in super-conductors.
A nucleus can be modeled as a composite particle if it is probed weakly using say electric and
magnetic fields. If you shoot a 40 MeV proton at it, the probe will see the inner structure invalidating
the composite particle model.
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Ex. 33) For atoms in a simple cubic structure, the maximum percentage of the total available volume
that can be occupied by the atoms is approximately
(A) 
( 
(C)
(D)60%
(E)70%
The idea is to stack solid spheres in the prescribed pattern and to compute the fraction of the total
volume that is in the spheres. A cube has eight corners and a side length s. A sphere of radius ½ s
can be centered on each corner. One eighth of each sphere will be in the cube times eight corner or
one sphere of radius ½ s per cube of side s.
(Prepare a sketch!)
4  (½ s)3 4 

fraction 

½
 0.52
3
3 s
3 8
3
The closest packing of hard spheres (74%) is achieved in the hexagonal close-packed and facecentered cubic configurations. Body-centered cubic is about 68 %.
simple cublic
body-centered cubic
face-centered cubic
hexangonal c.p.
MISCELLANEOUS TOPICS:
(34) Astro-Knowledge
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Fields Test Example Questions Guide
Ex. 34) Which of the following gives the distance to Andromeda (M31), the spiral galaxy nearest to
the Milky Way?
(A) 2 x 100
(2 x 102 
(C)2 x 104
(D)2 x 106
(E)2 x 108
The earth is 8 light minutes from the sun.
The sun is 4 light years from the nearest star.
Galaxies have lots of stars so Galaxies should be thousands of light years in dimension.
During a 1965 episode, science officer Spock remarked that it was 40,000 light years across the
Milky Way (our galaxy). Google says that the main disk of the Milky Way is about 105 light years in
diameter.
Galaxies should be separated by 10 to 100 galaxy diameters. Look for 10-100 x 105  (D)
The universe is 13-15 x 109 years of age and so is about 13-15 x 109 light years in dimension.
The separation of galaxies should be small compared to the extent of the universe.
More ASTRO facts:
(35) Math Methods
Ex. 35) Which of the following is the derivative with respect to x of the function x2 cos[3 x4 + 1]?
The function is the product of x2 and cos[3 x4 + 1] so by the product rule,
df
/dx = 2x cos[3 x4 + 1] + x2 d/dx(cos[3 x4 + 1])
Using the chain rule, d/dx(cos[3 x4 + 1]) = - sin[3 x4 + 1] (12 x3)
Combining terms: df/dx = 2x cos[3 x4 + 1] – 12 x5 sin[3 x4 + 1]
All students should be intimately familiar with every concept and detail presented in SP351/2.
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TO ADD
Kepler’s Laws
Astro
star cycle
sizes and times
create heavy elements
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