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Transcript
Physics 104
General Physics II Notes
© J Kiefer 2010
Table of Contents
Table of Contents .............................................................................................................. 1
I.
A.
Electromagnetism ..................................................................................................... 2
Electrostatics ............................................................................................................. 2
1. Electric Charge........................................................................................................ 2
2. Coulomb’s “Law” ................................................................................................... 3
3. Electric Field ........................................................................................................... 4
4. Gauss’ “Law” .......................................................................................................... 9
5. Electric Potential ................................................................................................... 14
6. Conductors and Insulators..................................................................................... 17
7. Capacitance ........................................................................................................... 18
B.
1.
2.
3.
4.
5.
6.
Electric Current ...................................................................................................... 21
Current .................................................................................................................. 21
Direct Current Circuits .......................................................................................... 24
Magnetic Fields ..................................................................................................... 28
Magnetic Induction ............................................................................................... 33
Alternating Current Circuits ................................................................................. 39
Maxwell’s Equations ............................................................................................ 43
II. Optics ....................................................................................................................... 46
A.
1.
2.
3.
Geometrical Optics ................................................................................................. 46
Wave Fronts & Light rays..................................................................................... 46
Mirrors .................................................................................................................. 48
Lenses ................................................................................................................... 50
1.
2.
3.
Physical Optics ........................................................................................................ 53
EM Waves ............................................................................................................. 53
Diffraction & Interference .................................................................................... 54
Quantized Waves .................................................................................................. 57
B.
III.
A.
Atoms ................................................................................................................... 60
Structure of the Atom ............................................................................................. 60
1. Bohr Model ........................................................................................................... 60
2. Nucleus ................................................................................................................. 61
B.
1.
2.
3.
Subatomic ................................................................................................................ 62
Leptons & Quarks ................................................................................................. 62
Hadrons ................................................................................................................. 62
Anti-Matter ........................................................................................................... 62
1
I.
Electromagnetism
A.
Electrostatics
1.
Electric Charge
a.
Charge
The physical quantity we call electric charge is a property of matter. The particles of
which all material objects are made have inertia (mass) and electric charge, among other
properties. In contrast to mass, however, electric charge occurs in two kinds, which are
called positive (+) and negative (-). The net charge on an object is simply the sum of all
the individual +/- charges present. Charged objects (objects with non-zero net charge)
exert equal and opposite forces on each other.
The SI unit of charge is the Coulomb (C).
b.
Quantized charge
Charge is a property of subatomic particles, such as the proton & electron. It is found
that isolated charges exist only in integer multiples of a constant number, e. Ions, or
other charged bodies, have an imbalance in the numbers of electrons & protons. The
electron has a charge of –e, while the proton has a charge of +e. In practice, larger
objects become charged by the addition or subtraction of electrons. We speak of charges
being present, or being in motion. In fact, the charges are charged particles of matter.
Fractional charges, that is, fractional multiples of e are never observed on particles that
can exist independently.
In SI units, e  1.602  10 19 C .
c.
Static charge
Static means unmoving. Electrostatics, then, refers to situations in which the distribution
of electric charge is fixed. The electric charges are not allowed to move. Either the net
electric force on each charge is zero, or other external forces balance the electric forces.
d.
induced charge separation
2
2.
Coulomb’s “Law”
a.
Point charges
The force that acts between two point charges is described by Coulomb’s “Law”.
kq1q 2
; the direction is along the radius joining the two charges.
r122
The force exerted on q1 by q2 is
kq q
F12  12 2 r̂12 . Notice that the
r12
unit vector r̂12 points from q2
toward q1. Since charge comes in
(+) and (-) kinds, the electrostatic
force may be either attractive or
repulsive. “Like charges repel,
unlike charges attract.”
Step 1: Diagram
y
F32
Draw a representative
sketch—done.
Q3=+65C
Q1=-86C
52 cm
Draw and label forces (only those on Q3).
Draw components of forces which are not along axes.
11
Step 2: Starting Equation
Step 3: Replace Generic Quantities by Specifics
y
F32
Q3=+65C
F k
32
F31
30 cm
=30º
Draw axes, showing
origin and directions— Q2=+50C
done.
The constant k is called the
Coulomb constant; in SI units
N  m2
.
k  9.0  10 9
C2
y
F31
30 cm
Draw and label relevant
quantities—done.
Q3Q 2
2
32
r
,
repulsive
=30º
Q1=-86C
Q2=+50C
x
F
k
32, y
52 cm
qq
F  k 12 2
12
r12
Q3 Q 2
r322
F32
Q3=+65C
F31
r32=30 cm
The magnitude is F 
=30º
Q1=-86C
Q2=+50C
x
52 cm
F
 0 (from diagram)
32, x
F32,y = 330 N and F32,x = 0 N.
Step 3 (continued)
Step 3: Complete the Math
attractive
QQ
F
  k 32 1 cos 
31, x
r31
(+ sign comes from
diagram)
QQ
F
 k 32 1 sin 
31, y
r31
y
Q3=+65C
r32=30 cm
F k 2 ,
31
r31
F32
The net force is the
vector sum of all the
forces on Q3.
F31
=30º
Q1=-86C
Q2=+50C
F32
Q3=+65C
30 cm
y
Q3Q1
F3
F31
=30º
x
Q1=-86C
Q2=+50C
52 cm
52 cm
F3x = F31,x + F32,x = 120 N + 0 N = 120 N
(- sign comes from diagram)
F3y = F31,y + F32,y = -70 N + 330 N = 260 N
F31,x = +120 N and F31,y = -70 N.
3
x
x
b.
Superposition
The total electrostatic force on q1 exerted by a number of other point charges is the vector
sum of the forces by the individual charges.
kq1qi
rˆ1i
r12i
Naturally, this vector sum is performed using the force components along the axes of a
selected reference frame.
F1  
3.
Electric Field
a.
Definition
The concept of an electric field is a means of representing the influence of a charge on
other, distant charges. The source charge gives rise to an electric field that extends
throughout space. Other charges in the field experience an electrostatic force.
Quantitatively, the electric field at a point is space is derived from the net electric force
on a small positive test charge, qo, located at that point. It is assumed that the existence
of the small test charge does not cause a rearrangement of the source charge(s).
F
E r   o
qo
Example: an electron moving with velocity v0 in the positive x
So, the field exists
direction enters a region of uniform electric field that makes a
right angle with the electron’s initial velocity. Express the
independently of the test
position and velocity of the electron as a function of time.
charge. Once the electric field
y
is defined, then the electric
v v a t v
- - - - - - - - - - - - force felt by any point charge
x v  v a t  0 E q t
placed in the field at any
m
E
v
1
location is
x  x v t  a t v t
2
F  qE .

