Physics 5 Midterm Exam, May 6, 2014
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Exam form A
1. (2 points) When a photon Compton scatters off of an electron that is at rest, is the frequency of the
outgoing photon larger, smaller, or the same as the frequency of the incoming photon?
The photon must lose energy since the electron is initially at rest. Less energy means smaller
frequency.
2. (8 points) Light acts sometimes like a wave and sometimes like a particle. For the following
situations, which of the two properties (wave or particle) best describes the behavior of light? (a)
The photoelectric effect. (b) Compton scattering. (c) Young’s double-slit experiment. (d) Atomic
spectra within the Bohr model.
(a) Particle. The photoelectric effect relies on the fact that light’s energy comes in quanta.
(b) Particle. Compton scattering corresponds to light scattering off electrons where the light behaves as a photon.
(c) Wave. The double-slit experiment probes the wave like interference of light.
(d) Particle. The quantized energy levels lead to emission and absorption of photons with only
specific energies.
3. (2 points) An electron is accelerated from rest through an electric-potential difference. Will increasing the difference in potential result in a longer or shorter de Broglie wavelength for the
accelerated electron?
A larger potential difference will give the electron more kinetic energy, and so more momentum.
Since the de Broglie wavelength is λ = h/p, that will correspond to a smaller wavelength.
4. (8 points) Two protons (with mass mp ) are initially moving with equal speeds in opposite directions. They continue to exist after a head-on collision that also produces another particle called a
neutral pion (of mass mπ ). If the protons and the pion are at rest after the collision, find the initial
speed of the protons in terms of mp and mπ .
We can solve this with conservation of energy. The initial energy of each of the protons is E =
γmc2 . After the collision, everything is at rest so the only energy is from the masses. Equating
initial and final energy we get.
2γmp c2 = 2mp c2 + mπ c2
So,
p
γ = (2mp + mπ )/2mp = 1/ 1 − v 2 /c2
Solving for v we get
v=c
q
1 − 4m2p /(2mp + mπ )2
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5. (15 points) A train of proper length L passes through a tunnel of proper length 2L. The train’s
speed, v, in the tunnel’s rest frame is such that γ = 2.5. Use a coordinate system in the reference
frame of the train that has its origin at the front of the train, and another coordinate system in the
reference frame of the tunnel that has its origin at the entrance of the tunnel, where the clocks of
the two frames measure zero when their origins overlap. Determine the time of two events, as
measured in each coordinate system, where event 1 is when the front of the train reaches the exit
of the tunnel, and event 2 is when the back of the train reaches the entrance of the tunnel. Express
these in terms of L, γ, and v.
We will call the tunnel’s frame S and the train’s frame S 0 and use the Lorentz transformations
between them.
The coordinates, (x1 , t1 ), of event 1 in S are at (2L, 2L/v), since speed is known in that frame.
(Note that we sometimes call this u, rather than v. The difference is just labelling, which you
should be able to see beyond.) In the S 0 frame, Event 1 happens at the front of the train, which is
the origin, so its coordinates are (0, t01 ). We can find t01 from the Lorentz transformation,
t01 = γ(t1 − ux1 /c2 ) = γ(2L/v − v2L/c2 ) =
2Lγ
(1 − v 2 /c2 ) = 2L/vγ
v
For Event 2, the coordinates in S are (x2 , t2 ) = (0, t2 ), where we can’t directly determine t2 . The
observer in the train, however, can easily determine the time because he just sees the entrance of
the tunnel move from the front of the train to the back at a speed of −v. So, Event 2 in S 0 , the
coordinates are (x02 , t02 ) = (−L, L/v). We can then find t2 from the Lorentz transformation.
t2 = γ(t02 + ux02 /c2 ) = γ(L/v − vL/c2 ) =
γL
L
(1 − v 2 /c2 ) =
v
vγ
This is very similar to the example problem done in class; only the numbers have been changed to
check for understanding.
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6. (5 points) In the figure below, three ships move relative to each other at speeds that are measured in
the reference frame of an observer on earth. When the ships are arranged as shown, ship A sends
a signal by emitting a pulse of light, with wavelength of 300 nm, toward ship B.
The three ships each measure the speed of the pulse and get vA , vB , vC , respectively. Rank these
speeds using <, =, or > signs.
They all measure the same speed, c, by the 2nd postulate of special relativity.
7. (5 points) If a proton and an electron have the same kinetic energy, which has the longer de Broglie
wavelength?
√
The kinetic energy is p2 /2m, so p = 2mK. The lower mass electron has a lower momentum
than the proton for the same kinetic energy, so it has a longer wavelength, λ = h/p.
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8. (10 points) An unknown element has a spectrum for absorption from its ground level with lines at
2.0 eV, 3.0 eV, and 8.0 eV. Its ionization energy is 9.0 eV.
