T
HE
W
AVE
N
ATURE OF
P
ARTICLES
39.1.
IDENTIFY
and
SET UP:
hh
pm
v
λ
==
.
For an electron,
31
9.11 10
kg
m
−
=×
.
For a proton,
27
1.67 10
kg
m
−
.
EXECUTE: (a)
34
10
31
6
6.63 10
J s
1.55 10
m
0.155 nm
(9.11 10
kg)(4.70 10 m/s)
−
−
−
×⋅
×
=
××
(b)
is proportional to
1
m
, so
31
10
14
e
pe
27
p
9.11 10
kg
(1.55 10
m)
8.46 10
m
1.67 10
kg
m
m
λλ
−
−−
−
⎛⎞
×
×
=
×
⎜⎟
×
⎝⎠
.
39.2.
IDENTIFY
and
SET UP:
For a photon,
hc
E
=
.
For an electron or proton,
h
p
=
and
2
2
p
E
m
=
, so
2
2
2
h
E
m
=
.
EXECUTE: (a)
15
8
9
(4.136 10
eV s)(3.00 10 m/s)
6.2 keV
0.20 10 m
hc
E
−
−
×
=
×
(b)
2
23
4
18
29
3
1
6.63 10
J s
1
6.03 10
J
38 eV
2
2(9.11 10
kg)
h
E
m
−
−
=
× =
(c)
31
e
27
p
9.11 10
kg
(38 eV)
0.021 eV
1.67 10
kg
m
EE
m
−
−
×
=
×
EVALUATE:
For a given wavelength a photon has much more energy than an electron, which in turn has more
energy than a proton.
39.3.
(a)
34
24
10
(6.63 10
J s)
λ
2.37 10
kg m s.
λ
(2.80 10
m)
p
p
−
−
−
=
⇒
=
×
⋅
×
(b)
22
4
2
18
31
(2.37 10
kg m s)
3.08 10
J
19.3 eV.
(
9
.
1
1
1
0
k
g
)
p
K
m
−
−
−
=
×
=
×
39.4.
λ
2
p
mE
34
15
27
6
19
(6.63 10
J s)
7.02 10
m.
2(6.64 10
kg) (4.20 10 eV) (1.60 10
J eV)
−
−
×
×
39.5.
IDENTIFY
and
SET UP:
The de Broglie wavelength is
.
v
In the Bohr model,
(/2)
,
n
mvr
n h
π
=
so
/(2
).
n
mv
nh
r
=
Combine these two expressions and obtain an equation for
in terms of
n
. Then
.
nn
rr
h
nh
n
ππ
EXECUTE: (a)
For
10
11
0
1,
2
with
0.529 10
m, so
nr
r
a
λπ
−
=
=
×
10
10
2 (0.529 10
m)
3.32 10
m
=
×
1
2;
r
=
the de Broglie wavelength equals the circumference of the orbit.
(b)
For
4
4,
2
/ 4.
2
04
0
so
16 .
n
rn
a r
a
10
9
00
2 (16
)/4
4(2
)
4(3.32 10
m)
1.33 10 m
aa
=
×
=
×
4
2/
4
;
r
=
the de Broglie wavelength is
4
n
=
times the circumference of the orbit.
EVALUATE:
As
n
increases the momentum of the electron increases and its de Broglie wavelength decreases. For
any
n
, the circumference of the orbits equals an integer number of de Broglie wavelengths.
39
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Chapter 39
39.6.
(a)
For a nonrelativistic particle,
2
,so
2
p
K
m
=
.
2
hh
p
Km
λ
==
(b)
34
19
31
11
(6.63 10
J s)
2(800 eV)(1.60 10
J/eV)(9.11 10
kg)
4.34 10
m.
−
−
×⋅
×
×
=
×
39.7.
IDENTIFY:
A person walking through a door is like a particle going through a slit and hence should exhibit wave
properties.
SET UP:
The de Broglie wavelength of the person is
= h/mv
.
EXECUTE: (a)
Assume
m
= 75 kg and
v
= 1.0 m/s.
= h/mv =
(6.626
×
10
–34
J
⋅
s)/[(75 kg)(1.0 m/s)] = 8.8
×
10
–36
m
EVALUATE: (b)
A typical doorway is about 1 m wide, so the person’s de Broglie wavelength is much too small
to show wave behavior through a “slit” that is about 10
35
times as wide as the wavelength. Hence ordinary objects
do not show wave behavior in everyday life.
39.8.
Combining Equations 37.38 and 37.39 gives
2
1.
pm
c
γ
=−
(a)
21
2
()
1
4
.
4
3
1
0
m
.
h
hmc
p
λγ
−
−=
×
(The incorrect nonrelativistic calculation gives
12
5.05 10
m.)
−
×
(b)
3
1
7
.
0
7
1
0
m
.
γ
−
×
39.9.
IDENTIFY
and
SET UP:
A photon has zero mass and its energy and wavelength are related by Eq.(38.2). An
electron has mass. Its energy is related to its momentum by
2
/2
Ep m
=
and its wavelength is related to its
momentum by Eq.(39.1).
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 Spring '06
 Buchler
 Physics, Electron, Conservation Of Energy, Heisenberg Uncertainty Principle, Photon, Uncertainty Principle, kg

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