1
1E q
x
ox
x
y
oy
y
o
y
0
2
o
ox
x
o
+ + + + + + + + + + + + +
y
y  yo  voyt  a y t 2  
 g t 2
2
2 m

To be completed at the blackboard.
The electric field is a vector
quantity. The SI units of
N
electric field are
.
C
32
4
Mathematically, the electric field is a point function, in that it specifies a magnitude and
direction at every point in space.
b.
Electric field lines
An electric field is represented pictorially using electric field lines. These are lines drawn
according to the following rules:
i) the electric field at any point is tangent to the electric field line.
ii) the spacing, or density, of the field lines is proportional to the magnitude of the electric
field. Where the magnitude is greater, the lines are more closely spaced.
iii) field lines originate from (+) charges and end on (-) charges.
c.
Field due to a point charge
Example: calculate the electric field at the electron’s distance
away from the proton in a hydrogen atom (5.3x10-11 m).
source point
r̂ +
q
E=k 2 rˆ
r
field point

q
Nm2 1.602 1019 C
N
E  k 2 r̂  9 109 2
r̂  5.13 1011 r̂
r
C 5.3 1011 m 2
C


For comparison, air begins to break down and conduct
electricity at about 30 kV/cm, or 3x106 V/m.
28
5
d.
Field due to a dipole source
(++)
(+-)
6
e.
Field due to a continuous line of charge
There may be electric charge distributed throughout a macroscopic object, or over its
surface. Then we speak of volume charge density,  (C/m3) or surface charge density,
 (C/m2) or linear charge density,  (C/m). The electric field at a point P due to the
charge densities is obtained by integrating over the charge distribution. Each charge
element, dq, makes a contribution dE to the net field at the point P. For a three
dimensional charge distribution,
  r  r 
E 
dxdydz
3
r  r
V
In practice, we decompose. For a simpler example, integrate over a continuous linear
charge distribution.
7
f.
Field due to an infinite sheet of charge
8
4.
Gauss’ “Law”
a.
Electric flux, 
Imagine that electric field lines represent the “flow” of something from one place to
another. Now, consider field lines passing through a possibly imaginary surface of area
A.
9
The field lines make an angle,  , with the surface normal, n̂ . We define the electric flux
to be   EA cos   E  Anˆ . Of course, the electric field may not be uniform over the
entire surface, so we break the surface into surface elements, and integrate.
   Ei  Ai nˆ i   E  dA
surface
If the surface is a closed surface, then unless there are source charges inside the surface,
the flux entering the closed surface equals the flux leaving, and the net flux through the
closed surface is zero.
10
b.
Gauss’s Theorem
 E  dA 
surface
  dV
enclosedvolume
Applied to electrostatics, Gauss’ “Law” says that
c  4 kQ 
Q
o
where Q is nothing more than the total net electric charge enclosed by the surface. The
C2
constant  o  8.85  10 12
is called the permitivity of free space (or of the
N  m2
vacuum). Gauss’ “Law” is particularly useful when the source charge distribution is
highly symmetric in shape. That such a case, the flux is easy to compute, and the E-field
can be obtained without carrying out an integral over the source charge distribution. For
instance, we can demonstrate the validity of Gauss’ “Law” for a point charge source.
c.
Examples
i) uniform spherical distribution of charge
11
Q
 E  da  
E 4 R 2 
E
o
Q
o
Q
4 o R 2
This seems almost too easy. It requires that we realize that the electric field will be the
same magnitude, and directed outward (or inward) all around at a specified distance from
the spherically shaped, uniform charge distribution.
ii) line of charge
12
iii) infinite sheet of charge
Notice that the electric field points upward and downward from the sheet of charge.
Therefore the flux goes out both the top and bottom of the cylindrical surface.
13
5.
Electric Potential
a.
Potential energy
The Coulomb’s “Law” is very similar to the Universal “Law” of Gravitation. The
electrostatic potential energy has a form similar to the gravitational potential energy
function.
kq q
U r12   1 2
r12
This claim can be verified by computing the work done by the electrostatic force as two
point charges are brought from infinite separation to a separation r12 . Like the
gravitational potential energy, the electrostatic potential energy of several (that is, n)
point charges is equal to the sum of pair-wise terms.
n 1
U 
n

kqi q j
rij
Notice the “triangular sum” that prevents us from counting any given pair of charges
twice.
i 1
j i 1
b.
Potential, or voltage
Another useful quantity is the potential energy per unit charge, or potential. Imagine one
of the n point charges in a system being a test charge, qo. We’d start counting from i = 0
and end with i= n – 1. The potential energy of the system would be
n2
n 1 kq q
i j
U  
i 0 j i 1 rij
14
Now consider the change in potential energy for the test charge qo if the test charge
moves from one position to another while all other charges remain fixed in place. The
potential energy changes U  U 2  U1 . The electric potential energy per unit charge at
U
the location of the test charge is V 
. The SI unit of electrostatic potential is the
qo
J
V
Volt (V). 1V  . In terms of voltage, the magnitude of the electric field is
. In this
C
m
connection, we can define a new unit of energy, the electron Volt (eV).
1eV  e  1V  1.6  10 19 J . It’s the energy gained by an electron when experiencing a
voltage change of one Volt.
c.
Voltage in a uniform E-field
For a displacement vector, s , in a uniform electric field, E , V   E  s . Save this for
later.
d.
Sources of electric potential
i) point charges
E r  
kq
rˆ
r2
r2
r2
r1
r1
r2
dr kq kq