(a) What would happen if infrared light, with wavelength of 2500 nm, shines on such an atom in
its ground state?
The energy of such a photon is E = hc/λ = 8×10−20 J = 0.5 eV. That is below the smallest
absorption energy (2.0 eV), so the infrared light does not interact with the atom at all.
(b) If an 8.0 eV photon is absorbed by such an atom that is in its ground state, what energies
could subsequently emitted photons have?
The atom would be excited into the 8.0 eV state. From there it could decay directly back down to
the ground state by emitting an 8.0 eV photon. It could also decay from the 8.0 to the 3.0 eV state
by giving off a 5.0 eV photon. From there it could decay to the ground state by giving off a 3.0
eV photon, or it could cascade through the 2.0 eV state by giving off a 1.0 eV and then a 2.0 eV
photon. There is a similar cascade from 8 to 2 to ground.
The full list of photon energies is then: 8, 6, 5, 3, 2, 1 eV. (Different exam forms had different
atoms, your specific photon list could differ.)
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9. (15 points) A particle with energy E is confined in a one-dimensional potential well that has potential energy U (x) = 0 for 0 < x < L and U (x) = ∞ otherwise. A solution to Schrodinger’s
equation for this situation is ψ(x) = Aeikx +Be−ikx for 0 < x < L and ψ(x) = 0 otherwise. Using
that solution, determine the energy levels of the particle in terms of L and physical constants. You
must show and explain all your work to get credit.
To plug ψ(x) into Schrodinger’s equation, we find the first and second derivative of ψ:
dψ/dx = ikAeikx − ikBe−ikx
d2 ψ/dx2 = −k 2 Aeikx − k 2 Be−ikx = −k 2 ψ(x)
Plugging this in Schrodinger’s equation gives:
~2 k 2
ψ = Eψ
2m
2 2
k
for 0 < x < L. So E = ~2m
. We can find k from the boundary condition that ψ(x) is continuous
at x = 0 and x = L. The x = 0 condition gives
ψ(0) = 0 = Aeik0 + Be−ik0 = A + B
So B = −A. Then applying the boundary condition at x = L we get
ψ(L) = 0 = A(eikL − e−ikL ) = A(cos kL + i sin kL − cos kL + i sin kL) = 2iA sin kL,
where I wrote out the cosines and sines within eikL . The ways to make sin kL = 0 is to have
k = nπ/L for any non-zero integer n. Plugging that into the energy obtained above, we get
E=
6
~2 n2 π 2
2mL2
10. (10 points) Suppose that a one-dimensional system can be described by a potential energy, in eV,
that is U (x) = ∞ for x ≤ −a, U (x) = −4 for −a < x < 0, U (x) = +1 for 0 ≤ x ≤ a,
and U (x) then rises linearly to infinity beyond x = a. Draw an appropriate wavefunction, which
would satisfy Schrodinger’s equation, for a particle with E = 2 eV confined in this potential well.
You need not calculate an algebraic solution. Rather, show graphically and explain in words the
important characteristics that the wavefunction should have.
The wavefunction needs to satisfy the following constraints; be zero where U = ∞; be continuous;
be a wave in the regions where E > U ; and be an exponential
where E < U < ∞.
p
√decay in regions
The wavelength of the waves is set by λ = h/p where p = 2mK = 2m(E − U ).
So ψ(x) = 0 for x < −a and then ψ(−a) = 0 to be continuous. For −a < x < 0, there is
a lot of kinetic energy so the wavelength is short. For 0 < x < a, there is less kinetic energy so the
wavelength is longer. It gets even longer for x > a and becomes an exponential when we reach
the point where U = E.
The drawing below satisfies these requirements. It is not the only option though.
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11. Suppose that a one-dimensional system can be described by a potential energy, in eV, that is
U (x) = x2 − 5 for x ≤ 0, U (x) = −5 for 0 < x < L, U (x) = +1 for L ≤ x < 2L, and
U (x) = ∞ for x ≥ 2L. Draw an appropriate wavefunction, which would satisfy Schrodinger’s
equation, for a particle with E = 2 eV confined in this potential well. You need not calculate
an algebraic solution. Rather, show graphically and explain in words the important characteristics
that the wavefunction should have.
The wavefunction needs to satisfy the following constraints; be zero where U = ∞; be continuous;
be a wave in the regions where E > U ; and be an exponential
where E < U < ∞.
p
√decay in regions
The wavelength of the waves is set by λ = h/p where p = 2mK = 2m(E − U ).
So ψ(x) = 0 for x > 2L and then ψ(2L) = 0 to be continuous. For L < x < 2L, there is
little kinetic energy so the wavelength is long. For 0 < x < L, there is more kinetic energy so
the wavelength is shorter. It gets longer again for x < 0 as the potential increases and becomes an
exponential when we reach the point where U = E.
The drawing below satisfies these requirements. It is not the only option though.
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