T
HE
W
AVE
N
ATURE OF
P
ARTICLES
39.1.
I
DENTIFY
and
S
ET
U
P
:
h
h
p
mv
λ
=
=
.
For an electron,
31
9.11
10
kg
m
−
=
×
.
For a proton,
27
1.67
10
kg
m
−
=
×
.
E
XECUTE
:
(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)
λ
−
−
−
×
⋅
=
=
×
=
×
×
(b)
λ
is proportional to
1
m
, so
31
10
14
e
p
e
27
p
9.11
10
kg
(1.55
10
m)
8.46
10
m
1.67
10
kg
m
m
λ
λ
−
−
−
−
⎛
⎞
⎛
⎞
×
=
=
×
=
×
⎜
⎟
⎜
⎟
⎜
⎟
×
⎝
⎠
⎝
⎠
.
39.2.
I
DENTIFY
and
S
ET
U
P
:
For a photon,
hc
E
λ
=
.
For an electron or proton,
h
p
λ
=
and
2
2
p
E
m
=
, so
2
2
2
h
E
m
λ
=
.
E
XECUTE
:
(a)
15
8
9
(4.136
10
eV
s)(3.00
10
m/s)
6.2 keV
0.20
10
m
hc
E
λ
−
−
×
⋅
×
=
=
=
×
(b)
2
2
34
18
2
9
31
6.63
10
J s
1
6.03
10
J
38 eV
2
0.20
10
m
2(9.11
10
kg)
h
E
m
λ
−
−
−
−
⎛
⎞
×
⋅
=
=
=
×
=
⎜
⎟
×
×
⎝
⎠
(c)
31
e
p
e
27
p
9.11
10
kg
(38 eV)
0.021 eV
1.67
10
kg
m
E
E
m
−
−
⎛
⎞
⎛
⎞
×
=
=
=
⎜
⎟
⎜
⎟
⎜
⎟
×
⎝
⎠
⎝
⎠
E
VALUATE
:
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)
2.37
10
kg m s.
(2.80
10
m)
h
h
p
p
−
−
−
×
⋅
=
⇒
=
=
=
×
⋅
×
(b)
2
24
2
18
31
(2.37
10
kg m s)
3.08
10
J
19.3 eV.
2
2(9.11 10
kg)
p
K
m
−
−
−
×
⋅
=
=
=
×
=
×
39.4.
2
h
h
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.
I
DENTIFY
and
S
ET
U
P
:
The de Broglie wavelength is
.
h
h
p
mv
λ
=
=
In the Bohr model,
(
/ 2
),
n
mvr
n h
π
=
so
/(2
).
n
mv
nh
r
π
=
Combine these two expressions and obtain an equation for
λ
in terms of
n
. Then
2
2
.
n
n
r
r
h
nh
n
π
π
λ
⎛
⎞
=
=
⎜
⎟
⎝
⎠
E
XECUTE
:
(a)
For
10
1
1
0
1,
2
with
0.529
10
m, so
n
r
r
a
λ
π
−
=
=
=
=
×
10
10
2
(0.529
10
m)
3.32
10
m
λ
π
−
−
=
×
=
×
1
2
;
r
λ
π
=
the de Broglie wavelength equals the circumference of the orbit.
(b)
For
4
4,
2
/4.
n
r
λ
π
=
=
2
0
4
0
so
16
.
n
r
n a
r
a
=
=
10
9
0
0
2
(16
)/4
4(2
)
4(3.32
10
m)
1.33
10
m
a
a
λ
π
π
−
−
=
=
=
×
=
×
4
2
/4;
r
λ
π
=
the de Broglie wavelength is
1
1
4
n
=
times the circumference of the orbit.
E
VALUATE
:
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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392
Chapter 39
39.6.
(a)
For a nonrelativistic particle,
2
, so
2
p
K
m
=
.
2
h
h
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.
I
DENTIFY
:
A person walking through a door is like a particle going through a slit and hence should exhibit wave
properties.
S
ET
U
P
:
The de Broglie wavelength of the person is
λ
= h/mv
.
E
XECUTE
:
(a)
Assume
m
= 75 kg and
v
= 1.0 m/s.
λ
= h/mv =
(6.626
×
10
–34
J
⋅
s)/[(75 kg)(1.0 m/s)] = 8.8
×
10
–36
m
E
VALUATE
:
(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.
p
mc
γ
=
−
(a)
2
12
(
)
1
4.43
10
m.
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 Spring '08
 Jabbour
 Materials Science And Engineering

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