Michichiii

PHOTONS, ELECTRONS, AND ATOMS

38.1.

38

IDENTIFY and SET UP: The stopping potential V0 is related to the frequency of the light by V0 =

h
φ
f − . The
e
e

slope of V0 versus f is h/e. The value fth of f when V0 = 0 is related to φ by φ = hf th .
EXECUTE: (a) From the graph, f th = 1.25 × 1015 Hz . Therefore, with the value of h from part (b),

φ = hf th = 4.8 eV .
(b) Fromthe graph, the slope is 3.8 × 10−15 V ⋅ s . h = (e)(slope) = (1.60 × 10−16 C)(3.8 × 10−15 V ⋅ s) = 6.1 × 10 −34 J ⋅ s
(c) No photoelectrons are produced for f < f th .

38.2.

(d) For a different metal fth and φ are different. The slope is h/e so would be the same, but the graph would be
shifted right or left so it has a different intercept with the horizontal axis.
EVALUATE: As the frequencyf of the light is increased above fth the energy of the photons in the light increases
and more energetic photons are produced. The work function we calculated is similar to that for gold or nickel.
IDENTIFY and SET UP: c = f λ relates frequency and wavelength and E = hf relates energy and frequency for a

photon. c = 3.00 × 108 m/s . 1 eV = 1.60 × 10 −16 J .
c 3.00 × 108 m/s
= 5.94 × 1014Hz
EXECUTE: (a) f = =
λ 505 × 10−9 m
(b) E = hf = (6.626 × 10 −34 J ⋅ s)(5.94 × 1014 Hz) = 3.94 × 10−19 J = 2.46 eV
(c) K = 1 mv 2 so v =
2
38.3.

2K
2(3.94 × 10−19 J)
=
= 9.1 mm/s
m
9.5 × 10−15 kg

3.00 × 108 m s
= 5.77 × 1014 Hz
5.20 × 10−7 m
h 6.63 × 10 −34 J ⋅ s
= 1.28 × 10−27 kg ⋅ m s
p= =
5.20 × 10 −7 m
f=

c

=

λ

λ

38.4.

E = pc = (1.28 × 10−27 kg ⋅ m s)(3.00 × 108 m s) = 3.84 × 10−19 J = 2.40 eV.
energy
hc
. 1 eV = 1.60 × 10−19 J . For a photon, E = hf = . h = 6.63 × 10−34 J ⋅ s.
IDENTIFY and SET UP: Pav =
t
λ
EXECUTE: (a) energy = Pavt = (0.600 W)(20.0 × 10−3 s) = 1.20 × 10 −2 J = 7.5 × 1016 eV

(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
= 3.05 × 10−19 J = 1.91 eV
652 × 10−9 m
λ
(c) The number of photons is the total energy in a pulsedivided by the energy of one photon:
1.20 × 10−2 J
= 3.93 × 1016 photons .
3.05 × 10 −19 J/photon
EVALUATE: The number of photons in each pulse is very large.
IDENTIFY and SET UP: Eq.(38.2) relates the photon energy and wavelength. c = f λ relates speed, frequency and
wavelength for an electromagnetic wave.
E (2.45 × 106 eV)(1.602 × 10−19 J/1 eV)
= 5.92 × 1020 Hz
EXECUTE: (a) E = hf so f ==
6.626 × 10−34 J ⋅ s
h
c 2.998 × 108 m/s
= 5.06 × 10−13 m
(b) c = f λ so λ = =
f
5.92 × 1020 Hz
(c) EVALUATE: λ is comparable to a nuclear radius. Note that in doing the calculation the energy in MeV was
converted to the SI unit of Joules.
(b) E =

38.5.

hc

=

38-1

38-2

38.6.

Chapter 38

λth = 272 nm . c = f λ .

IDENTIFY and SET UP:

h = 4.136 × 10−15 eV ⋅ s.
EXECUTE: (a) f th =

38.7.

c

λth

12
mvmax = hf − φ . At the threshold frequency, f th , vmax → 0.
2

3.00 × 108 m/s
= 1.10 × 1015 Hz .
272 × 10−9 m
eV ⋅ s)(1.10 × 1015 Hz) = 4.55 eV .

=

(b) φ = hf th = (4.136 × 10−15
12
(c) mvmax = hf − φ = (4.136 × 10−15 eV ⋅ s)(1.45 × 1015 Hz) − 4.55 eV = 6.00 eV − 4.55 eV = 1.45 eV
2
EVALUATE: The threshold wavelength depends onthe work function for the surface.
hc
12
IDENTIFY and SET UP: Eq.(38.3): mvmax = hf − φ =
− φ . Take the work function φ from Table 38.1. Solve
2
λ
for vmax . Note that we wrote f as c / λ .
EXECUTE:

12
(6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
− (5.1 eV)(1.602 × 10−19 J/1 eV)
mvmax =
2
235 × 10−9 m

12
mvmax = 8.453 × 10 −19 J − 8.170 × 10−19 J = 2.83 × 10−20 J
2
2(2.83 × 10 −20J)
vmax =
= 2.49 × 105 m/s
9.109 × 10−31 kg

38.8.

EVALUATE: The work function in eV was converted to joules for use in Eq.(38.3). A photon with λ = 235 nm
has energy greater then the work function for the surface.
hc
IDENTIFY and SET UP: φ = hf th =
. The minimum φ corresponds to the minimum λ .

λth

EXECUTE:
38.9.

φ=

hc

λth

=

(4.136 × 10

−15

eV ⋅ s)(3.00...