381
P
HOTONS
,
E
LECTRONS
,
AND
A
TOMS
38.1.
IDENTIFY
and
SET UP:
The stopping potential
V
0
is related to the frequency of the light by
0
h
Vf
ee
φ
=−
.
The
slope of
V
0
versus
f
is
h/e
.
The value
f
th
of
f
when
0
0
V
=
is related to
by
th
hf
=
.
EXECUTE: (a)
From the graph,
15
th
1.25 10 Hz
f
=×
. Therefore, with the value of
h
from part (b),
th
4.8 eV
hf
==
.
(b)
From the graph, the slope is
15
3.8 10
V s
−
×⋅
.
16
15
34
( )(slope)
(1.60 10
C)(3.8 10
V s)
6.1 10
J s
he
−−
−
×
×
⋅
=
×
⋅
(c)
No photoelectrons are produced for
th
ff
<
.
(d)
For a different metal
f
th
and
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 frequency
f
of the light is increased above
f
th
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.
38.2.
IDENTIFY
and
SET UP:
cf
λ
=
relates frequency and wavelength and
Eh
f
=
relates energy and frequency for a
photon.
8
3.00 10 m/s
c
.
16
1 eV
1.60 10
J
−
.
EXECUTE: (a)
8
14
9
5.94 10 Hz
505 10 m
c
f
−
×
=
×
×
(b)
34
14
19
(6.626 10
J s)(5.94 10 Hz)
3.94 10
J
2.46 eV
f
×
⋅
×
=
×
=
(c)
2
1
2
Km
v
=
so
19
15
22
(
3
.
9
4
1
0
J
)
9.1 mm/s
9.5 10
kg
K
v
m
−
−
×
=
×
38.3.
8
14
7
3.00 10 m s
5.77 10 Hz
λ
5.20 10 m
c
f
−
×
=
×
×
34
27
7
27
8
19
6.63 10
J s
1.28 10
kg m s
λ
5.20 10
m
(1.28 10
kg m s) (3.00 10 m s)
3.84 10
J
2.40 eV.
h
p
Ep
c
−
−
−
=
×
⋅
×
×
⋅
×
=
×
=
38.4.
IDENTIFY
and
SET UP:
av
energy
P
t
=
.
19
1 eV
1.60 10
J
−
.
For a photon,
hc
f
.
34
6.63 10
J s
h
−
⋅
.
EXECUTE: (a)
32
1
6
av
energy
(0.600 W)(20.0 10 s) 1.20 10 J
7.5 10 eV
Pt
×
=
×
=
×
(b)
34
8
19
9
(6.63 10
J s)(3.00 10 m/s)
3.05 10
J
1.91 eV
652 10 m
hc
E
−
−
−
×⋅ ×
= ×
=
×
(c)
The number of photons is the total energy in a pulse divided by the energy of one photon:
2
16
19
1.20 10 J
3.93 10 photons
3.05 10
J/photon
−
−
×
×
.
EVALUATE:
The number of photons in each pulse is very large.
38.5.
IDENTIFY
and
SET UP:
Eq.(38.2) relates the photon energy and wavelength.
=
relates speed, frequency and
wavelength for an electromagnetic wave.
EXECUTE: (a)
f
=
so
61
9
20
34
(2.45 10 eV)(1.602 10
J/1 eV)
5.92 10 Hz
6.626 10
J s
E
f
h
−
−
××
=
×
(b)
=
so
8
13
20
2.998 10 m/s
5.06 10
m
c
f
−
×
=
×
×
(c) EVALUATE:
is comparable to a nuclear radius. Note that in doing the calculation the energy in MeV was
converted to the SI unit of Joules.
38
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Chapter 38
38.6.
IDENTIFY
and
SET UP:
th
272 nm
λ
=
.
cf
=
.
2
max
1
2
mv
hf
φ
=−
.
At the threshold frequency,
th
f
,
max
0
v
→
.
15
4.136 10
eV s
h
−
=×
⋅
.
EXECUTE: (a)
8
15
th
9
th
3.00 10 m/s
1.10 10 Hz
272 10 m
c
f
−
×
==
=
×
×
.
(b)
15
15
th
(4.136 10
eV s)(1.10 10 Hz)
4.55 eV
hf
−
×
⋅
×
=
.
(c)
21
5
1
5
max
1
(4.136 10
eV s)(1.45 10 Hz)
4.55 eV
6.00 eV
4.55 eV
1.45 eV
2
mv
hf
−
=
−
=
×⋅
×−=−=
EVALUATE:
The threshold wavelength depends on the work function for the surface.
38.7.
IDENTIFY
and
SET UP:
Eq.(38.3):
2
max
1
.
2
hc
mv
hf
φφ
=−= −
Take the work function
from Table 38.1. Solve
for
max
.
v
Note that we wrote
f
as
/
.
c
EXECUTE:
34
8
2
19
max
9
1
(6.626 10
J s)(2.998 10 m/s)
(5.1 eV)(1.602 10
J/1 eV)
2
235 10 m
mv
−
−
−
×
×
×
91
9
2
0
max
1
8.453 10
J
8.170 10
J
2.83 10
J
2
mv
−−
−
−×
=
×
20
5
max
31
2(2.83 10
J)
2.49 10 m/s
9.109 10
kg
v
−
−
×
×
×
EVALUATE:
The work function in eV was converted to joules for use in Eq.(38.3). A photon with
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 Spring '11
 Shaefer
 Atom, Energy, Photon, ev, Wien

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