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Name: ____________________________________________________ Student ID: ___________________________ Section #: _________

Physics 208 Final Exam Dec. 21, 2007 Print your name and section clearly above. If you do not know your section number,

write your TA’s name. Your final answer must be placed in the box provided. You must show all your work

to receive full credit. If you only provide your final answer (in the box), and do not show your work, you will receive very few points.

Problems will be graded on reasoning and intermediate steps as well as on the final answer. Be sure to include units, and also the direction of vectors.

You are allowed two 8½ x 11” sheets of notes and no other references. The exam lasts exactly 120 minutes.

Coulomb constant:

!

ke

= 9.0 "109

N #m2

/C2

Speed of light in vacuum:

!

c = 3"108m /s Problem 1: _______ / 20

Permittivity of free space:

!

"o

=1/4#ke

= 8.85 $10%12C2/N &m

2 Permeability of free space:

!

µo

= 4" #10$7T %m /A Problem 2: _______ / 25 Planck’s constant:

!

h = 6.626 "10#34J $ s

!

= 4.1357 "10#15eV $ s , h = h /2% Problem 3: _______ / 20

Bohr radius:

!

ao

= 0.053nm Bohr magneton:

!

µB

= 5.788 "10#5eV /T Problem 4: _______ / 20 Atomic mass unit:

!

1 u =1.66054 "10#27kg

!

= 931.494 MeV /c2 Problem 5: _______ / 25

Electron mass:

!

me = 9.11"10#31kg

!

= 0.00055u = 0.51 MeV /c2 Problem 6: _______ / 10

Proton mass:

!

mp =1.67262 "10#27kg

!

=1.00728u = 938.28 MeV /c2

Neutron mass:

!

mn =1.67493"10#27kg

!

=1.00866u = 939.57 MeV /c2 TOTAL: _______ / 120

!

hc =1240eV " nm

!

1eV =1.602 "10#19J

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1) [20 pts, 4 pts each] Multiple choice. Show work/explanation for full credit. i) A parallel-plate 1 µF capacitor is connected to a 2 V battery while the plates are being

pulled apart, doubling the separation. The 2V battery stays connected the entire time. How does the charge after compare to the charge before pulling the plates apart

a. two times bigger b. two times smaller c. four times bigger d. four times smaller e. same

ii) A positively-charged particle is moving as shown parallel to a wire carrying a positive

current I in the direction shown. The direction of the force on the particle due to the current in the wire is

a. down b. up c. right d. left e. out of page f. into page

iii) A quantum particle in a box can have only these energies: E1=1 eV, E2=4 eV, and

E3=9 eV. Which of these photons could be emitted from the box. a. 310 nm b. 138 nm c. 413 nm d. 572 nm e. both a & b

I v

Explanation/Work:

!

Q = CV and

!

C = "oA /d , so at constant

voltage, the charge decreases by a factor of two since the capacitance goes down by a factor of two

Explanation/Work

!

" = hc /E =1240eV # nm /3eV = 413nm

1240eV # nm /5eV = 248nm

1240eV # nm /8eV =155nm

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iv) Radioactive

!

9

18F , used in positron emission tomography (PET), is observed to emit a

positron from the nucleus. What isotope does it become after the decay? a.

!

7

14N

b.

!

8

18O

c.

!

10

18Ne

d.

!

9

19F

e.

!

9

17F

v) You are in a room that is 3 meters high. You hold a lens below a light bulb on the ceiling, forming an image of the bulb directly below it on the floor. What focal length lens could form the largest image on the floor?

a. 0.2 m b. 0.5 m c. 1.0 m d. 2.0 m e. 3.0 m

Explanation/Work: Positron emitted means proton changes to neutron. This means that the element now has 8 protons. This is oxygen.

Explanation/Work: Only the 0.2m and 0.5m lenses can make an image on the floor. Of these should use shortest focal length to get biggest image.

!

1/i +1/o =1/ f

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2) [25 points, 5 pts each] Short-answer questions Explain your answers! a) The binding energy of 235U is 7.3 MeV/nucleon. Suppose that each 235U nucleus in

1 kg breaks apart (fissions) into two smaller nuclei, one with 120 nucleons and the other with 115 nucleons. Each of these smaller nuclei has a binding energy of 8.3 MeV/nucleon. There are about ~2.5x1025 U atoms in 1 kg.

Calculate the energy released in the fission of 1 kg of 235U. b) A parallel-plate capacitor has two square plates, each 10cmX10cm, with a

separation of 0.1 cm. The top plate has charge density 10µC/m2, and the bottom plate has charge density -10µC/m2. The charge is uniformly distributed over the plates. What is the force on a -0.2 µC point-like particle 0.02 cm above the bottom plate?

