Review for Final Exam
Schedule
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Review Session on Monday at 4:00 PM in
101 UPL if it extends beyond 6:00 PM it
will move to 707 Keen.
Final exam – Tuesday, April 23 at 7:30 AM
in 101 Carroway building.
This Power Point presentation contains 36
slides – most likely I won’t get to more
than about 1/3 of them. Make sure you
understand the concepts and problems
presented here.
B-Field Due to Currents

Electric currents produce magnetism.


B = oI/(2r) (due to a long straight wire)
Direction from the right-hand rule.
B
I
B
I
Curl your fingers
as if following B.
Your thumb is in
the direction of
the current.
Example: Mass Spectrometer.
F = qvB
F = ma
q,v
R
ma = qvB
a = qvB/m
qvB/m = v2/R  R = mv/(qB)
B is out of the
page.
Example: Mass Spectrometer.
F = qvB
q,v
F = ma = mv2/R = qvB
mv = qBR or
R
p = qBR
½mv2 = (mv)2/(2m) = p2/(2m)
Energy = ½mv2 = ½(qBR)2/m
B is out of the
page.
q < 0 !!
Forces Between Currents
I1
F12
B2
F21
I2
F12/L = I1B2
= I1oI2/(2r)
= o I1 I2/(2r)
B2
I1
F12
B2
Forces Between Currents
B1
I1
F12
B1
F21
I2
F21
I2
B1
Major Concepts

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



Battery (voltages)
Ohm’s Law:
V = IR
Resistance:
R = V/I (in ohms “”)
Resistivity:
R = L/A ( in /m)
Power:
P = I2R (resistors)
Power:
P = VI (batteries or resitors)
Major Concepts



Series Circuit

Current must go through all resistors

R T = R 1 + R2 + R 3 + …
Parallel Circuit

Current is divided between the resistors

1/RT = 1/R1 + 1/R2 + 1/R3 + …
Terminal Voltage


Accounts for the resistance within the battery.
Vterminal = V – Ir(internal)
Current in a Simple Circuit
A
V=8V
+
-
R = 24 
C
Arrow shows the direction
that positive charges move.
V = IR  I = V/R
I = 8 V/24 
B
I = 0.33 Amp
Current at A = 0.33 Amps
Current at B = 0.33 Amps
Current at C = 0.33 Amps
Current is the
same everywhere
in the circuit!
Energy in a Simple Circuit
B
V=8V
+
-
A
R = 24 
D
Recall: I = 1/3 Amp
Consider the energy of a
proton moving through the
circuit. (recall: q = +1 e)
Energy(A) = qVA = 8 eV
C
Proton loses energy
moving from B to C.
It gains energy moving
from D to A.
Energy(B) = qVB = 8 eV
Energy(C) = qVC – q(IR)
= 8eV – 1e * ( (1/3)*24 V )
= 0 eV
Energy(D) = qVD = 0 eV
Energy in a Simple Circuit
B
V=8V
+
-
A
R = 24 
D
C
An electron loses energy
moving from C to B.
It gains energy moving
from A to D.
Recall: I = 1/3 Amp
Consider the energy of a
proton moving through the
circuit. (recall: q = -1 e)
Energy(D) = qVA = -8 eV
Energy(C) = qVB = -8 eV
Energy(B) = qVC – q(IR)
= -8eV – -1e*((1/3)*24 V )
= 0 eV
Energy(A) = qVD = 0 eV
Circuit Analysis

Kirchhoff’s Rules:

Loop Rule:




The sum of the voltage drops around any closed loop is
zero.
Conservation of Energy
V = 0
Junction Rule:



The net current into and out of any point in a circuit is
zero.
Charge conservation.
I = 0
Complex Circuits
R1
A
V
I1
C
F
R2
H
B
G
I2
I3
E
R1 = 10 , R2 = 3 , R3 = 6 
and V = 24 Volts
R3
D
1. I1 goes from A to B
and from E to H.
2. I2 goes from B to F
to G to E.
3. I3 goes from B to C
to D to E.
Complex Circuits
R1
A
V
I1
C
F
R2
H
B
I2
I3
1. Junction Rule at B:
I1 - I2 - I3 = 0
R3
2. Loop Rule: ABFGEHA
-I1R1 – I2R2 + V = 0
G
E
D
3. Loop Rule: ABCDEHA
-I1R1 – I3R3 + V = 0
Complex Circuits
R1
A
V
I1
C
F
R2
H
B
G
I2
I3
I1 - I2 - I3 = 0
-I1R1 – I2R2 + V = 0
R3
So: -10I1 – 3I2 + 24 = 0
-I1R1 – I3R3 + V = 0
E
D
So: -10I1 – 6I3 + 24 = 0
Solving the equations.
I1 - I2 - I3 = 0
10I1 + 3I2 - 24 = 0
10I1 + 6I3 - 24 = 0
10I1 + 3I2 - 24 = 0
- ( 10I1 + 6I3 - 24 = 0 )
3I2 – 6I3 = 0
3I2 = 6I3
I2 = 2I3
Solving the equations.
I1 - I2 - I3 = 0
I2 = 2I3
I1 - 2I3 - I3 = 0
I1
- 3I3 = 0
I1 = 3I3
10I1 + 6I3 - 24 = 0  10*3I3 + 6I3 - 24 = 0
OR: 36I3 - 24 = 0  I3 = 24/36 = 0.667 A
I1 = 3I3  I1 = 3* 0.667 A = 2.0 A
Solving the equations.
I1 - I2 - I3 = 0
I2 = 2I3
I3 = 24/36 = 2/3 A
I1 = 3I3  I1 = 3* 2/3 A = 2 A
I2 = 2I3 = 2 * 2/3 = 4/3 A
Check the Equations


