Problem A Copyright reserved

NAME ______________________________________ DATE _______________ CLASS ____________________
Atomic Physics
Problem A
QUANTUM ENERGY
PROBLEM
Free-electron lasers can be used to produce a beam of light with variable
wavelength. Because the laser can produce light with wavelengths as long
as infrared waves or as short as X rays, its potential applications are much
greater than for a laser that can produce light of only one wavelength.
If such a laser produces photons of energies ranging from 1.034 eV to
620.6 eV, what are the minimum and the maximum wavelengths corresponding
to these photons?
SOLUTION
Given: E1 = 1.034 eV
= (1.034 eV)
1.60 × 10 −19 ⎯ e
JV ⎯ = 1.65 × 10−19 J
E 2 = 620.6 eV
= (620.6 eV)
1.60 × 10 −19 ⎯ e
JV ⎯ = 9.93 × 10−17 J
h = 6.63 × 10 −34 J•s
c = 3.00 × 10 8 m/s
Unknown: l min = ? l max = ?
Use the equation for the energy of a quantum of light. Use the relationship
between the frequency and wavelength of electromagnetic waves.
E = hf
f = ⎯ l
c ⎯
Substitute for f in the first equation, and rearrange to solve for wavelength.
Copyright © Holt, Rinehart and Winston. All rights reserved.
E = ⎯ h c
⎯
l
l = ⎯ h c
⎯
E
Substitute values into the equation.
(6.63 × 10 −34 J•s)(3.00 × 10 8 m/s)
l max = ⎯⎯⎯⎯
(1.65 × 10 −19 J)
l max = 1.21 × 10 −6 m
l max =
1210 nm
(6.63 × 10 −34 J•s)(3.00 × 10 8 m/s)
l min = ⎯⎯⎯⎯
(9.93 × 10 −17 J)
l min = 2.00 × 10 −9 m
l min =
2.00 nm
Problem A 173

NAME ______________________________________ DATE _______________ CLASS ____________________
ADDITIONAL PRACTICE
1. In 1974, IBM researchers announced that X rays with energies of
1.29 × 10 −15 J had been guided through a “light pipe” similar to optic
fibers used for visible and near-infrared light. Calculate the wavelength
of one of these X-ray photons.
2. Some strains of Mycoplasma are the smallest living organisms. The wavelength
of a photon with 6.6 × 10 −19 J of energy is equal to the length of
one Mycoplasma. What is that wavelength?
3. Of the various types of light emitted by objects in space, the radio signals
emitted by cold hydrogen atoms in regions of space that are located between
stars are among the most common and important. These signals
occur when the “spin” angular momentum of an electron in a hydrogen
atom changes orientation with respect to the “spin” angular momentum of
the atom’s proton. The energy of this transition is equal to a fraction of an
electron-volt, and the photon emitted has a very low frequency. Given that
the energy of these radio signals is 5.92 × 10 −6 eV, calculate the wavelength
of the photons.
4. The camera with the fastest shutter speed in the world was built for research
with high-power lasers and can expose individual frames of film
with extremely high frequency. If the frequency is the same as that of a
photon with 2.18 × 10 −23 J of energy, calculate its magnitude.
5. Wireless “cable” television transmits images using radio-band photons
with energies of around 1.85 × 10 −23 J. Find the frequency of these
photons.
6. In physics, the basic units of measurement are based on fundamental physical
phenomena. For example, one second is defined by a certain transition
in a cesium atom that has a frequency of exactly 9 192 631 770 s −1 . Find
the energy in electron-volts of a photon that has this frequency. Use
the unrounded values for Planck’s constant (h = 6.626 0755 × 10 −34 J•s)
and for the conversion factor between joules and electron volts (1 eV =
1.602 117 33 × 10 −19 J).
7. Consider an electromagnetic wave that has a wavelength equal to 92 cm,
a length that corresponds to the longest ear of corn grown to date. What
is the frequency corresponding to this wavelength? What is its photon
energy? Express the answer in joules and in electron-volts.
8. The slowest machine in the world, built for testing stress corrosion, can
be controlled to operate at speeds as low as 1.80 × 10 −17 m/s. Find the
distance traveled at this speed in 1.00 year. Calculate the energy of the
photon with a wavelength equal to this distance.
Copyright © Holt, Rinehart and Winston. All rights reserved.
174
Holt Physics Problem Workbook

