PHYS2265 Modern Physics/
PHYS2627 Introductory Quantum Physics
2014/15 Semester 2 Conceptual Questions 2
1. By considering how to enable objects of unequal temperature to reach the same common
temperature, explain why the electromagnetic waves enclosed in a cavity has the same
temperature as that of the cavity walls.
Answer:
We could put objects of unequal temperature in contact so that they could exchange
energy. They will do so until their temperatures are equal. Similarly, the cavity walls
are in contact (being able to exchange energy) with the electromagnetic wave within
the cavity. When they cease to exchange (net) energy, they are necessarily at the same
temperature.
2. What assumptions did Planck make in dealing with the problem of blackbody radiation?
Discuss the consequences of these assumptions.
Answer:
Planck made two assumptions: (1) Energy of radiation oscillator is quantized and (2)
they emit or absorb energy in discrete irreducible packets. Here the “oscillator” refers
to the atoms or molecules that made up the wall of the blackbody cavity. These assumptions contradict the classical ideas that energy is continuously divisible.
3. Consider the following characteristics of the photoelectric effect:
(i)
(ii)
(iii)
(iv)
The generation of photoelectrons.
The existence of a threshold frequency.
The photoelectric current increases with increasing light intensity.
The photoelectric current is independent of the anode-cathode potential difference
∆V for ∆V > 0.
(v) The photoelectric current decreases slowly as ∆V becomes much more negative.
(vi) The stopping potential is independent of the light intensity.
(vii) The photoelectric current appears almost instantly when the light is turned on.
Which of these cannot be explained by classical physics?
Answer:
(ii), (vi) and (vii) cannot be explained by classical physics.
4. Below figure shows a typical current-versus-potential difference graph for a photoelectric
effect experiment.
I
V
Vstop
0
On the figure, draw and label curves for the following situations:
(i) The light intensity (i. e. power per area) is increased.
(ii) The light frequency is increased.
(iii) The cathode work function is increased.
It is assumed that no other parameters of the experiment are changed in each case.
Answer:
I
(i)
(ii)
(iii)
Vstop
V
0
(i) Maximum current ↑
(ii) |Vstop | ↑ and maximum current ↓ (∵ number density of photon decreases if the
frequency of light increases with fixed intensity)
(iii) |Vstop | ↓
5. Suppose we produce X-rays not by smashing electrons into targets but by smashing
protons, which are far more massive (but each with the same magnitude of charge as an
electron). If the same accelerating potential difference were used for both, how would
the cutoff wavelengths of the two X-ray spectra compare? Explain your answer.
Answer:
No difference. The cutoff wavelength corresponds to all the kinetic energy of the incoming particle going to one photon. If the same accelerating potential difference is used,
the protons would have the same maximum kinetic energy as the electrons.
2
6. Why is it important to produce X-ray tubes with high accelerating voltages that are
also able to withstand high electric currents?
Answer:
A higher electric current means more number of electrons hitting the metal target in
the X-ray tube per unit time. So it is proportional to the rate of X rays produced in
the tube since a X-ray photon is produced by the collision of an electron with an atom
in the target. Hence, the X-ray tubes with high accelerating voltages must be also able
to withstand high electric currents for the production of X rays beam with high enough
intensity.
7. Must Compton scattering take place only between X-rays and free electrons? Can
radiation in the visible region (say, a green light) undergo Compton scattering with a
free electron?
Answer:
The wavelength of visible light (e.g. λgreen ∼ 10−6 m) is much larger than the Compton
wavelength of the electron (λe = h/me c ∼ 10−12 m). For visible light, the Compton
shift is negligibly small compared with its wavelength. So the change in wavelength of
the visible light due to the Compton scattering would be too small to be measured.
8. Why is the Compton effect unimportant in the transmission of television and radio
waves?
Answer:
Recall that the maximum wavelength shift of a photon in a Compton scattering is
2h/me c = 0.004852 nm. It is very much smaller than the wavelength of television and
radio waves (∼1 m to 100 m). These waves have to undergo numerous times of Compton
scattering for its wavelength to be shifted by a significant amount, which is unlikely to
occur. Thus Compton effect is unimportant in the transmission of television and radio
waves.
9. An isolated atom can emit a photon and the atom’s internal energy drops. This process
is called spontaneous emission. Can an isolated electron emit a photon? Why or why
not?