I1 = 2 A, I2 = 4/3 A, I3 = 2/3 A
I1 - I 2 - I3 = 0


10I1 + 3I2 - 24 = 0

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2 – 4/3 – 2/3 = 0 
10*2 + 3*4/3 – 24 = 0 
10I1 + 6I3 - 24 = 0

10*2 + 6*2/3 – 24 = 0 
Balancing the Energy

P(in) = VI1 = 24 * 2 = 48 W
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P1 = I12R1 = 22*10
= 40 W
P2 = I22R2 = (4/3)2*3 = 16/3 W
P3 = I32R3 = (2/3)2*6 = 8/3 W
PT = 40 + 16/3 + 8/3 = 48 W 
Reflected Light
The angle of incidence equals
the angle of reflection.
in
out
Refracted Light
air
The direction changes when the
light moves from material to
another.
glass
The change depends on the
material.
in
Snell’s Law determines the angles:
out
N1 sin 1 = N2 sin 2
The index of refraction, N, is the
ratio of the speed of light in
vacuum to the speed of light in the
material.
N = c/v
c = speed of light in vacuum.
v = speed of light in material.
Lenses
1
2
object
1 – Parallel Ray
2 – Central Ray
3 – Focal Ray
3
image
Optic
Axis
Lens Equation
1
1
1
— = — + —
F
O
I
F
image
object
O
I
Optic
Axis
Diverging Lens
1
3
object
Optic
Axis
image
1 – Parallel Ray
2 – Central Ray
3 – Focal Ray
2
Diverging Lens
1
3
object
Optic
Axis
image
F = - 20 cm
O = 50 cm
Use the Lens Equation
2
Diverging Lens w/ Lens Equation
1
1
1
— = — + —
F
O
I
F = - 20 cm O = 50 cm
1/(-20) = 1/50 + 1/I
 1/I = -1/20 – 1/50 = -5/100 – 2/100
1/I = -7/100  I = -100/7 = - 14.28 cm
m = -I/O = - (-14.28 cm)/( 50 cm ) = + 2/7
m > 0  Image is upright.
I < 0  image is virtual.
Atomic Model





Electrons moved around nucleus only in
certain stable orbits.
Stable orbits are those in which an integral
number of wavelengths fit into the diameter
of the orbit (2rn = n)
They emitted (absorbed) light only when
they changed from one orbital to another.
Orbits have quanta of angular momenta. L
= nh/2
Orbit radius increases with energy rn = n2
r1 (r1 = .529 x 10-10 m)
Atomic Energy Levels

En =
Z2/n2
Ionized atom
E1
n=3

Hydrogen

En = - 13.6 eV/n2
- 3.4 eV
E = - 13.6 eV
n=2
n=1
Emission & Absorption

Energy is conserved.




E = Eatom = Ef - Ei
Photon energy = hf = hc/
Absorption  photon disappears a electron in the
atom changes from a lower energy level to a higher
energy level.
Emission  an electron in atom goes from higher
energy level to a lower energy level. This change
in energy is the energy of the photon.
Heisenberg Uncertainty Principle

Impossible to know both the position and the momentum
of a particle precisely.

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A restriction (or measurement) of one, affects the other.
x p  h/(2)
Similar constraints apply to energy and time.

E t  h/(2)
EXAMPLE: If an electron's position can be measured to an accuracy of
1.96×10-8 m, how accurately can its momentum be known?
x p  h/(2)  p = h/(2x)
p = 6.63x10-34 Js /(2 1.96x10-8 m) = 5.38 x 10-27 N s
Nuclear Decay Rates

N = -No  t

Number of decaying nuclei (in a given
time t), depends on:




Number of remaining nuclei, No
Nuclear decay constant for that type of
nuclei, 
Number of nuclei remaining after a
time t:
N = No e-t
Nuclear Decay Rates
Nuclear Decay
Nuclei Remaining
1000.0
At t =  N is 1/e
(0.368) of the
original amount
800.0
600.0
400.0
200.0
0.0
0.0
1.0
2.0
3.0
Time(s)
4.0
5.0
Example
10
5
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

B ( 01 n , 24 He ) 37 Li
1
0
n +
10
5
B 
10
7
3
Li +
4
2
He
B ( n, )7 Li
Start with 5 protons  end up with 5 protons.
Start wit 11 baryons  end up with 11 baryons.
Q-value: (Mass Energy of Final State – Mass Energy of Initial
State)
Mn = 1.008665 u + M(B) = 10.012936 u = 11.021601 u
M = 4.002602 u + M(L) = 7.016003 u = 11.018605 u
Q = M (in MeV) = Mf – Mi = 11.018605 u - 11.021601 u
Q = -0.002996 u = -0.002996 u * 931.5 MeV/u = -2.79 MeV
Negative Q-value means we get extra energy out of the reaction.
Example: Nuclear Power
1.
Suppose that the average power consumption, day and night, in a typical
house is 340 W. What initial mass of 235U would have to undergo fission
to supply the electrical needs of such a house for a year? (Assume 200
MeV is released per fission.)
Energy used = rate * time = 340 W * 3.15 x 107 s/yr = 1.07 x 1010 J
Energy from each nucleus is:
200 MeV/nuclei = 200 MeV /nuclei * 1.60 x 10-13J/MeV = 3.2 x 10-11
J/nuclei
Number of nuclei required = Energy/(Energy per nucleus)
Number of nuclei = 1.07 x 1010 J/3.2 x 10-11 J/nuclei = 3.34 x 1020 nuclei
Total mass of 235U = number of nuclei * mass/nucleus
Mass = 3.34 x 1020 nuclei * 235 amu * 1.66 x 10-27 kg/amu = 1.31 10-4 kg