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NOTE:
In all Questions choose the answer that is the closest!!
(hc = 1240 eVnm, c = 3 x 108m/s) also ʃysin2ydy = y2/4 – ysin2y/4 – cos2y/8
Question I. (40 pts) Particle-Wave Duality
1. (5 pts) If the wavelength of an electron is equal to the wavelength of a proton, then
(A) the speed of the proton is greater than the speed of the electron,
(B) the speeds of the proton and the electron are equal,
(C) the Kinetic energy of the proton is greater than the Kinetic energy of the electron
(D) the speed of the proton is less than the speed of the electron
(E) Both (a) and (d) are correct.
Since the wavelengths are equal, their momenta are also equal. Thus because the proton
has a much bigger mass than the electron, the proton moves slower than the electron.
2.(5 pts) The muon is a subatomic particle with a mass, m , that is 200 times the electron
mass, me , so that m = 200 me . Assume a muon and an electron have the same energy.
Compared to the electron, the De Broglie wavelength of the muon is
(A)greater
(B)equal to (C) less than (D) not comparable in any way
3. (5 pts) Particle – wave duality is observed by Young’s double-slit experiment, what
theoretical work gave the greatest support to particle-wave duality?
(A) Schrodinger’s cat (B) Heisenberg’s uncertainty principle
(C) Compton scattering
(D) the J. J. Thompson cathode rays experiemnt
(E)
none of (A) – (D)
4. (5 pts) What is the photon energy in electron volts (eV) for light of wavelength
450 nm?
(A) 2.76
(B) 5.62
(C) 1240.03 (D) 620.00
(E) 3.54
E = hf = hc/ = 1240 eVnm/450 nm = 2.76 eV
5. (5 pts) The deuteron is composed of one proton and one neutron bound together. If the
mass of the proton is taken to be 940 MeV/c2, assume the deuteron’s mass is twice that of
a proton (that is, the mass of the neutron is equal to the mass of the proton). What is the
deuteron’s Compton wavelength (in units of pm)?
(A) 0.0007
(B) 7.0
(C) .07
(D) 659
2
6
Chc/2mpc = 1240 eVnm/[2(940x10 eV)] = 0.0007pm
(E) 3.5
6. (5 pts) A proton and an electron have equal kinetic energies. It follows that the
wavelength of the proton is
(A) greater than the wavelength of the electron,
(B) equal to the wavelength of the electron,
B PHYS 123A
Exam III
June 1, 2016
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(C) less than the wavelength of the electron.
(D) cannot be determined with the information given.
For Questions 7 and 8. An x-ray photon of wavelength 6 pm makes a head on collision
with an electron initially at rest, so that the scattered photon goes in a direction opposite
to that of the incident photon.
7. (5 pts) How much longer (in units of pm) is the wavelength of the scattered photon
than the wavelength of the incident photon?
(A) 2.43
(B) 1.22
(C) 4.86
(D) 10
(E) 93
This is example 34-3 on page 1180 of your text
8. (5 pts) What is the kinetic energy (in units of keV) of the recoiling electron from
conservation of energy?
(A) 2.43
(B) 1.22
(C) 4.86
(D) 93
(E) 10
This is example 34-3 on page 1180 of your text
Question II (15 pts) De Broglie, Photoelectric Effect, Wave Functions
9. (5 pts) Find the De Broglie wavelength (in units of meters) of a 1.00 x 10-6 g particle
moving with a speed of 1.00 x 10-6 m/s.
(B) 6.63x10-19
(C) 6.63x10-7(D) 3.32x10-7 (E) none of A-D
(A) 3.32x10-19
This is example 34-4 on page 1182 of your text.
2 
x
sin  2  of a particle
L  L
confined within a 1-D box of length L. What value of x, other than x = 0 or L, will yield
the probability density P(x) = 0 within the valid range of x?
(A) L/3
(B) L/4
(C) L/2
(D) 2L
(E) L/8
10. (5 pts) Given the n = 2 state wave function  2  x  
P(x) = x)2 = 2sin2[2x/L]/L  for P(x) = 0, 2x/L = m  x =mL/2. Since the well
is only of width L, x cannot be larger than L. Thus m = 1 and x = L/2.
11. (5 pts) The work function for Cadmium is 3.68 eV. Find the threshold frequency (in
units of 1012 Hz) for the photoelectric effect to occur when monochromatic
electromagnetic radiation is incident on the surface of a sample of Cadmium.
(A)0.8
(B) 890
(C) 98
(D) 9800
(E) 0.68
.
B PHYS 123A
Exam III
June 1, 2016
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Question III (25 pts) Lecture Free Response: Show all work and explain in enough
detail so Prof. Buck understands your thinking and methodology! Use back of page
if needed.