2
r2 r1
r1 r
V    E  ds    Edr  kq
Evidently, for a point charge source, V 
kq
. For a number of point charge sources, the
r
kqi
, where ri is the distance from qi to the field point, the
ri
point in space at which we are computing the potential.
potential is added up: V  
15
ii) a line of charge
For a continuous charge distribution, we have to integrate. The fact that the potential is a
scalar quantity makes the integration a little easier than the integration of the electric
field, which is a vector.
kdq
dV 
r
ds
V  k
for line or a ring of charge.
r
da
V  k
for a surface or layer of charge.
r
dv
V  k
for a volume charge distributi on.
r
For instance, to compute the potential due to a line of charge. . .
e.
Equipotential lines
By definition, dV  E  ds  E x dx  E y dy  E z dz . Solve as it were for the field
V
V
V
, Ey  
, & Ez  
. In other words,
x
z
y
V ˆ V ˆ V ˆ
E 
i
j
k  V
x
y
z
An equipotential is a curve or surface in space along which the potential is the same
value. Equipotentials have the following additional properties:
components. E x  
i) the electric field is perpendicular to equipotentials
16
ii) equipotential lines or surfaces never cross each other—if they did so it would imply
that there were two values of potential at the same point in space.
6.
Conductors and Insulators
a.
Conducting materials; insulating materials
In conducting materials, there are charged particles that are free to move in response to an
external electric field, or in response to electric forces among themselves. These are
metals for instance, or electrolytic solutions, or plasma. [Plasma is a gas composed of
charged particles—electrons and ions.]
Insulating materials lack the free charged particles, therefore they will not conduct
electric current.
b.
Static charges & E-fields and conductors
Since charges are free to move within a conductor, they do so until achieving
equilibrium. For instance, if an external electric field is imposed upon a conductor, the
free (+/-) charges move in opposite directions in response, until the secondary electric
field the separated charge carriers create is equal but opposite to the external field.
Therefore the net static electric field inside a conductor is zero. By a similar token, if
excess charge resides on a conductor, it will spread out in such a way as to maximize the
separation between like charges. Therefore, excess charge resides on the outside surface
of a conductor. Further, the static electric field near the surface of the conductor must be
everywhere perpendicular to the surface, else a tangential component of the field would
cause the free charges to move until the net tangential component is zero.
17
Finally, consider a charge +Q inside a hollow conducting shell. Negative charge will be
drawn to the inside surface, and positive charge driven to the outside surface, both equal
in magnitude to Q. The electric field inside & outside of the shell is just that of a point
charge source, while the net field within the thickness of the conducting shell is zero.
7.
Capacitance
a.
Definition of capacitance
C
Q
V
1C
. The capacitance is computed by
1V
finding the voltage between the two conductors, then dividing by the magnitude of the
charge on one of the conductors. This is easy only for simple geometries, like flat plates
or spheres. The voltage is gotten by knowing the electric field and computing
V    E  ds , where ds is the displacement of one conductor from the other.
The SI unit of capacitance is the Farad (F). 1F 
Example—two parallel conducting
plates.
Assume that that the E-field is
perpendicular to the plates, all over the
plates. This means we are neglecting
edge-effects, where the field lines curve
outward past the edges of the plates. We
18
might also say that we are assuming that the two plates are infinite in area. In such a
case, the E-field is uniform between the plates.
Q
  EA 
o
E
Q
A o
V  Ed  E 
V
d
Set the two expressions for the E-field equal, and solve for
Q
.
V
V
Q

d A o
Q A o

C
V
d
Notice that the capacitance of the two parallel plates depends on the geometry of the
plates--area and separation.
c.
Energy storage
In order to separate the +Q and –Q charges on the two plates, work must be done. Let’s
say we transfer a small amount of charge, dq from one plate to another that already has an
excess charge q on it. The work done is dW  Vdq . To go from 0 to Q, the total work is
Q
Q
q
1 Q2 1
dq 
 QV  U . This is a change in electrostatic potential
C
2 C
2
0
0
energy, which we regard as being stored on or in the capacitor.
W   Vdq  
It is convenient to associate the stored energy with the electric field that “fills” the space
between the plates. We define the energy density to be the energy per unit volume.
1 o A 2 2
1

E d
CV 2
2 d 
U
1
2
u


 oE2
Ad
Ad
Ad
2
19
d.
Dielectric breakdown in a capacitor
The constant  o is the permitivity of the vacuum, or of “free space.” Material
substances, particularly insulators, called also dielectric materials, have permitivities
greater than  o . If the space between the plates is filled with an insulating substance,
A
then the capacitance is increased. C 
. More charge can be stored at the same
d
voltage. However, the E-field polarizes the molecules of the insulating substance. If the
E-field is large enough, it ionizes those molecules, and the substance becomes
conductive, leading to a spark discharge. The dielectric has broken down.
e.
Combinations of capacitors
i) series
For capacitors connected in series, the charge on each capacitor is the same Q. If it
weren’t, charge would flow until it was.
V AB  V1  V2
 1
Q Q
1  Q
 

 Q 
C1 C 2
 C1 C 2  Ceq
1
1
1
The equivalent capacitance of the series combination is


 .
Ceq C1 C 2
V AB 
20
ii) parallel
For capacitors connected in parallel, it’s the voltage that is the same on each of the
capacitors.
Q Q
V AB  V1  V2  1  2
C1 C2
The total charge stored on the capacitors is Q  Q1  Q2 . Therefore,
Q
Q
V AB 

C1  C 2 C eq
C eq  C1  C 2  
Sing it out: “capacitors in series have the same charge; capacitors in parallel have the
same voltage.”
B.
Electric Current
1.
Current
Electric current is charged particles in motion.
a.
Charge carriers
Charge carriers are charged particles whose motion is the electric current. In a metal
wire, the charge carriers are electrons, in electrolytic solutions the charge carriers are
ions, in particle beams the particles may be electrons, protons or ionized atoms. We
imagine the particles passing through a virtual surface area, A. Then the current is
defined as the amount of charge passing per unit time.
Q
dQ
I

t
dt
21
In terms of the number of particles passing through the surface,
dQ
I
 nqv d A
dt
n is the number of particles per unit volume
q is the charge on each particle (assuming for the moment that they are all the same.
vd is the average velocity of the particles.
A is the cross sectional area through which the particles are passing.
For instance, in Copper, n  8.5  10 22 electrons/ cm 3 and v d  2.5  10 4 m / s .
If we divide by that cross sectional area, we obtain the current density, J 
The SI unit of electric current is the Ampere’ (A). 1A  1
I
.
A
C
s
Current density is actually a vector quantity. The direction by convention is (+) in the
direction that (+) particles are moving. The movement of (-) particles from right to left is
equivalent to (+) particles moving from left to right.
b.
Ohm’s “Law”
In some conductive materials, it is found experimentally that
J E
wherein  is the conductivity of the conductor. This is called ohmic or linear behavior.
Consider a section of conducting wire.
If the E-field is uniform throughout the wire, then the voltage from one end of the wire to
the other is V  E . Substituting for the E-field in the current density,
22
I
V

A


V 
I  RI
A

We define the resistance of the wire, R 
. Notice that the conductivity is a property
A
of the substance, while the resistance is a property of the particular piece of that
substance.
J
The SI unit of resistance is the Ohm (  ). 1  1
V
.
A
To make it more confusing, we also sometimes use the resistivity,  
1

. It turns out
that the resistivity (or the conductivity) is a function of temperature.
For metals in particular, usually we expect    o 1   T  To  . The  and the To
would be parameters characteristic of a particular metal, such as Copper or Aluminum,
etc.
c.
Power
When a current passes through a voltage drop, the charge carriers are losing potential
energy. The time rate of change in the potential energy is
U
Q
P