Answer:
No. This is the reverse of a completely inelastic collision. Kinetic energy would have
to increase, meaning that internal energy would have to decrease. The photon has
zero internal energy, and the electron’s internal energy cannot be changed as it is a
fundamental particle.
3
10. A beam of photons passes through a block of matter. What are the three ways discussed
in this chapter that the photons can lose energy in interacting with the material?
Answer:
Photoelectric effect, Compton scattering, and pair production.
11. You have a monoenergetic source of X-rays of energy 84 keV, but for an experiment you
need 70 keV X-rays. How would you convert the X-ray energy from 84 to 70 keV?
Answer:
To convert the X-ray energy from 84 to 70 keV, the X-ray photons need to lose 14
keV energy. The Compton scattering is one of the chief means by which X-rays lose
energy when they pass through matter. Moreover, a scattered X-ray photon typically
lose energy in the range of few eVs to 1MeV. So we can convert the X-ray energy by
passing the X-ray beam on a block of matter. Then we adjust the scattering angle until
the energy of the scattered X-ray photons have lost the desired amount of energy.
12. The intensity of a beam of light is increased but the frequency of the light is unchanged.
As a result,
(i)
(ii)
(iii)
(iv)
the photons travel faster.
each photon has more energy.
the photons are larger.
there are more photons per second.
Which of these (perhaps more than one) are true? Explain your answer.
Answer:
Only (iv) is true. A more intense light delivers more light quanta to the surface and
ejects more photoelectrons per second in the photoelectric effect. Here, similarly, a
greater intensity means more photons per second.
4
PHYS2265 Modern Physics/
PHYS2627 Introductory Quantum Physics
2014/15 Semester 2 Worked Example 2
1. (a) Verify that E = E0 ei(kx−ωt) is a solution to the wave equation
∂ 2E
1 ∂ 2E
=
∂x2
c2 ∂t2
provided that ω = kc. (Hint: Use the Euler formula eiθ = cos θ + i sin θ.)
(b) The wave equation in a medium such as glass with index of refraction n is given by
∂ 2E
n2 ∂ 2 E
=
∂x2
c2 ∂t2
What must be the relationship between the wavelength λ and frequency f so that
that E = E0 ei(kx−ωt) is a solution of this wave equation?
Solution:
(a) Note that
d
d iθ
e = (cos θ + i sin θ) = − sin θ + i cos θ = ieiθ
dθ
dθ
i(kx−ωt)
Plugging E = E0 e
, we have
∂ 2E
d
d ikx dkx
−iωt
= E0 e
(e ) ·
∂x2
dx dkx
dx
d
= E0 e−iωt
(ikeikx )
dx
d ikx dkx
−iωt
= ikE0 e
(e ) ·
dkx
dx
−iωt
ikx
= ikE0 e
(ike )
2
= −k E
Similarly, we can show that
ω2
1 ∂ 2E
=
−
E
c2 ∂t2
c2
Hence,
∂ 2E
1 ∂ 2E
=
∂x2
c2 ∂t2
ω2
⇒ −k 2 E = − 2 E
c
⇒ ω = kc (∵ k, c, ω > 0)
(b) For E = E0 ei(kx−ωt) to be the solution, it requires
∂ 2E
n2 ∂ 2 E
=
∂x2
c2 ∂t2
n2 ω 2
⇒ −k 2 E = − 2 E
c
⇒ nω = kc (∵ k, c, ω > 0)
2πc
⇒ 2πf n =
λ
c
⇒ f=
nλ
2. The average intensity of an electromagnetic wave in vacuum is:
1
2
I = c0 Em
2
where Em is the amplitude of the electric field and 0 = 8.85 × 10−12 C2 /N · m2 is the
permittivity of free space. The electric field of an EM wave in vacuum has a peak value
of 0.0218 V/m. What is the average power at which this wave carries across unit area?
Solution:
The peak value of the electric field of an EM wave is equal to the amplitude Em of the
field. The average power at which this wave carries energy across unit area is given by
its average intensity:
1
2
I = c0 Em
2
1
= (3.00×108 m/s)(8.85×10−12 C2 /N·m2 )(0.0218 V/m)2
2
= 6.31×10−7 W/m2
3. Estimate the surface temperature of the Sun using the following information. The Sun’s
radius is Rs = 7.0 × 108 m. The average Earth-Sun distance is d = 1.5 × 1011 m. The
power per unit area (at all frequencies) from the Sun is measured at the Earth to be
Id = 1400 W/m2 . Take the approximation that the Sun is a blackbody.
