CHAPTER 39 INSIDE ATOMS AND NUCLEI
ActivPhysics
can help with these problems: Activities 17.3–17.6
Section 391: Toward the Quantum Theory
Useful numerical values of Planck’s constant, in SI and atomic units, are:
h
=
×
⋅
=
×
⋅
=


6 626
10
4136
10
34
15
.
.
J
s
eV
s
1240 eV
nm c
⋅
/ ,
and
=
=
×
⋅
=
×

h
=
2
1055
10
6 582
34
p
.
.
J
s
10
197 3
16

⋅
=
⋅
eV
s
MeV
fm c
.
/ .
Problem
1.
Find the energy in electronvolts of (a) a 1.0MHz radio photon, (b) a 5 0
10
14
.
×
Hz optical photon, and (c) a
3 0
10
18
.
×
Hz Xray photon.
Solution
The energy of a photon (Equation 391) is
E
hf
g
=
,
where
h
is Planck’s constant. For the frequencies given, the photon
energies in atomic units are (a) 4.14 neV (b) 2.07 eV, and (c) 12.4 keV.
Problem
2.
What is the wavelength of a 6.5eV photon? In what spectral region is this?
Solution
l
g
=
=
=
⋅
=
c f
hc E
=
=
=
(
) (
.
)
.
1240
6 5
191
eV
nm
eV
nm (We used Equation 391 to relate the frequency to the photon
energy and atomic units for
hc
). This photon is in the ultraviolet region of the electromagnetic spectrum.
Problem
3.
A microwave oven uses electromagnetic radiation at 2.4 GHz. (a) What is the energy of each microwave photon? (b) At
what rate does a 625W oven produce photons?
Solution
(a)
E
hf
g
m
=
=
×
⋅
=
×


( .
)( .4
)
.
.
),
4136
10
2
9 93
159
10
15
24
eV
s
GHz
eV (or
J
as in Problem 1. (b)
P=
E
g
=
(
/ ) ( .
/
)
.
625
159
10
3 94
10
24
26
J s
J photon
=
×
=
×

photons/s is the power output in photons.
Problem
4.
A red laser at 650 nm and a blue laser at 450 nm emit photons at the same rate. How do their total power outputs
compare?
Solution
The ratio of the photon energies is
E
E
f
f
blue
red
blue
red
red
blue
=
=
=
=
=
=
=
=
l
l
650 450
1.44. Since photons are emitted at the
same rate, this is also the ratio of the power outputs.
Problem
5.
Find the rate of photon production by (a) a radio antenna broadcasting 1.0 kW at 89.5 MHz, (b) a laser producing
1.0 mW of 633nm light, and (c) and Xray machine producing 0.10nm X rays with a total power of 2.5 kW.
Solution
The rate of photon emission is the power output (into photons) divided by the photon energy,
P
P
P
=
=
=
E
h f
hc
g
l
=
=
. For
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CHAPTER 39
915
the devices specified, rates are (a) 1
6 626
10
89 5
169
10
34
28
1
kW
J
s
MHz
s
=
( .
.
)
.
,
×
⋅
×
=
×


(b) (
)
1
633
mW
nm
×
÷
( .
/ )
.
,
6 626
10
3
10
318
10
34
8
15
1
×
⋅
×
×
=
×


J
s
m s
s
and (c) 1.
s
26
10
18
1
×

.
Problem
6.
Calculate the wavelengths of the first three lines in the Lyman series for hydrogen.
Solution
From Equation 393,
l
=

=

=


[
(
)]
(
)
R
n
R
n
n
H
H
1
1
1
122
2
1
1
2
2
=
=
nm, 103 nm, and 97.2 nm for the first three
n
values in
the Lyman series,
n
=
2
3
,
, and 4. (Note:
R
H
=

0 01097
1
.
(
)
nm
or
R
H

=
1
912
.
nm, which is the Lyman series limit.)
Problem
7.
Which spectral line of the hydrogen Paschen series (
)
n
2
3
=
has wavelength 1282 nm?
Solution
The wavelengths in the Paschen series for hydrogen are given by Equation 393, with
n
n
2
1
3
4
5
6
=
=
and
.
,
,
, . . .
Thus,
l
=

(
)
(
).
1
9
9
1
2
1
2
=
=
R
n
n
H
If one substitutes
l
=
1282 nm and
R
H
=

0 01097
1
.
(
)
,
nm
one finds 9
0 01097
1
2
n
=
×
.
1282
9
1
2
(
),
n

or
n
1
9
141
141
9
5
=
×

=
(
. ) (
.
)
,
=
so this is the second line in the series.
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