V  IV
t
t
For ohmic devices, or materials, V = RI, therefore
V2
2
P  IV  I R 
R
This is the power dissipated by the current flowing through a resistance.
23
[Keep in mind that the V is a voltage difference between two locations, even though we
don’t use the  .]
On the other hand, a device which increases the potential energy of the charge carriers is
called an electromotive force, or EMF. Batteries, generators, & solar cells are sources of
EMF.
2.
Direct Current Circuits
Direct current is current that flows in one direction at all times.
a.
Internal resistance of a battery
The material of which a battery is made has electrical resistance of its own. Therefore,
the terminal voltage, the voltage available to a circuit, is V  EMF  Ir , where r is the
internal resistance of the battery. If R is the total resistance of the external circuit, then
we can write EMF  IR  Ir  I R  r  .
b.
Kirchhoff’s Rules
We have a system, or procedure as it were, to find the current flowing in different
branches of an electrical circuit. The approach is to apply conservation principles to the
flow of charge.
i) the total current entering a junction must equal the total current leaving the junction.
ii) the sum of potential differences around any closed circuit loop is zero.
In practice:
i) assume some directions for the currents.
ii) form loop equations by applying Kirchhoff’s Rules to closed circuit loops.
iii) when tracing a loop, remember to multiply by (-1) when traversing a device opposite
to the assumed current flow.
iv) if a battery is traversed from (+) to (-) terminals, then V  V rather than V.
v) if you are solving for currents, the algebraic signs of your answers will tell you the
correct directions of the current flows.
24
c.
Resistances in series and in parallel
series
I1 = current through R1.
I2 = current through R2.
I = current through the battery.
We choose to traverse the loop clockwise.
i) I1  I 2  I , since there are no branches.
ii) EMF  I 2 R2  I1 R1  0
rearrange
EMF  I R1  R2   IReq
The equivalent resistance is Req  R1  R2 .
parallel
i) I  I1  I 2
ii) loop 1
loop 2
loop 3
EMF  I 2 R2  0
I1 R1  I 2 R2  0
EMF  I1 R1  0
We want the two currents, so solve the first & third loop equations.
EMF
EMF
I2 
and I 1 
R2
R1
If follows then that
25
I  I1  I 2 
 1
EMF EMF
1  EMF
 

 EMF  
R1
R2
Req
 R1 R2 
1
 1
1 
 .
The equivalent resistance is Req   
 R1 R2 
d.
Example
26
e.
Charging & discharging a capacitor
Charging a capacitor requires a battery.
When the switch, S, is closed, current flows and charge accumulates on the capacitor.
The loop equation is
q
EMF   IR  0
C
EMF
Notice that we traverse the capacitor in reverse. Initially, I  I o 
, but as q
R
q
 EMF and I  0 .
increases, the current I decreases until
C
d 
q

 EMF   IR   0
dt 
C

I
d
 R I 0
C
dt
d
I

dt
RC
t
I  I o e RC
t
t
dq
EMF RC
 I o e RC 
e
dt
R
t
t
q
EMF RC
e dt
R 0
q
 t

EMF
RC  e RC  e 0 
R


t
t




q  EMF  C 1  e RC   Qmax 1  e RC 




27
We define the time constant (or the decay constant) to be   RC . When t = RC, then
q  Qmax 1  e for a charging capacitor.
On the other hand, if we envision discharging a capacitor through a resistance,
t
Q
RC
q  Qmax e , and when t = RC, then q  max  0.37Qmax .
e
3.
Magnetic Fields
Moving charges generate magnetic fields; magnetic fields exert forces on moving
charges.
a.
Biot-Savart “Law”
A current element, or a short segment of electric current, Id  generates a contribution to
the magnetic field at the point O:
 d   rˆ
dB  o I
4
r2
Wb
The permeability of free space is  o  4  10 7
.
Am
The total magnetic field at the point O is obtained by integrating along the current path.
For example, consider a circular loop of current, and find the magnetic field at a point on
the axis of the loop.
28
The SI unit of magnetic field is the Tesla (T). 1T  1
N s
C m
b.
Ampere’s “Law”
In highly symmetric situations, we have something similar to Gauss’ “Law”.
 B  d   o I
The I is the current passing through the interior of a closed loop
Example, a long, straight wire.
At a distance R from the long wire, B
has the same magnitude all around the
wire, and the B-field is parallel to the
d  also all around the wire.
 B  d 
2
 BRd   2BR
0
2BR   o I
B
o I
2R
29
Example, inside a long coil of radius a.
We can think of a coil as number of current loops, side by side. We’ll assume that the
magnetic field inside the coil is straight and uniform.

 B  d   B  d  B
0
B   o nI
n
o I

Now, if the fields and currents are not steady, then we generalize Ampere’s “Law” by
adding another term.
d
 B  d    o I   o  o dt
 is the electric flux through the integration loop.
B
We define a kind of effective current associated with the time-changing flux, called the
displacement current.
d
JD  o
dt
c.
Magnetic force on moving charges
The magnetic force on a single charged particle is
ˆj
iˆ
kˆ
F  qv  B  q  v x
vy
vz
Bx
By
Bz
q is the charge on the particle
v is the velocity of the particle
B is the externally applied magnetic field
Since the force is perpendicular to the velocity, the magnetic force on the moving charge
does no work. The external magnetic field can change only the direction of the velocity,
but not its magnitude.
Now, if a number of charge carriers are moving along a conductive wire, each will
experience a force in an external magnetic field. The wire will experience a force that is
the total of all the forces on the individual charge carriers.
30
F  q v d  B nA
n is the density of charge carriers in motion
q is the charge on each carrier
A is the cross-sectional area of the wire
 is the length of the wire, immersed in the B-field
vd is the drift velocity of the charge carriers
F  qnAv d  B  I   B
Let’s say that the conducting wire is not straight. Then the force on each short segment
of current-carrying wire is
dF  Ids  B
b
F  I  ds  B
a
If the B-field is uniform, then we can integrate before taking the cross product.
b 
F  I   ds   B  I    B ,
a 
where   is the vector pointing straight from the point a to the point b.
b 
If the current follows a closed loop, in a uniform magnetic field, then   ds  =0, since the
a 
point a and the point b are the same point. Therefore, the net magnetic force on a closed
current loop in a uniform magnetic field is zero.
d.
Magnetic torque on current loops
On the other hand, a uniform magnetic field does exert a torque on a closed current loop.
Consider a rectangular loop carrying a current, I, in a uniform magnetic field.
31
Add up the forces on each side of the rectangle. The sides of length “a” experience no
force, since   B  0 . The sides of length “b” experience equal but opposite forces,
because the directions of the current is opposite, relative to the magnetic field. However,
there is a torque about the axis through the middle of the loop.
a
a
a
 F2  2 IbB  abIB  AIB
2
2
2
o
If the angle is not 90 , then more generally,
  IAB sin 
where  is the angle the B-field makes with the normal to the plane of the loop. The
direction of the torque is obtained by the right-hand-rule.
  F1
Define the magnetic moment of a current loop M  IA , with the direction gotten by the
right-hand-rule. Then   M  B .
32
e.
Potential energy of a current loop in a magnetic field
If a current-carrying loop rotates in a magnetic field, the magnetic field will do work.
2
2
1
1
1
1
W   d   MB sin d  MB  sin d
W  MB  cos  2  cos 1    U
U  M  B
4.
Magnetic Induction
a.
Magnetic flux
Similarly to electric flux, we define the magnetic flux to be  
 B  dA .
surface
Over a closed surface,
C 
 B  dA  0
surface
This means that there are no isolated magnetic monopoles. We never find isolated north
or south magnetic poles.
b.
Faraday’s “Law” of Induction
Consider a conduction loop in a magnetic field. There will be a certain amount of
magnetic flux through the loop. If the flux changes with time, an EMF is induced in the
conducting loop proportional to the time-rate-of-change of the flux.
33
d
dt
d
loopE  ds   dt
EMF  
An electric field is created by a changing magnetic flux.
In a uniform B-field, EMF  
B, A, and  may be changing.
d
BA cos   . Any combination of the three factors
dt
i) changing B
B  Bo sin  t ,
EMF  
dA
 0, and   0
dt
d
 ABo sin  t    ABo cos  t
dt
ii) changing A
d
dA
( BA )   B
  Bv
dt
dt
EMF Bv
I