Problem Workbook Solutions
Additional Practice A
Atomic Physics
Givens
1. E = 1.29 × 10 −15 J
C = 3.00 × 10 8 m/s
h = 6.63 × 10 −34 J•s
Solutions
l = ⎯ h c
⎯ =
E
(6.63 × 10 −34 J•s)(3.00 × 10 8 m/s)
⎯⎯⎯⎯
1.29 × 10 −15 J
l = 1.54 × 10 −10 m = 0.154 nm
2. E = 6.6 × 10 −19 J
C = 3.00 × 10 8 m/s
h = 6.63 × 10 −34 J•s
3. E = 5.92 × 10 −6 eV
C = 3.00 × 10 8 m/s
h = 6.63 × 10 −34 J•s
l = ⎯ h c
⎯ =
E
l = 3.0 × 10 −7 m
l = ⎯ h c
⎯ =
E
l = 0.210 m
(6.63 × 10 −34 J•s)(3.00 × 10 8 m/s)
⎯⎯⎯⎯
6.6 × 10 −19 J
(6.63 × 10 −34 J•s)(3.00 × 10 8 m/s)
⎯⎯⎯⎯
(5.92 × 10 −6 eV)(1.60 × 10 −19 J/eV)
II
4. E = 2.18 × 10 −23 J
h = 6.63 × 10 −34 J •s
E = hf
f = ⎯ E h ⎯
−23
2.18
× 10
J
f = ⎯⎯
6 .63
× 10−
34 J •s
f = 3.29 × 10 10 Hz
Copyright © Holt, Rinehart and Winston. All rights reserved.
5. E = 1.85 × 10 −23 J
h = 6.63 × 10 −34 J •s
6. f = 9 192 631 770 s −1
h = 6.626 0755 × 10 −34 J •s
1 eV = 1.602 117 33 × 10 −19 J
7. l = 92 cm = 92 × 10 −2 m
c = 3.00 × 10 8 m/s
h = 6.63 × 10 −34 J •s
h = 4.14 × 10 −15 eV •s
f = ⎯ E −23
h ⎯ =⎯ 1.85
× 10
J
⎯ −34
= 2.79 × 10 10 Hz
6.63
× 10
J/ s
E = hf
E =
(6.626 0755 × 10 −34 J •s)(9 192 631 770 s −1 )
⎯⎯⎯⎯⎯
1.602 117 33 × 10 −19 J/eV
E = 3.801 9108 × 10 −5 eV
f = ⎯ l
c ⎯
f = ⎯ 3. 8
00
× 10
m/s
⎯
92
× 10−
2 m
f =
3.3 × 10 8 Hz = 330 MHz
E = hf
Section Two—Problem Workbook Solutions II Ch. 21–1

Givens
Solutions
E = (6.63 × 10 −34 J •s)(3.3 × 10 8 Hz)
E =
2.2 × 10 −25 J
E = (4.14 × 10 −15 eV •s)(3.3 × 10 8 Hz)
E = 1.4 × 10 −6 eV
II
8. v = 1.80 × 10 −17 m/s
∆t = 1.00 year
l =∆x
c = 3.00 × 10 8 m/s
h = 6.63 × 10 −34 J •s
∆x = v∆t
∆x = (1.80 × 10 −17 m/s)(1.00 year)
⎯ 365 .25
days
24 h
⎯
1 year
⎯⎯ 1 day
⎯ 36 00s
⎯
1 h
∆x =
E = hf = ⎯ h c
⎯
l
E =
5.68 × 10 −10 m
hc
= ⎯⎯
∆ x
(6.63 × 10 −34 J •s)(3.00 × 10 8 m/s)
⎯⎯⎯⎯
5.68 × 10 −10 m
E = 3.50 × 10 −16 J
Additional Practice B
1. hf t = 4.5 eV
KE max = 3.8 eV
h = 4.14 × 10 −15 eV•s
f = [KE max + hf t ] [3.8
eV
+ 4.
5 eV]
=⎯⎯ h 4
−15
= 2.0 × 10 15 Hz
.14
× 10 eV•
s
2. hf t = 4.3 eV
KE max = 3.2 eV
h = 4.14 × 10 −15 eV •s
3. hf t,Cs = 2.14 eV
hf t,Se = 5.9 eV
h = 4.14 × 10 −15 eV •s
c = 3.00 × 10 8 m/s
KE max = 0.0 eV for both
cases
KE max = hf − hf t
f = ⎯ KE max + hf
⎯ t
h
3.
2 eV + 4.3
eV
f = ⎯⎯
4.
−15
14
× 10 eV
•s
f = 1.8 × 10 15 Hz
a. KE max = hf − hf t = 0.0 eV = ⎯ h c
⎯ − hf t
l
hc
l = ⎯⎯
h ft
hc
(4.14 × 10 −15 eV •s)(3.00 × 10 8 m/s)
l Cs = ⎯⎯ = ⎯⎯⎯⎯
hf t, Cs
2.14 eV
l Cs =
5.80 × 10 −7 m = 5.80 × 10 2 nm
Copyright © Holt, Rinehart and Winston. All rights reserved.
hc
(4.14 × 10 −15 eV •s)(3.00 × 10 8 m/s)
b. l Se = ⎯⎯ = ⎯⎯⎯⎯
hf t, Se
5.9 eV
l Se = 2.1 × 10 −7 m = 2.1 × 10 2 nm
II Ch. 21–2
Holt Physics Solution Manual