F= Value Units Direction

E = Value Units

-10µC/m2

+10µC/m2

-0.2µC 0.1cm

0.02cm

Binding energy of 235U is 7.3 MeV/nucleon Binding energy of element with 120 is 8.3 MeV/nucleon. This is a binding energy difference of 1 MeV/nucleon. There are 235 nucleons, so the energy released is about 235 MeV per nucleus.

!

Ereleased

= 235 "106eV( ) 1.6 "10#19J /eV( ) 2.5 "1025atoms( ) = 9.4 "1014 J

The electric field is

!

V /d =Q

Cd=

Q

"oA /d( )d=Q

"oA=#

"o

=

!

10"5C /m

2

8.85 #10"12C2/N $m

2=1.13#10

6N /C

!

F = qE = 2 "10#7C( ) 1.13"106N /C( ) = 0.22N

E-field points from top to bottom plate, so force is up.

Page 5

c) A hydrogen atom is in the n=4, l=3 state. How many orientations of the orbital angular momentum are possible?

d) Experiments have shown that a human eye is sensitive enough to detect 1 milli-

second (0.001s) flashes of 50 femto-Watt (50x10-15 W) green (500 nm) light. How many photons are in this flash of light? e) A 2 cm diameter conducting wire is carrying a

current of 200 A. The current is uniformly distributed throughout the volume of the wire. Calculate the force on a +10µC charge moving at a speed of 5 m/s in the direction shown.

[Hint: Use Ampere’s law]

Magnitude Direction Units

F =

# orientations = Value

2 cm

1 cm

+10µC V=5m/s

I=200 A

x y

z

Since ml ranges from –l to l by integers, there are 7 possible orientations.

Ampere’s law says that

!

r B • d

r l " = µ

oI

enclosed. Taking

the path bounding the surface which I cuts to be a circle of radius r=2 cm, we find

!

B2" r= µoI , or

!

B =4" #10$7T %m /A( ) 200A( )

2" 0.02m( )= 2 #10$3T . The force

is then

!

F = qvB = 10"5C( ) 5m /s( ) 2 #10"3T( ) =1#10"7N .

By the right-had rule, the direction is up (z-direction)

# photons = Value

The total energy deposited in the eye is (50x10-15 J/s) (0.001s)=5x10-17J

Green light has

!

hc /" =1240eV # nm /500nm = 2.48eV / photon

Ephoton = 2.48eV( ) 1.6 $10%19J /eV( ) = 3.97 $10%19J / photon

So there are 5x10-17J / (3.97x10-19 J/photon) = 126 photons in the flash

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3) [20 pts, 4 pts each] A magnet and coil of wire are shaken so that the magnet slides all the way back and forth through the coil, as shown in the drawing below. The ends of the coil are connected to each other, so that current can flow in the coil. a) The graph below shows the magnetic flux through the coils as a function of time as it

slides back and forth. Sketch the current in the coil as a function of time. The vertical CURRENT axis and the horizontal TIME axis have arbitrary units.

b) Does the magnitude of the maximum current depend on how fast you shake it?

Explain

!

"

TIME

!

+1 mT "m2

!

"1 mT #m2

TIME

CU

RR

ENT

0

Since the current is proportional to the emf, which is proportional to the time rate-of-change of the flux, the faster you shake it, the bigger the current.

N S

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c) Does the magnitude of the total amount of charge flowing through the coil on each shake (one trip of magnet through coil in one direction) depend on how fast you shake it? Explain.

d) If the coil has N=100 turns, and a resistance of R=10Ω, what is the magnitude of the

total charge that flows through the coil each time the magnet moves through it? Hint: the flux through the coil changes by

!

2 mT "m2

= 0.002 T "m2( ) each time the

magnet passes through the coil. Each shake produces a current

!

N d" /dt( ) /R , and hence total charge

!

N"# /R = 100( ) 0.002T $m2( ) /10% = .02C passes through the coil. e) Flashlights without batteries use this technique to charge a super-capacitor, which

then powers the flashlight. It also has a circuit element so that during shaking the capacitor is charged, but never discharged.

Suppose that the charge through the coil is 0.03 C for each shake, and all this charge is stored on the capacitor. Then how many shakes are needed to charge the 1.0 F capacitor to 3 volts?

Charging 1.0 F to 3 V requires Q=CV = 3 C of charge. So 3C/0.03C=100 shakes are

required. The problem ignores the voltage increase of the capacitor. In a real circuit, the capacitor voltage would increase as it charges up, and the current in the coil would decrease so that less charge is delivered to the capacitor on each shake.

The total amount of charge does not depend on how fast you shake. The current does, but the current lasts for a shorter amount of time. Mathematically,

!

I = d" /dt( ) /R , and

!