R
R
Now, the moving bar experiences a magnetic force because of the current.
EMF  
34
B 2 2
v
R
Lenz’s “Law” told us the direction of the induced current. The observed fact is that the
induced current is always directed in such a way as to counteract or compensate for
changes in the original external magnetic flux. The net flux through a conducting loop is
equal to the sum of the external magnetic flux and the induced magnetic flux.
F  IB  
iii) changing 
Imagine a conducting loop with area A, spinning with angular velocity  
d
, in a
dt
uniform B-field.
The induced EMF is
d
d
  BA cos  
dt
dt
d
EMF   BA sin  
dt
EMF  BA  sin 
EMF  
The induced current that results in the loop is
EMF BA 
BA 
I

sin  
sin  t 
R
R
R
The current is alternating, flowing back and forth. Of course, if there are N loops,
forming a coil, then everything is multiplied by N.
c.
Inductance
Let us say that the magnetic flux through a conducting loop is changing. The change in
magnetic flux in turn generates an induced EMF. That induced EMF drives an induced
current. That induced current generates a new magnetic
field, so two magnetic fields are present: the original
externally applied field and the field due to the induced
current in the loop. Lenz’s “Law” tells us the direction of
that induced current. It is observed that the induced current
is always directed so as to counteract or compensate for
changes in the original external magnetic field. If the
35
external magnetic flux is decreasing, the induced B-field will be in the same direction as
the external field. If the external magnetic flux is increasing, the induced B-field will be
opposite to the external B-field.
Consider what happens when the current in a loop or circuit is changing. A circuit loop
comprises a loop of current, which generates a magnetic field. If the current changes, the
magnetic flux through the circuit loop changes.
Here is the chain of events:
i) close the switch—the current increases from 0.
ii) the magnetic field due to this current also increases.
iii) the magnetic flux through the current loop increases, inducing an EMF in the circuit.
(this is the self-induced EMF.)
iv) according to Lenz’s “Law”, the self-induced EMF opposes the current increase.
v) when the current reaches equilibrium at
EMFbattery
R
, the magnetic flux is constant
thereafter. The induced EMF vanishes.
vi) similarly when the switch is opened.
We define the inductance, L, of a conducting loop or circuit as
d
dI
EMF  
 L
dt
dt
For a coil of N identical loops, with the current I in each loop the inductance is L 
EMF
. Its value depends on the geometry of the loop(s).
dI
dt
V s
The SI unit of inductance is the Henry(H). 1H  1
.
A
More generally, L  
36
N
.
I
d.
Energy storage in magnetic fields
i) RL-circuit
EMF  IR  EMFL  0
dI
0
dt
EMF
L dI
I 
0
R
R dt
EMF  IR  L
EMF 
1  e
I
R 
R
t
L




Let us multiply the circuit loop equation by the current, I.
dI
I  EMF  I 2 R  LI
0
dt
These three terms have units of power, which suggests that we might define an energy
stored in the inductor by integrating.
dU m
dI
 LI
dt
dt
Um
I
0
0
 dU m   LIdI
Um 
1 2
LI
2
37
Specifically for a coil, L   o n 2 A , and inside the coil (a long coil) B   o nI . This
allows us to substitute in Um for the I in terms of B.
1
B2
U m LI 2 
A
2
2 o
The energy density is the energy divided by the volume enclosed inside the coil.
U
B2
um  m 
 B2
A 2 o
ii) LC-circuit
We assume that there is no resistance in the circuit. Initially, the capacitor is charged
2
Qmax
with Qmax. The total energy in the circuit is U = UC + Um. At t = 0, I = 0 and U 
.
2C
When the switch is closed, the capacitor begins to discharge. The current increases to a
maximum Imax at t = t’, at which time U
Q
dI
L
0
C
dt
Q
d 2Q
L 2 0
C
dt
2
d Q
1

Q
2
LC
dt
There are not many functions that are proportional to their own second derivative. One is
Q
 t 
 t 
dQ
the cosine. Q  Qmax cos
 . In turn, the current is I 
  max sin 
 .
dt
LC
 LC 
 LC 
What we see is that the energy sloshes back and forth between the electric field in the
capacitor and the magnetic field in the inductor. That is, the circuit oscillates. The
1
frequency of oscillation is  
. The circuit can be “tuned” by adjusting either L or
LC
C, usually the latter. This electrical circuit has exactly the same “equation of motion” as
a harmonic oscillator.
38
iii) RLC-circuit
The resistor dissipates energy when the current is not zero:
dU
 I 2 R .
dt
d 2Q
dQ Q
L 2 R
 0
dt C
dt
1


 Rt
 1  R 2  2 
 t
Q  Qmax e 2 L cos 

 LC  2 L   

 


