Homework Set 12: Solutions
Due: Friday, November 19, 2010
Chapter 27: Questions
(2) 6 points
If energy is radiated by all objects, why can we not see them in the dark?
The difficulty with seeing objects in the dark is that although all objects emit radiation, only a small
portion of the electromagnetic spectrum can be detected by our eyes. Usually objects are so cool that
they only give off very long wavelengths of light (infrared), which are too long for our eyes to detect.
(7) 6 points
UV light causes sunburn, whereas visible light does not. Suggest a reason.
UV light causes sunburns and visible light does not due to the higher frequency and energy carried by
the UV photons. Our skin is sensitive to these frequencies and absorbs the energy readily, which can
ionize atoms/molecules and cause damage to cells.
(14) 8 points
(20) 8 points
What is the difference between a photon and an electron? Be specific: make a list.
Property
Photon
Electron
Mass
None
9.11 × 10−31 kg
Charge
None
−1.60 × 10−19 C
Speed
3 × 108 m/s
< 3 × 108 m/s
Is it possible for the de Broglie wavelength of a “particle” to be greater than the dimensions
of the particle? To be smaller? Is there any direct connection?
The de Broglie wavelength of a “particle’ is λ = h/p. Since this expression does not depend upon the
dimensions of the particle, we conclude that there is no direct connection between the two.
(28) 8 points
Suppose we obtain an emission spectrum for hydrogen at very higher temperature (when
some of the atoms are in excited states), and an absorption spectrum at room temperature,
when all atoms are in the ground state. Will the two spectra contain identical lines?
The spectral lines of hydrogen found at room temperature will be identical to some of the lines found
at high temperature. However, there will be more lines at higher temperature. These lines correspond
to the electrons making transitions to the higher energy excited states that are only accessible at higher
temperature.
Chapter 27: Problems
(12) 8 points
(I) A typical gamma ray emitted from a nucleus during radioactive decay may have an
energy of 300 keV. What is its wavelength? Would we expect significant diffraction of this
type of light when it passes through an everyday opening, like a door?
The wavelength of this ray is
λ=
c
hc
(6.63 × 10−34 J · s)(3.00 × 108 m/s)
=
=
= 4.14 × 10−3 nm.
f
E
(1.60 × 10−19 J/eV)(300 × 103 eV)
Significant diffraction occurs when the opening is on the order of the wavelength. Thus there would not
be significant diffraction through the doorway.
(24) 14 points
(II) The threshold wavelength for emission of electrons from a given surface is 350 nm.
What will be the maximum kinetic energy of ejected electrons when the wavelength is (a)
280 nm, (b) 360 nm?
(a) The cutoff frequency f0 occurs when hf0 = ϕ. Solving for ϕ, we find that
ϕ = hf0 =
hc
(6.63 × 10−34 J · s)(3.00 × 108 m/s)
=
= 5.68 × 10−19 J = 3.55 eV.
λ0
350 × 10−9 m
Therefore, using hf = Kmax + ϕ, we find that the maximum kinetic energy of ejected electrons is
Kmax
=
=
hc
(6.63 × 10−34 J · s)(3.00 × 108 m/s)
−ϕ=
− 5.68 × 10−19 J
λ
280 × 10−9 m
1.42 × 10−19 J = 0.89 eV.
(b) Because this frequency is below the cutoff frequency, no electrons will be ejected.
(38) 8 points
(I) What is the wavelength of a neutron (m = 1.67 × 10−27 kg) traveling at 6.5 × 104 m/s?
The de Broglie wavelength of the neutron is given by λ = h/p. The neutron’s speed is non-relativistic,
so p = mv. Therefore
λ=
(44) 14 points
6.63 × 10−34 J · s
h
=
= 6.11 × 10−12 m.
mv
(1.67 × 10−27 kg)(6.5 × 104 m/s)
(II) Calculate the de Broglie wavelength of an electron in your TV picture tube if it is
accelerated by 30,000 V. Is it relativistic? How does its wavelength compare to the size of
the “neck” in the rub, typically 5 cm? Do we have to worry about diffraction problems
blurring our picture on the screen?
Using E0 = mc2 , rest energy of an electron is approximately 0.511 MeV. From conservation of energy,
K = e∆V = 30.0 × 103 eV = 4.8 × 10−15 J.
Since this is 6% of m0 c2 , the electron is relativistic. Using K = (γ − 1)me c2 and solving for γ we find
that
K
30.0 × 103 eV
γ=
+
1
=
+ 1 = 1.059,
me c 2
0.511 × 106 eV
which corresponds to a speed of v = 0.328c. Therefore, the de Broglie wavelength of an electron in your
TV picture tube is
λ=
h
6.63 × 10−34 J · s
h
=
=
= 6.98 × 10−12 m.
p
γme v
(1.059)(9.11 × 10−31 kg)(9.85 × 107 m/s)
As a result, diffraction effects are negligible.
(48) 10 points
(I) For the three hydrogen transitions indicated below, with n being the initial state and n0
being the final state, is the transition an absorption or an emission? Which is higher, the
initial state energy of the final state energy of the atom? Finally, which of these transitions
involves the largest energy photon? (a) n = 1, n0 = 3 (b) n = 6, n0 = 2 (c) n = 4, n0 = 5.
Transitions (a) and (c) involve an electron moving to a higher energy state (absorption). As a result, the
state energy increases.
Transition (b), on the other hand, involves an electron moving to a lower energy state (emission). As a
result, the state energy decreases.
Using En = −13.6/n2 eV, we find that transition (a) involves the largest energy photon.
(56) 10 points
(II) In the Sun, an ionized helium (He+ ) atom makes a transition from the n = 6 state
to the n = 2 state, emitting a photon. Can that photon be absorbed by hydrogen atoms
present in the Sun? If so, between what energy states will the hydrogen atom jump?
Singly ionized helium is like hydrogen, except that there are two positive charges (Z = 2) in the nucleus.
We can use Bohr’s result for the hydrogen atom, if we replaced e2 by Ze2
En = −
22 (13.6 eV)
54.4 eV
Z 2 (13.6 eV)
=−
=−
.
2
n
n2
n2
The energy of the photon from the n = 6 to n = 2 transition is
1
1
∆E = E6 − E2 = −(54.4 eV)
− 2 = 12.1 eV.
62
2
Because this is the energy difference for the n = 1 to n = 3 transition in hydrogen, the photon can be
absorbed by a hydrogen atom which will jump from n = 1 to n = 3.