review for final exam. schedule review session on monday at 4:00 pm in 101 upl if it extends beyond...
TRANSCRIPT
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Review for Final Exam
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Schedule 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.
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B-Field Due to Currents
Electric currents produce magnetism. B = oI/(2r) (due to a long straight wire) Direction from the right-hand rule.
BI
BI
Curl your fingers as if following B. Your thumb is in the direction of the current.
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Example: Mass Spectrometer.
F = qvB
F = ma
ma = qvB
a = qvB/m
qvB/m = v2/R R = mv/(qB)
R
q,v
B is out of thepage.
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Example: Mass Spectrometer.
F = qvB
F = ma = mv2/R = qvB
mv = qBR or p = qBR
½mv2 = (mv)2/(2m) = p2/(2m)
Energy = ½mv2 = ½(qBR)2/m
R
q,v
B is out of thepage.
q < 0 !!
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Forces Between Currents
B2
B2
I1
F2
1
F1
2
I2
B2
F1
2
I1
F12/L = I1B2
= I1oI2/(2r)
= o I1 I2/(2r)
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Forces Between Currents
I1
F2
1
F1
2
I2
B1
B1
B1
F2
1
I2
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Major Concepts 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)
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Major Concepts Series Circuit
Current must go through all resistors RT = R1 + R2 + R3 + …
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)
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Current in a Simple Circuit
V = 8 V+-
R = 24
Arrow shows the direction that positive charges move.V = IR I = V/R
I = 8 V/24
I = 0.33 Amp
A
B
C
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!
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Energy in a Simple Circuit
V = 8 V+-
R = 24
Recall: I = 1/3 AmpConsider the energy of a proton moving through the circuit. (recall: q = +1 e)
Energy(A) = qVA = 8 eV
Energy(B) = qVB = 8 eV
Energy(C) = qVC – q(IR) = 8eV – 1e * ( (1/3)*24 V ) = 0 eVEnergy(D) = qVD = 0 eV
B
C
D
A
Proton loses energy moving from B to C. It gains energy moving from D to A.
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Energy in a Simple Circuit
V = 8 V+-
R = 24
Recall: I = 1/3 AmpConsider 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 eVEnergy(A) = qVD = 0 eV
B
C
D
A
An electron loses energy moving from C to B. It gains energy moving from A to D.
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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
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Complex CircuitsR1
R2
V
R3
I1
I2 I3
F
G
A B C
DE
H
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.
R1 = 10 , R2 = 3 , R3 = 6
and V = 24 Volts
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Complex CircuitsR1
R2
V
R3
I1
I2 I3
A B C
DE
F
G
H
1. Junction Rule at B:
I1 - I2 - I3 = 0
2. Loop Rule: ABFGEHA
-I1R1 – I2R2 + V = 0
3. Loop Rule: ABCDEHA
-I1R1 – I3R3 + V = 0
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Complex CircuitsR1
R2
V
R3
I1
I2 I3
A B C
DE
F
G
H
I1 - I2 - I3 = 0
-I1R1 – I2R2 + V = 0
So: -10I1 – 3I2 + 24 = 0
-I1R1 – I3R3 + V = 0
So: -10I1 – 6I3 + 24 = 0
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Solving the equations.
I1 - I2 - I3 = 0
10I1 + 3I2 - 24 = 0
10I1 + 6I3 - 24 = 0 3I2 – 6I3 = 0
3I2 = 6I3
I2 = 2I3
10I1 + 3I2 - 24 = 0
- ( 10I1 + 6I3 - 24 = 0 )
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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
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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
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Check the Equations I1 = 2 A, I2 = 4/3 A, I3 = 2/3 A I1 - I2 - I3 = 0
2 – 4/3 – 2/3 = 0 10I1 + 3I2 - 24 = 0
10*2 + 3*4/3 – 24 = 0 10I1 + 6I3 - 24 = 0
10*2 + 6*2/3 – 24 = 0
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Balancing the Energy P(in) = VI1 = 24 * 2 = 48 W
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
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Reflected Light
The angle of incidence equals the angle of reflection.
in out
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Refracted LightThe direction changes when the light moves from material to another.
The change depends on the material.
Snell’s Law determines the angles:
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.
in
out
air glass
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Lenses
OpticAxis
1
2
3
1 – Parallel Ray2 – Central Ray3 – Focal Ray
object
image
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Lens Equation
OpticAxis
1 1 1— = — + —F O I
object
image
O I
F
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Diverging Lens
OpticAxis
1
2
3
1 – Parallel Ray2 – Central Ray3 – Focal Ray
object image
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Diverging Lens
OpticAxis
1
2
3
object image
F = - 20 cmO = 50 cmUse the Lens Equation
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Diverging Lens w/ Lens Equation
1 1 1— = — + —F O I1/(-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.
F = - 20 cm O = 50 cm
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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 (2rn = 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)
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Atomic Energy Levels
En = Z2/n2 E1
Hydrogen En = - 13.6 eV/n2
Ionized atom
E = - 13.6 eV
- 3.4 eV
n = 1
n = 2
n = 3
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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.
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Heisenberg Uncertainty Principle
Impossible to know both the position and the momentum of a particle precisely.
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/(2x)
p = 6.63x10-34 Js /(2 1.96x10-8 m) = 5.38 x 10-27 N s
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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
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Nuclear Decay Rates
Nuclear Decay
0.0
200.0
400.0
600.0
800.0
1000.0
0.0 1.0 2.0 3.0 4.0 5.0
Time(s)
Nu
cle
i R
em
ain
ing
At t = N is 1/e (0.368) of the original amount
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Example
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)
B105 Li73 He4
2n10 + +
B105 Li73He4
2n10( ), B10 Li7n( ),
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.
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Example: Nuclear Power1. 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/nucleusMass = 3.34 x 1020 nuclei * 235 amu * 1.66 x 10-27 kg/amu = 1.31 10-
4 kg