This is exactly the same equation we obtain for the damped harmonic oscillator—a
harmonic oscillator with a dissipative force, such as friction.
5.
Alternating Current Circuits
a.
Alternating current & voltage
v  Vmax sin  t 
i
v Vmax

sin  t   I max sin  t 
R
R
39
1.5
current
1
0.5
0
-0.5
-1
-1.5
0
100
200
300
400
500
600
700
800
time
RMS
2
P  i 2 R  RI max
sin 2  t 
 P  R  I
I rms 
I max
Vrms 
Vmax
2
max
2
RI max
sin  t  
2
2
2
2
We use the rms quantities to describe a.c., along with the frequency, f. For instance, the
standard electrical residential supply is Vrms = 115 V at f = 60 Hz.
b.
Reactance & impedance
Reactance and impedance are generalized concepts of load resistance.
i) reactance
Consider an inductor circuit.
vL
di
0
dt
Vmax sin  t   L
Solve for the current.
40
di
0
dt
t
Vmax
sin  t dt
L
0
 di  
Vmax
V


cos t    max sin   t  
L
L 
2
Notice the phase shift between the voltage and the induced current. Define inductive
reactance to be:
V
X L  max   L
I max
Reactance has units of Ohms. The voltage on the inductor is written
v L  I max X L sin  t  .
The current in the inductor lags behind the applied EMF.
iL  
Similarly for a capacitor, we define a capacitive reactance, X C 
1
.
C
Q  CVmax sin  t 
dQ


 CVmax  cos t   CVmax  sin   t  
dt
2

vC  I max X C sin  t 
The voltage on the capacitor lags behind the current into the capacitor.
i
Defining reactances in this way puts resistances and capacitances on the same footing in
the relationship between voltage and current. I.e., the magnitudes of the voltages on the
circuit elements are written as
VC  I max X C
VL  I max X L
VR  I max R
ii) impedance
Consider an RLC circuit, with an alternating EMF.
41
The loop equation is v  v L  vC  v R . We’d like to write a simple form for the “effective
resistance” of the circuit. Vmax  I max Z . The Z is to be called the impedance.
v  Vmax sin  t
v L  I max X L sin  t
IL 
Vmax


sin   t  
XL
2



v R  Vmax sin   t  
2

 

vC  I max X C sin   t   
2 2

2
Vmax
 VL  VC   VR2
2
Vmax  VR2  VL  VC   I max R 2   X L  X C   I max Z
2
2
In general, the current in an a.c. circuit lags the applied EMF.
c.
Transformers
Consider a circuit having two coils, a source EMF, and a load resistance as shown.
v1 = input voltage (primary)
v2 = output voltage (secondary)
N1 = number of turns (loops) on primary coil
N2 = number of turns on secondary coil
R = load resistance
42
By Faraday’s “Law,” v1   N 1
d
in the primary coil. The change in flux induces a
dt
voltage in the secondary coil.
v  N
d
  N 2  1   2 v1  i2 R
dt
 N1  N1
The voltage can be stepped up or stepped down, according to the ratio of the number of
N
turns in the coils. v2  2 v1
N1
v2   N 2
From the point of view of the primary coil,
N1
i2 R
N2
For an ideal transformer, we would expect v1i1  v2i2 by virtue of conservation of energy.
v1 
N
N
v
N
v N
v1  i1 1 R 1  i1 1 R 1 1  i1  1
N 2 v2
N 2 v1 N 2
 N2
This is an equivalent resistance “seen” by the source EMF.
2

 R  i1 Req

6.
Maxwell’s Equations
In the mid nineteenth century, Maxwell brought together Gauss’ “Laws” for the electric
& magnetic fields with Faraday’s “Law” of Induction & Ampere’s “Law” into a set of
equations—Maxwell’s Equations.
Q
 E  da   o
 B  da  0
 B  ds  
o
I   o o
d
dt
d
dt
These describe the relations between electric & magnetic fields. Further, they can be
combined to form a wave equation for either the electric or magnetic field.
2E
 2 E   o o 2
t
The apparent fact that the wave speed in this equation equals the measured speed of light
indicates that light is an electromagnetic wave. We’ll illustrate this by working
backwards to show that electromagnetic plane waves in the vacuum are consistent with
Maxwell’s Equations.
 E  ds  
In the vacuum, Q = 0 and I = 0. The plane wave propagates in the x-direction, the E-field
is in the y-direction and the B-field is in the z-direction.
43
Begin with Faraday’s “Law”, substituting for the magnetic flux on the right-hand-side.
The  is along the y-axis.
d
 E  ds   dt
E y  dE y   E y    d Bz dx 
dt
E y
B
dx  dx z
x
t
E y
B
 z
x
t
Similarly, substitute for the electric flux on the right-hand-side
of the generalized Ampere’s “Law”. This time the  is along
the z-axis. (The current, I, is zero.)
 B  ds   
o
o
d
dt
B z   B z  dB z    o  o
d E y dx 
dt
E y
B z
dx   o  o dx
x
t
E y
B z
  o o
x
t

Take the derivative of the result from Faraday’s “Law”, and substitute for
Ampere’s “Law”.
44
B z
from
x
B
  E y