Solution:
Approximated as a blackbody, the intensity radiated by the Sun with surface temperature Ts is given by the Stefan-Boltzmann law:
Is = σTs4
which is the power emitted by the Sun across unit area of its surface. Instead of this
quantity, the power from the Sun across unit area of the Earth’s surface — the intensity
Id — is given in this problem. So we need the connection between Is and Id to find the
surface temperature of the Sun Ts . This comes from the conservation of energy:
2
Is · 4πRs2 = Id · 4πd2
d2
⇒ Is = Id 2
Rs
Using the Stefan-Boltzmann law again, we obtain the surface temperature
)1/4
1/4 (
Id d2
(1400 W/m2 )(1.5×1011 m)2
Ts =
=
= 5802 K
σRs2
[5.671×10−8 W/(m2 ·K4 )](7.0×108 m)2
4. (a) Show that the Planck formula can be expressed in terms of the wavelength λ as:
dU 8πV hc
1
=
dλ (ehc/λkB T − 1) · λ5
(b) Show that the Wien’s displacement law can be derived from the spectral energy
density dU/dλ in part (a).
Solution:
(a) From the relation c = f λ, we find that
c
λ2
dλ
=− 2 =−
df
f
c
Substituting it into the Planck formula gives:
dU dλ
hc/λ
8πV c 2
·
·
= hc/λk T
B
dλ df
(e
− 1) c3
λ
2
dU
λ
8πV h
1
⇒
· −
= hc/λk T
· 3
B
dλ
c
(e
− 1) λ
dU 8πV hc
1
⇒ = hc/λk T
· 5
B
dλ
(e
− 1) λ
(b) To find the wavelength λmax for which the spectral energy density dU/dλ in (a)
attains maximum, we consider the condition
d dU =0
dλ dλ 1
1
1
5
hc ehc/λkB T
· hc/λk T
· 5 − hc/λk T
· 6 =0
⇒
2
2
B
B
λ kB T
(e
− 1) λ
(e
− 1) λ
hc
⇒
− 5 + 5e−hc/λkB T = 0
λkB T
Note that the solution of this equation must be the absolute maxima of dU/dλ since
this function has a concave shape. Then we can find the value of λmax is given by
λmax T = 2.897×10−3 m·K
which is the Wien’s displacement law.
3
5. An object of mass m = 2 kg is attached to a “massless” spring of spring constant
k = 25 N/m. The spring is stretched A = 0.40 m from its equilibrium position and
released.
(a) Find the total energy and frequency of oscillation according to classical calculations.
(b) Assume that the energy of the system is quantized. Find the quantum number, n,
for this system.
(c) How much energy would be carried away by 1-quantum change?
Solution:
(a) The total energy of a simple harmonic oscillator with amplitude A is
1
1
E = kA2 = (25 N/m)(0.40 m)2 = 2.0 J.
2
2
The frequency of oscillation will be
s
r
1
k
1 25 N/m
f=
=
= 0.563 Hz.
2π m
2π
2 kg
(b) If the energy E of the system is quantized, then the quantum number n is given by
E = nhf
⇒ 2.0 J = n(6.626×10−34 J · s)(0.563 Hz)
⇒ n = 5.36×1033
(c) The energy carried away by one quantum change in energy would be
E = hf = (6.626×10−34 J · s)(0.563 Hz) = 3.73×10−34 J.
6. Consider a potassium surface that is at a distance d = 75 cm away from a bulb of power
P = 100 W . Suppose that the energy radiated by the bulb is 5% of the input power.
Treating each potassium atom as a circular disk of radius r = 0.05 nm, determine the
time required for an atom to absorb an amount of energy equal to its work function of
φ = 2.0 eV (1 eV = 1.602 × 10−19 J), according to the wave picture of light.
Solution:
Treating the bulb as a point source, the power Pb (= 0.05P ) radiated by the bulb can be
regarded as spreading on a sphere of area Ad = 4πd2 . So the intensity of light incident
on the potassium surface is equal to
I=
(0.05×100 W)
Pb
=
= 0.707 W/m2 .
