Review for Final Exam

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.

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.

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.

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 !!

Forces Between Currents

B2

B2

I1

F2

1

F1

2

I2

B2

F1

2

I1

F12/L = I1B2

= I1oI2/(2r)

= o I1 I2/(2r)

Forces Between Currents

I1

F2

1

F1

2

I2

B1

B1

B1

F2

1

I2

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)

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)

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!

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.

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.

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 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

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

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

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 )

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 - 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

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

Reflected Light

The angle of incidence equals the angle of reflection.

in out

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

Lenses

OpticAxis

1

2

3

1 – Parallel Ray2 – Central Ray3 – Focal Ray

object

image

Lens Equation

OpticAxis

1 1 1— = — + —F O I

object

image

O I

F

Diverging Lens

OpticAxis

1

2

3

1 – Parallel Ray2 – Central Ray3 – Focal Ray

object image

Diverging Lens

OpticAxis

1

2

3

object image

F = - 20 cmO = 50 cmUse the Lens Equation

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

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)

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

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.

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

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

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

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.

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