 z
x  x
t
2Ey
x
2
2Ey
x 2
2Ey
x 2




E y 
 B z
 B z



     o  o
x t
t x
t 
t 
2Ey
  o o
t 2
2
1  Ey
 2
c t 2
This is the wave equation, describing waves traveling with speed c 
1
 o o
. This
number turns out to be equal to the observed speed of light, namely c  3 10 8
45
m
.
s
II.
Optics
A.
Geometrical Optics
1.
Wave Fronts & Light rays
a.
Pictorial representations
There are two alternative ways of depicting waves in a drawing. The first is to draw the
wave fronts. Wave fronts are surfaces over which the phase of the wave is the same.
That is, the disturbance is the same in direction and magnitude everywhere on the wave
front. Think of ripples in a water surface. The second way to depict the wave motion is
to draw arrows showing the direction in which the wave is moving—these are rays. In
the case of light, we speak of light rays.
b.
Plane waves & spherical waves
In a plane wave (not plain!), the wave fronts are straight; the rays point all in the same
direction. Here, the waves are propagating from left to right:
Spherical waves are diverging from a centrally located point source. The wave fronts are
concentric spherical shells propagating outward from the central point. The rays radiate
outward equally in all directions. Equally well we might have waves converging on a
central point rather than radiating outward. In the case of spherical waves, the energy
transported by each wave front, as it were, is spread out over an ever-larger spherical
surface. Therefore, the intensity, or power per unit area, decreases by the square of the
distance from the source. This is the famous inverse-square law, which applies to
anything that radiates equally in all directions from a point source.
46
c.
Interface
Light travels in a straight line in a uniform medium. At an interface between media of
differing properties, reflection and refraction will occur. The reason for this is the change
in wave speed that occurs between one medium and another. The index of refraction of a
medium is defined as the ratio of the speed of light in the vacuum to the speed of light in
the medium.
c
n
v
Notice that the angles are measured with respect to the normal at the interface.
“Law” of Reflection: the angle of incidence equals the angle of reflection  a   r .
Snell’s “Law” of Refraction: na sin  a  nb sin  b .
d.
Images
An image is formed where light rays converge, or appear to converge. In the former case
it’s called a real image; in the latter case, it’s called a virtual image. The technique of
locating the image by following the path of the light rays is called ray tracing.
47
Rules for tracing rays
Light rays radiate from every point on the source object. We follow, or trace, the path of
two or three representative rays from a couple of representative points on the source
object. The following rules are valid for light rays close to or nearly parallel to the
optical axis. These are called paraxial rays.
i) incident rays parallel to the optical axis pass through the focal point after
encountering the interface.
ii) incident rays passing through the focal point before encountering the interface
leave the encounter parallel to the optical axis.
iii) rays encountering the interface at the optical are not refracted.
2.
Mirrors
a.
Flat, or plane mirror (not plain!)
The “Law” of Reflection says that the angle of reflection equals the angle of incidence.
By similar triangles, we can say that
y
 1.0
y
A plane mirror forms an erect, virtual image, with a magnification of 1.0. To an
observer, the light rays appear to come from a source behind the mirror—that is the
location of the virtual image.
s  s
y  y
m
b.
Spherical
A curved surface can form a real image in front of the mirror.
48
R = radius of curvature of the spherical mirror
c = the center of curvature
f = the focal point (and focal length, from the center of the mirror)
y = the height of the object, from the optical axis
y’ = the height of the image, from the optical axis
s = the object distance, from the center of the mirror
s’ = the image distance, from the center of the mirror
Note that y, y’ are (+) above the axis, (-) below. Likewise, f, s, s’ are (+) on the object
side of the mirror.
y
s
   0. A
y
s
negative magnification signifies a real inverted image. It can be shown geometrically
that
1 1 1 2
   .
s s f R
With a convex mirror, the reflected rays diverge, so the image is virtual and erect: y’ > 0
s' y'
but s’ < 0. The lateral magnification is M     0 .
s y
By similar triangles, we can see that the lateral magnification, M 
49
3.
Lenses
a.
Spherical interface
At a curved surface between media of differing indices of refraction.
Location of the image
By inspection,     u and   u ' ' .
h
h
h
tan u 
tan u ' 
tan  
s 
s '
R 
For paraxial rays, sin  tan  angle in radians, and   0 . Therefore, approximately,
n  n '  '
n  u   n' '
n
 '    u 
n'
n
  u ' '  u '   u 
n'
n'  n' u ' n  nu
n'n   n' u ' nu
n'n  h  n' h  n h
R
s'
s
n n' n' n
 
s s'
R
Knowing s, n, n’, and R, we can determine s’. This expression works also for a mirror,
with n = n’.
Lateral magnification
Ina similar vein, we can determine the magnification by a curved interface.
50
Once more, for paraxial rays,  and   are small, sin   tan    and
sin    tan      .
n  n '  '
y
y'
n   n'
s
s'
y'
n s'
m 
y
n' s
b.
Thin lenses
A lens is formed by a combination of two curved surfaces. A thin lens is one in which
the radii of curvature are much greater than the separation of the two surfaces (the
thickness of the lens). Here is a double convex lens.
The distances s, s’, R, and f are (+) on the sides(s) that the real light rays travel (as
opposed to virtual rays). By similar triangles,
y
y'
y'
s'
y
y'
y'
s' f
   M    and

 
s
s'
y
s
f
s' f
y
f
Set ‘em equal.
51
s'
s ' f

s
f
f
f
 1
s
s'
1 1 1
 
s f s'
1 1 1
 
f s s'
For a converging lens, f > 0; for a diverging lens, f < 0. However, a converging lens can
produce a virtual image, if s < f. the lens shown below is a double concave lens, with a
negative focal length.

c.
Lens makers equation
The Lens Maker’s Equation relates the focal length to the curvatures of the two surfaces
and to the indices of refraction on either side of the lens and inside the lens. We derive
the equation by tracing an arbitrary ray through the two surfaces in sequence. In effect,
the image formed by the first surface serves as the object for the second surface.
n1 n n  n1
n n1 n1  n
. At the second surface,
.




s1 s'1
R1
s 2 s' 2
R2
Notice this—we are assuming that the index of refraction is the same on both sides of the
lens. This need not be the case.
At the first surface,
For a thin lens, s2 = -s’1.
n1 n
n  n1 n1  n
n n

  1 

s1 s '1 s 2 s ' 2
R1
R2
 1
n1 n1
1 

 n  n1  

s1 s ' 2
 R1 R2 
52
B.
Physical Optics
1.
EM Waves
a.
Reflection & refraction & dispersion
Why waves refract when crossing the interface between two media can be shown by
looking at wave fronts. Consider a plane wave approaching a straight interface at an
angle.
The part of the wave front that encounters the boundary first changes speed, while the
part of the wave front not yet meeting the boundary continues at its original speed. As a
result, the wave front is bent, or the direction of propagation is altered.
On the other hand, the reflected wave front has not changed speed, so the angle of
reflection is the same as the angle of incidence.
53
b.
Energy & intensity
In a region of space where electric and magnetic fields exist, the energy density is
1
1
E
u  oE2 
B 2   o E 2 because B   E  o  o . As the EM-wave advances,
2
2 o
c
energy is transported, dU  udV   o E 2 Acdt . dV is the volume swept out as the wave
advances a distance dx = cdt; A is a cross sectional area through which the wave passes.

dU
EB
The energy flow per unit area per unit time is S 
. If we
  o cE 2  E 2 o 
Adt
o o
take the directions of the vectors into account, we write the Poynting Vector
 1  
S
EB
o
The intensity of the EM-radiation is the time average of S. For sinusoidal waves, the time
1
average of the sine2 or cosine2 is . Therefore,
2
E B
E2
I  S av  max max  max .
2 o
2 o c
2
The SI units of intensity are W/m .
c.
Polarization
In this course, we’ll mention just a qualitative description of polarization. The EM-wave
is a transverse wave. The electric and magnetic fields oscillate in planes perpendicular to
the direction of propagation. If the direction if the electric field is constant, that is always
along the y-axis while the wave propagates along the x-axis, for instance, then the wave is
said to be linearly polarized. Polarizing materials allow E-fields oscillating along one
axis to pass through while blocking E-fields of other orientations. Some surfaces reflect
E-fields that are tangent to the surface but absorb fields normal to the surface.
Circular polarization occurs when the E-field rotates like the second hand of a clock as it
oscillates.
Ordinary light waves from a light bulb or the Sun are randomly polarized—the E-field is
not constantly in the same direction.
2.
Diffraction & Interference
a.
Diffraction
54
When light rays encounter a barrier, they scatter and spread out. Therefore, for instance,
sound goes around corners, and shadows are not sharp-edged. The angle of diffraction is
related to the wavelength, so the larger the wavelength, the greater is the diffraction.
sin   
Passing through a narrow opening, the diffraction depends on the size of the opening
relative to the wavelength. The same is true of small objects, such as dust grains or water
droplets. If the wavelength is much smaller than the size of the opening/object, little
diffraction occurs.
Now consider more closely light passing through a narrow opening. After passing
through, the light spreads out from all parts of the opening (say it’s a narrow slit). Light
rays from different parts of the slit will superimpose and interfere with each other. Take
two representative rays interfering at the point P. They will interfere destructively if the
difference in their path lengths to point P equals an integer multiple of one-half
wavelength. The peak of one wave will coincide with the trough of the other.
sin  