2
Ad
4π(0.75 m)
Thus the power incident on each potassium atom of area AK = πr2 is
PK = IAK = (0.707 W/m2 )[π(5×10−11 m)2 ] = 5.55×10−21 W
4
By the wave picture of light, the time interval to absorb energy of E = 2.0 eV equals
∆t =
(2.0 eV)(1.602×10−19 J/eV)
E
=
= 57.7 s.
PK
(5.55×10−21 W)
In the calculation, it has been assumed that all the incident energy has been absorbed.
Since, with a wave picture, much of the incident energy would be reflected, the actual
time would be in excess of 57.6 s. Thus a wave picture of electromagnetic radiation
predicts an emission time many orders of magnitude larger than the experimentally
observed time of less than 10−9 s.
7. When light of wavelength λ1 = 450 nm is incident on potassium, photoelectrons with
stopping potential Vs,1 = 0.52 V are emitted. If the wavelength is changed to λ2 =
300 nm, the stopping potential is Vs,2 = 1.90 V. Using only these numbers together with
the values of the speed of light c and the electron charge e,
(a) find the work function Φ of potassium and
(b) compute a value of Planck’s constant h.
Solution:
(a) According to the given information, we have
eVs,1 =
hc
− Φ,
λ1
eVs,2 =
hc
− Φ.
λ2
Combining these equations gives:
e(Vs,2 λ2 − Vs,1 λ1 )
(λ1 − λ2 )
(1.602×10−19 C)[(1.90 V)(300×10−9 nm) − (0.52 V)(450×10−9 nm)
=
(450×10−9 nm − 300×10−9 nm)
= 3.59×10−19 J
Φ=
(b) Plugging back the value of the work function Φ obtained in (a), we have
(eVs,1 + Φ)λ1
c
[(1.602×10−19 C)(0.52 V) + 3.59×10−19 J](450×10−9 nm)
=
(2.998×108 m/s)
= 6.64×10−34 J · s
h=
8. An electron with kinetic energy of K = 20 keV is brought to rest in a collision with a
heavy nucleus.
(a) Calculate the frequency f of the photon produced by the collision. Assume the
energy transfers to the heavy nucleus is negligible.
5
(b) Show that momentum is not conserved in such collision.
Solution:
Let Eγ and pγ be the energy and momentum of the photon produced, and Ee and Ee0
be the initial and final total energy of the electron.
(a) Applying the conservation of energy, we have
Ee = Eγ + Ee0
⇒ K + me c2 = hf + me c2
⇒ K = hf
3
⇒ 20×10 eV = (4.136×10−15 eV · s)f
⇒ f = 4.84×1018 Hz
(b) The initial momentum pfinal of the system (nucleus+electron+photon) is just the
momentum of the electron pe which can be found from
p
Ee = K + me c2 = (pe c)2 + (me c2 )2
⇒ (0.02 MeV + 0.511 MeV)2 = (pe,i c)2 + (0.511 MeV)2
⇒ pinitial = pe = 0.144 MeV/c
But the final momentum pfinal of this system is the momentum of the photon (∵
the electron comes to rest after the collision) which is given by
pfinal = pγ = hf /c = K/c = 0.02 MeV/c.
The excess momentum is absorbed by the nucleus that stops the electron. As the
nucleus is much heavier than the electron, its change in energy is negligible.
9. An electron accelerated to kinetic energy E = 50 keV in an X-ray tube has two successive
collisions in being brought to rest in the target, emitting two bremsstrahlung photons
in the process. The second photon has wavelength 0.095 nm longer than the first one.
(a) What are the wavelengths of the two photons?
(b) What is the kinetic energy of the electron after the emission of the first photon?
Solution:
(a) Let ∆K1 and ∆K2 be the change in kinetic energy of the electron due to the emission
of the first and second photons, which have wavelengths λ1 and λ2 , respectively. It
is given that:
λ2 = λ1 + 0.095 nm,
∆K1 + ∆K2 = 50×103 eV.
Applying the conservation of energy in the two collisions, we have
∆K1 =
hc
,
λ1
6
∆K2 =
hc
.