This is based on number 8 of HW 7
Relativity
A particle of mass 2.43 MeV/c2 and kinetic energy 12.15 MeV collides with a stationary
particle of mass 12.15 MeV/c2. After the collision, the particles stick together.
(a) Find the speed of the first particle before the collision.
8a seenKey
0.986 c
(b) Find the total energy of the first particle before the collision.
8b seenKey
14.6 MeV
(c) Find the initial total momentum of the system.
8c seenKey
14.4 MeV/c
(d) Find the total kinetic energy after the collision.
8d seenKey
4.2 MeV
(e) Find the mass of the system after the collision.
8e seenKey
22.5 MeV/c2
B PHYS 123A
Exam III
June 1, 2016
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Monochromatic
light source
VII. [20 points total] A photoelectric effect experiment is performed
using monochromatic light that is incident on electrode B, as
shown. Electrodes A and B are made of the same type of metal.
Initially, there is no battery connected in the circuit. It is
observed that there is a constant non-zero current I0 in the circuit
as measured by the ammeter A.
A
Evacuated
tube
B
Electrodes
I0 A
18. [4 pts] Suppose the spacing between the electrodes were
reduced, as shown. The minimum absolute value of the
voltage difference across the electrodes that would cause a
zero current would:
A. increase.
A
B
B. decrease.
A
C. remain the same.
D. There is not enough information to answer.
The voltage described above is the stopping voltage Vstop. Since this is the potential difference
needed to prevent an electron emitted with the maximum kinetic energy KEmax from reaching
electrode A, KEmax = e|Vstop|. Since neither KEmax nor the electron charge e depends on the
distance between the plates, the stopping voltage would remain unchanged.
A battery is added to the original apparatus as shown. (The
positive terminal of the battery is on the left side as viewed in the
diagram.) After the experiment has been running for a long
time, the ammeter reads a constant current of I1.
19. [4 pts] The number of electrons ejected from electrode B
per second would:
A. increase.
A
I1 A
B
Battery
B. decrease.
C. remain the same.
D. There is not enough information to answer.
The model developed in tutorial includes the idea that one photon interacts with one electron. A
photon can be ejected from electrode B if the energy of the photon is greater than the work
function of the metal. The potential difference across the electrodes affects electrons once they
are already ejected, so adding a battery will not affect the number of electrons ejected.
Physics 123A, Spring 2008
Final Exam
WO-UWA123A082T-EF(PEF)mc_sol.doc
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20. [4 pts] The absolute value of the new current, |I1| is:
A. greater than that of the original current, |I0|.
B. less than that of the original current, |I0|.
C. equal to that of the original current, |I0|.
D. There is not enough information to answer.
The electrons that are ejected from B will
be attracted toward electrode A since it is
at a higher potential than electrode B due
to the battery. Thus, more of the ejected
electrons will reach electrode A, and hence
the absolute value of the current will
increase.
The light source is now directed at electrode A, as shown. No
other changes to the experiment are made. After the experiment
has been running for a long time, the ammeter now reads a
constant current of I2.
21. [4 pts] The absolute value of the new current, |I2|, is:
A
A. greater than that of the previous current, |I1|.
B. less than that of the previous current, |I1|.
B
Battery
I2 A
C. equal to that of the previous current, |I1|.
D. There is not enough information to answer.
Since the electrodes are made of the same type of metal, the work functions of both electrodes
are the same; thus, the number of electrons ejected from electrode A is the same. However,
since the electrons must now reach electrode B, which is at a lower potential than electrode A,
fewer of the ejected electrons will reach electrode B because they will be repelled.
The light source is re-directed at electrode B and a (closed)
switch is added to the circuit, as shown. An additional change is
then made to the apparatus. After this additional change, it is
observed that no electrons are ejected from electrode B. (The
diagram at right shows the apparatus before this additional
change is made.)
22. [4 pts] Which of the following changes to the apparatus
could account for the observed change?
A. The intensity of the light is changed by some non-zero
factor (without changing the frequency of the light).
Before additional change
A
A
B
Switch
B. The voltage of the battery and/or the orientation of the battery is changed.
C. The switch is opened.
D. More than one of the above could account for the observed change.
E. None of the above could account for the observed change.
As described in question 20, a photon can be ejected from electrode B if the energy of the
incident photon is greater than the work function. Increasing the intensity of light in the manner
described will not change the energy of each photon. Changing the voltage and/or the
orientation of the battery will not affect the work function (which is a property of the metal.)
Opening the switch will cause the current to drop to zero (due to the open circuit), but electrons
would still be ejected from electrode B. Thus, none of the above could account for the observed
change.
Physics 123A, Spring 2008
Final Exam
WO-UWA123A082T-EF(PEF)mc_sol.doc