Q = Idt = 1/R( )N d" /dt( )dt = N#" /R$$

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4) [20 pts, 5 pts each] A soap film (n = 1.33) is contained within a rectangular wire frame in air (n = 1.0). The frame is held at an angle so that the film drains downwards and forms a wedge with flat faces as shown in the figure. The thickness of the film at one end is essentially zero as shown in the figure. The film is viewed in reflected white light with near-normal incidence. The first violet (λ1 = 420 nm) interference bright band is observed a distance x1 = 4.00 mm (shown in figure) from the zero-thickness edge of the film.

a) What is the thickness of the wedge at the first violet (λ1 = 420 nm) band? Solution: Bright bands are observed at 2nt =(m+1/2)λ, so the first band (m=0) t = λ/4n for the various colors. The first violet band corresponds to t1=λ1/(4n) = 78.9 nm

xx

x1 x2

t1 = Value Units

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b) What is the thickness of the wedge at the first red (λ2 = 680 nm) band? Solution: Bright bands are observed at 2nt =(m+1/2)λ, so the first band (m=0) t = λ/4n for the various colors. The first red band corresponds to t2=λ2/(4n) = 127.8 nm c) What is the x-distance of the first red (λ2 = 680 nm) interference band from the zero-thickness edge? (This distance is labeled x2 in the figure) Solution: Bright bands are observed at 2nt =(m+1/2)λ, so the first band (m=0) t = λ/4n for the various colors. The first violet band corresponds to t1=λ1/(4n) and the first red band to t2=λ2/(4n). Given the geometry of the 2 triangles with one corner in common: x1/x2 = t1/t2 = => x2 = x1(t2/t1) = x1(λ2/λ1) = 4x680/420= 6.48 mm d) Suppose oil (n=1.5) is used for the film rather than soap. Would the bands of colors be more closely spaced in x or farther apart in x? Circle the correct answer and explain. Since oil has a higher index of refraction, the required thicknesses will be less, and so

the bands of colors will be spaced more closely together.

x2 = Value Units

Closer together Farther apart

t2 = Value Units

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5) [25 points, 5 points each] A piece of charcoal containing 50 g of carbon is found in some ruins of an ancient city. The sample shows a 14C activity of R = 200 decays/min. The half-life of 14C is 5,730 years. Assume that the ratio of 14C to 12C in our atmosphere has a constant value of 1.3 x 10-12. a) What is the number of 14C nuclei in the piece of charcoal immediately after the tree

was cut down and burned? (there are 5.0 x 1022 nuclei in 1 gm of 12C) b) When

!

6

14C decays it produces

!

7

14N .

What type of decay is this (alpha, beta-, beta+, or gamma)? Explain. c) Calculate the decay rate r for 14C.

Decay type=

r =

Value Units

N(C14) =

Value

# 14C nuclei = (1.3 x 10-12 14C/12C)(50 gm)( 5.0 x 1022 12C/gm)=3.3x1012

Here a neutron changes into a proton, increasing the positive charge. So an electron must have been emitted to conserve charge. This is beta- decay.

Decay rate is related to half-life as

!

r =ln2

T1/ 2

= 3.84 "10#12s#1

Here

!

T1/ 2

= 5730yrs( ) 365day / yr( ) 24hr /day( ) 60min/hr( ) 60s /min( ) =1.8 "1011s

Page 11

d) What was the initial activity Ro of the charcoal immediately after the tree was cut down and burned? e) How long has the tree from which the charcoal came been dead? For this part assume an initial activity Ro=630 decays/min.

!

R = Roe"rt

= Ro

1

2

#

$ % &

' (

t / t1/2

) lnRo

R=

t

t1/ 2

ln2

!

t

t1/ 2

= ln630

200/ln2 =1.655" t = 9485yrs

Ro= Value Units

# years = Value

Initial activity is

!

N14C( )r = 3.25 "10

12( ) 3.84 "10#12 /s( ) =12.48decays /s = 749decays /min

Page 12

6) [10 pts, 5 pts each] This problem talks about a quantum dot, As we did in class, model this as a one-dimensional particle in a box.

The energy levels are given by

!

En

= n2h2/8m

eL2 , with wavefunctions

!

" x( ) =2

Lsin n#

x

L

$

% &

'

( )

for a particle confined in a box between x=0 and x=L. a) For the n=2 state, draw the wavefunction and the probability density P(x) for the

particle. ( P(x)dx is the probability of finding the particle in the interval dx )

Ψ

x x=L

Ψ=0 =0

Prob

x=0

x x=L x=0

P=1

P=0

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b) You need a quantum dot that emits blue light (λ=450 nm).

You are offered a 1 nm dot. Is a 1 nm dot too big or too small to get a 450 nm photon from the n=2->n=1 transition? Circle the correct answer and explain.

Too big Too small

The energy of the photon for the n=2-> n=1 transition is

!

En= 2 " En=1

, so its wavelength is

!

hc / E2" E

1( ) = hc / 22 "12( )h2 /8mL2 =8

3mcL

2/h

!

=8

3

mc2

hcL2 =8

3

0.51"106eV

1240eV # nm1nm( )

2

=1096nm . This is too long a

wavelength, so a 1 nm dot is too big.