λ2
Combining the above equation gives
hc(λ1 + λ2 )
λ1 λ2
(1240
eV
·
nm)(2λ
1 + 0.095 nm)
⇒ 50×103 eV =
λ1 (λ1 + 0.095 nm)
2
⇒ 500000λ1 + 2270λ1 − 117.8 = 0
⇒ λ1 = 0.0309 nm or λ1 = −0.0763 nm (rejected as λ1 > 0)
∆K1 + ∆K2 =
Then the wavelength of the photon emitted in the second collision is
λ2 = λ1 + 0.095 nm = 0.0309 nm + 0.095 nm = 0.1259 nm
(b) The electron comes to rest after the second collision. So its kinetic energy after the
emission of the first photon is given by
∆K2 =
hc
1240 eV · nm
= 9.85 keV
=
λ2
0.1259 nm
10. A 0.3 MeV X-ray photon makes a head-on collision with an electron initially at rest.
(a) Using conservation of energy and momentum, find the recoil velocity of the electron
u in terms of the speed of light c.
(b) Check the velocity obtained in (a) agrees with the value determined from the Compton’s formula.
Solution:
(a) Using conservation of energy, with E and E 0 denoting the energy of the original
and scattered photon, we have
me c2
E + me c2 = E 0 + p
1 − u2 /c2
0.511 MeV
⇒ 0.3 MeV + 0.511 MeV = E 0 + p
1 − u2 /c2
The photon’s momentum is pγ = hf /c = E/c. So conservation of momentum yields
E
E0
me u
+0=− + p
c
c
1 − u2 /c2
0.3 MeV
E0
0.511 MeV u
⇒
=− +p
·
c
c
1 − u2 /c2 c2
Simultaneous solution of the energy and momentum equation gives u = 0.651c.
7
(b) For head-on collision, the scattering angle θ = 180◦ . According to the Compton’s
formula,
2h
h
(1 − cos 180◦ ) =
me c
me c
hc
Eme c2
hc
0
⇒ E = 0 =
=
λ
λ + 2h/(me c)
(2E + me c2 )
λ0 − λ =
Rearranging the conservation of energy formula found in (a), we obtain
s
s
u
m2e c4
m2e c4 (2E + me c2 )2
= 1−
1
−
=
c
[(E + me c2 ) − E 0 ]2
[(E + me c2 )(2E + me c2 ) − Eme c2 ]2
Putting in the known energies (all in units of MeV) yields:
s
u
(0.511)2 [2(0.3) + 0.511]2
= 1−
= 0.651
c
{(0.3 + 0.511)[2(0.3) + 0.511] − (0.3)(0.511)}2
which agrees with the velocity obtained in (a).
11. An X-ray source of unknown wavelength is directed at a carbon sample. An electron is
scattered with a speed of u = 4.5×107 m/s at an angle of θ = 60◦ relative to the motion
of the original X-ray photon. Determine the wavelength λ of the X-ray source.
Solution:
Let the scattered X-ray photon has wavelength λ0 moving in the direction at angle φ
relative to the motion of the original X-ray photon.
During the scattering, the lost in the energy of the X-ray photon is converted to the
kinetic energy of the scattered electron K. So we have
!
hc hc
1
− 1 me c2 =
− 0
K= p
2
2
λ
λ
1 − u /c
"
!
#2
h
1
h2
=
− 1 me c
⇒
− p
02
λ
λ
1 − u2 /c2
!
!2
h2 2h
1
1
= 2−
· p
− 1 me c + p
− 1 m2e c2
2
2
2
2
λ
λ
1 − u /c
1 − u /c
In addition, applying the conservation of momentum for the directions parallel and
perpendicular to the motion of the original X-ray photon yields
h
me u
h
=p
cos θ + 0 cos φ,
λ
λ
1 − u2 /c2
8
me u
h
0= p
sin θ − 0 sin φ.
λ
1 − u2 /c2
Combining the above equations, we obtain
!2
!2
me u
h
me u
−p
cos θ + p
sin θ
λ
1 − u2 /c2
1 − u2 /c2
!
