Solutions for HandIn Set #8
PHYS 212  Spring 2007
A54 — Phtotoelectric Effect
Application of simple classical theories of the photo
electric effect suggests the following:
•
The energy of emitted electrons should depend on
the intensity of the light; this is contradicted by
experiment.
•
There should be an observable time delay for an
electron to absorb enough energy to be emitted
by lowintensity light; electrons are observed to be
emitted “instantaneously.”
A55 — Kinetic energies of emitted electrons
This is a twopart problem; both parts use the equation
governing the photoelectic effect:
K
max
=
hf

φ.
The information about the results of the experiment with
700nm light allows us to determine the work function
φ
.
Rearranging the equation above gives
φ
=
hf

K
max
=
hc
λ

K
max
=
1240 eV
·
nm
700 nm

0
.
25 eV
= 1
.
77 eV

0
.
25 eV
= 1
.
52 eV
.
For the 400nm light we have
K
max
=
hf

φ
=
hc
λ

φ
=
1240 eV
·
nm
400 nm

1
.
52 eV
= 3
.
1 eV

1
.
52 eV
= 1
.
58 eV
.
A59 — Electron in a box
(a) The relationship between kinetic energy and momen
tum is the same as it is in classical mechanics:
K
=
p
2
2
m
,
so
p
=
√
2
mK
=
q
2
×
5
.
11
×
10
5
eV/
c
2
×
75 eV
= 8
.
75 keV/
c,
or equivalently
p
=
√
2
mK
=
p
2
×
9
.
1
×
10

31
kg
×
75 eV
×
1
.
6
×
10

19
J/eV
= 4
.
67
×
10

24
kg
·
m/s
.
(b) Using the de Broglie relationship gives
λ
=
h
p
=
hc
pc
=
1240 eV
·
nm
8
.
75
×
10
3
eV
= 0
.
142 nm
or equivalently
λ
=
h
p
=
6
.
63
×
10

34
J
·
s
4
.
67
×
10

24
kg
·
m/s
= 1
.
42
×
10

10
m
(c)
Second Excited State
λ
b
Low probability
(d) From the sketch of the wavefunction it is clear that
the box is 1.5 wavelengths long, or
b
= 1
.
5
×
λ
= 1
.
5
×
0
.
142 nm = 0
.
212 nm
.
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A60 — WhirlaTune and quantization
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 Spring '08
 Ladd
 Energy, Kinetic Energy, Photon, ev, kmax

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