a
 
a
m
sin  
2
2
2
where m  1,  2,  3,  . If L  a , then  is small.
ay
y
m
mL
sin   tan    m 
 ym 
L
2R
2
a
These are the locations of the dark fringes.
55
b.
Two-slit interference
Once again, we consider the difference in path length for rays from the centers of the two
slits.
  r1  r2  d sin 
For destructive interference to occur, at the location y,
1

   m   , m  0,  1,  2,  3,  .
2

For constructive interference at the point y,
  m , m  0,  1,  2,  3,  .
Take the latter case,   m  d sin  m If L  d , then
y
d sin  m  d tan  m  d m  m
L
mL
ym 
d
These are the locations of uniformly spaced bright fringes. However, notice that a
diffraction pattern is superimposed on the interference pattern. This occurs because each
of the two slits has a finite width.
56
The locations of the diffraction and interference fringes depend upon the wavelength.
Therefore, the fringes of light that is not monochromatic will be separated, or dispersed
into a spectrum. Passing light through two or more slits in this way disperses the colours,
just as does a prism, though by a different physical mechanism.
3.
Quantized Waves
a.
Photoelectric effect
When light is shone on a metal plate, electrons in the metal absorb energy and may be
ejected (or liberated) from the metal. These are called photoelectrons. The ejection of
electrons from a metal by light is called the photoelectric effect. Investigation of the
photoelectric effect reveals the following.
57
The number of photoelectrons depends on the intensity of the light, but if the frequency is
below a certain threshold, fo, then no photoelectrons are produced, no matter how great
the intensity. Increasing the intensity, at a fixed frequency, increases the number of
photoelectrons, but not their average kinetic energy (that’s how much energy they have
left after escaping from the metal). On the other hand, as shown in the graph above, if
the frequency of the incident light is increased, then the maximum kinetic energy of the
photoelectrons is larger.
The conclusions are, i) light comes in photons with energy E = hf and ii) a photon is
wholly absorbed or not absorbed at all and iii) a photon is not shared between two
electrons; if the photon delivers more energy than is needed to liberate a single electron,
the single liberated electron has greater kinetic energy left over.
b.
Photons
hc
hc h
 , where p is the momentum of the photon.

E
p
A photon is a particle that has no mass, but does have energy and momentum. Having no
mass, it travels at the speed of light. There are other apparently massless particles, called
neutrinos, which also travel at the speed of light. While no massive particle can travel at
or above the speed of light, the massless particles must travel at the speed of light, no
faster and no slower. The photon has the properties of waves—polarization &
diffraction, etc., yet has properties of particles as well—has momentum and kinetic
energy and undergoes collisions like particles.
E  hf 
c.

deBroglie
Material particles exhibit wave-like properties, too, especially diffraction and
interference. For instance, electron diffraction. . . . .
58
DeBroglie proposed that a wavelength can be associated with a matter particle, such that
h
  , where p is the particle’s translational momentum. The observed pattern formed
p
by diffracted electrons matches this wavelength.
Evidently, the distinction between quantized waves and material particles is not a real
distinction. For macroscopic massive objects, the wavelength is very, very short, so we
do not observe wave-like effects with our unaided sight.
59
III. Atoms
A.
Structure of the Atom
1.
Bohr Model
a.
Orbits
PreQuantum Mechanics model for the Hydrogen atom.
i. stable circular orbits
ii. Coulomb’s “Law” between the proton & electron
iii. quantized orbital angular moment, based on standing deBroglie waves fitted in
the circumference of the orbit—get discrete energy levels.
iv. transitions occur upon absorption or emission of a photon, E  hf .
Applying Newton’s 2nd “Law” to the electron in its uniform circular orbit,
 ke2
v2
ke2
2
.


m

v

e
r
me r
r2
We need the orbital speed for the angular momentum, which is set equal to an integer
h
multiple of  
.
2
ke2
n 2 2
2
L  me vr  me r
 me rke  n  rn 
me r
me ke2
This determines the “allowed” orbits for the electron. Next, the total mechanical energy
of the electron is
1
ke2 1  ke2  ke2
1 ke2
 
.
E  me v 2 
 me 

2
r
2  me r  r
2 r
Finally, we substitute the allowed orbital radii into the expression for the energy to obtain
the discrete energy levels.
m ke2
1 ke2
1
1 me k 2 e 4
1
En  
  ke2 e2 2  
 2 E1
2 2
2 rn
2
2 n 
n 
n
For other atoms, having more electrons, of course the energy levels are more numerous
and their spectra more complicated. Still, the spectrum of each chemical element or
compound is unique.
b.
Spectra
When the electron makes a transition from one orbit to another, its energy increases
(decreases), and a photon is absorbed (emitted). The wavelength of that photon is
determined by the magnitude of the energy change.
60
 1
1
E  E1  2  2
 n 2 n1
f 
1





E1  1
E
1 
 2  2 

h
h  n 2 n1 
E1  1
1 
 2  2 
hc  n 2 n1 
Experimentally, the emission spectrum of the Hydrogen atom had been observed to
follow the following pattern.
 1
1
1 
 R 2  2 

 n2 n1 
The Rydberg Constant was evaluated by fitting to the data: R  1.097  10 7 m 1 . The
Bohr Model for the Hydrogen atom exactly matches the experimental observations of the
m e4
Hydrogen spectrum. R  2e 3
8 o h c
However, subsequent more detailed observations of atomic spectra revealed features that
the Bohr model did not include. Only the quantum mechanical description of the atom
can “explain” all the features of atomic spectra. The picture of an atom as a tiny solar
system is not in the end wholly realistic. However, the Bohr Model introduces the notion
of discrete energy levels for the electrons in atoms.
2.
Nucleus
a.
Nucleons
protons & neutrons
strong nuclear force
stability, instability
b.
Radioactivity
alpha, beta, gamma
decay
penetration
dose
61
B.
Subatomic
1.
Leptons & Quarks
a.
Leptons
b.
Quarks
c.
2.
Hadrons
a.
Mesons
b.
Baryons
c.
3.
Anti-Matter
62