!2
1
1
p
− 1 me c + p
− 1 m2e c2
1 − u2 /c2
1 − u2 /c2
h2
=
λ02
⇒
h2 2h
·
−
λ2
λ
me u
m2e u2
h2 2h
p
·
cos
θ
+
−
λ2
λ
(1 − u2 /c2 )
1 − u2 /c2
"
#
(u/c) cos θ
h
p
⇒ λ=
−1
me c (1 − 1 − u2 /c2 )
=
Putting h/me c = 0.002426 nm, we find that
"
#
h
(u/c) cos θ
p
−1
λ=
me c (1 − 1 − u2 /c2 )
(
)
[(4.5×107 m/s)/(2.998×108 m/s)] cos 60◦
−12
p
= (2.426×10 m)
−1
[1 − 1 − [(4.5×107 m/s)2 /(2.998×108 m/s)]2 ]
= 1.36×10−11 m
12. A gamma-ray photon changes into a proton-antiproton pair. After creation, each of the
pair moves off at 0.6c perpendicular to the motion of the photon.
(a) Ignore momentum conservation, find the wavelength λ of the photon.
(b) Assume that this interaction occurs as the photon encounters a lead plate and that a
lead nucleus participates in momentum conservation. What fraction of the photon’s
energy must be absorbed by the lead nucleus?
Solution:
(a) Both the proton and antiproton (each with mass mp = 1.673×10−27 kg) move at
speed v = 0.6c after creation. Applying the conservation of energy gives:
hc
2mp c2
=p
λ
1 − v 2 /c2
⇒
6.626×10−34 J · s
2(1.673×10−27 kg)(2.998×108 m/s)
p
=
λ
1 − (0.6c2 )/c2
⇒ λ = 5.28×10−16 m
(b) The mass of a lead nucleus is
mPb = 207u × 1.661×10−27 kg/u = 3.44×10−25 kg.
9
Since the total momentum of the proton-antiproton pair is zero, the momentum
of the photon becomes the momentum of the lead nucleus. If uPb is the resultant
speed of the lead nucleus, we have
mPb vPb
h
=p
2
λ
1 − vPb
/c2
hc
⇒ vPb = p
(mPb cλ)2 + h2
(6.626×10−34 J)(2.998×108 m/s)
=p
[(3.44×10−25 kg)(2.998×108 m/s)(5.28×10−16 m)]2 + (6.626×10−34 J)2
= 3.65×106 m/s
So this lead nucleus has kinetic energy
mPb c2
− mPb c2
K=p
2
2
1 − vPb /c
(3.44×10−25 kg)(2.998×108 m/s)2
=p
− (3.44×10−25 kg)(2.998×108 m/s)2
1 − (3.65×106 m/s)2 /(2.998×108 m/s)2
= 2.29×10−12 J
(In fact, since vPb c, we will get the same results here if we use the classical
formula to find the speed and kinetic energy of the lead nucleus.)
As the energy of the photon is equal to
Eγ =
hc
(6.626×10−34 J · s)(2.998×108 m/s)
= 3.76×10−10 J.
=
−16
λ
(5.28×10 m)
Thus lead absorbs
K
2.29×10−12 J
=
= 0.609%.
Eγ
3.76×10−10 J
of the photon’s energy.
13. An electron with kinetic energy of Ke = 5 MeV undergoes annihilation with a positron
that is at rest, producing two photons. One of the photons travels in the direction of
the incident electron. Calculate the energy of each photon.
Solution:
Let E1 and E2 be the energy of the first and second photon and pe be the momentum
of the electron.
The second photon must travel parallel (denoted as = +1) or anti-parallel (denoted as
= −1) to the first photon so that the momentum would be conserved in the transverse
direction. By conservation of momentum,
E2
E1
+
c
c
⇒ E1 + E2 = pe c
pe =
10
Substituting for pe c from
K E + m e c2 =
p
(pe c)2 + (me c2 )2 ,
we obtain
p
(KE + me c2 )2 − (me c2 )2
p
= (5 MeV + 0.511 MeV)2 − (0.511 MeV)2 = 5.49 MeV
E1 + E2 =
Conservation of energy requires
E1 + E2 = KE + me c2 + me c2 = 5 MeV + 2(0.511 MeV) = 6.02 MeV
Substituting for E1 in the momentum equation gives
−0.53 MeV = ( − 1)E2 .
Therefore must be taken equal to −1, so that the second photon travels in the opposite
direction from the first. The energies of the photons are then found to be
E1 = 5.76 MeV, E2 = 0.27 MeV.
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