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Photons, Electrons, and Atoms
385
38.20.
(a)
Equating initial kinetic energy and final potential energy and solving for the separation radius
r
,
19
14
6
00
1
(92 ) (2 )
1
(184) (1.60 10
C)
5.54 10
m.
44
(
4
.
7
8
1
0
J
C
)
ee
r
K
ππ
−
−
×
==
=
×
×
PP
(b)
The above result may be substituted into Coulomb’s law, or, the relation between the magnitude of the force
and the magnitude of the potential energy in a Coulombic field is
61
9
14
(4.78 10 eV) (1.6 10
J ev)
13.8 N.
(5.54 10
m)
K
F
r
−
−
××
=
×
38.21.
(a) IDENTIFY:
If the particles are treated as point charges,
12
0
1
.
4
qq
U
r
π
=
P
SET UP:
1
2
qe
=
(alpha particle);
2
82
=
(gold nucleus);
r
is given so we can solve for
U
.
EXECUTE:
19
2
92
2
1
3
14
(2)(82)(1.602 10
C)
(8.987 10 N m /C )
5.82 10
J
6.50 10
m
U
−
−
−
×
=×
⋅
=
×
×
13
19
6
5.82 10
J(1 eV/1.602 10
J)
3.63 10 eV
3.63 MeV
U
−−
×
=
(b) IDENTIFY:
Apply conservation of energy:
11
2 2
.
KU KU
+= +
SET UP:
Let point 1 be the initial position of the alpha particle and point 2 be where the alpha particle
momentarily comes to rest. Alpha particle is initially far from the lead nucleus implies
1
r
≈∞
and
1
0.
U
=
Alpha
particle stops implies
2
0.
K
=
EXECUTE:
Conservation of energy thus says
13
5.82 10
J
3.63 MeV.
KU
−
×
=
(c)
2
1
2
Km
v
=
so
13
7
27
22
(
5
.
8
2
1
0
J
)
1.32 10 m/s
6.64 10
kg
K
v
m
−
−
×
=
×
×
EVALUATE:
/
0.044,
vc
=
so it is ok to use the nonrelativistic expression to relate
K
and
v
. When the alpha
particle stops, all its initial kinetic energy has been converted to electrostatic potential energy.
38.22.
(a)
,
(b)
For either atom, the magnitude of the angular momentum is
2
h
=
34
2
1.05 10
kg m s.
−
×⋅
38.23.
IDENTIFY
and
SET UP:
Use the energy to calculate
n
for this state. Then use the Bohr equation, Eq.(38.10), to
calculate
L
.
EXECUTE:
2
(13.6 eV)/
,
n
En
=−
so this state has
13.6/1.51
3.
n
In the Bohr model.
Ln
=
U
so for this state
34
2
33
.
1
6
1
0
k
g
m
/
s
.
L
−
⋅
U=
EVALUATE:
We will find in Section 41.1 that the modern quantum mechanical description gives a different
result.
38.24.
IDENTIFY
and
SET UP:
For a hydrogen atom
2
13.6 eV
n
E
n
.
hc
E
λ
Δ=
, where
E
Δ
is the magnitude of the
energy change for the atom and
is the wavelength of the photon that is absorbed or emitted.
EXECUTE:
41
(13.6 eV)
12.75 eV
EE E
⎛⎞
− =
−
−
=
+
⎜⎟
⎝⎠
.
15
8
(4.136 10
eV s)(3.00 10 m/s)
97.3 nm
12.75 eV
hc
E
−
×
=
Δ
.
15
3.08 10 Hz
c
f
×
.
38.25.
IDENTIFY:
The force between the electron and the nucleus in
3+
Be
is
2
2
0
1
,
4
Ze
F
r
=
P
where
4
Z
=
is the nuclear
charge. All the equations for the hydrogen atom apply to
3+
Be
if we replace
2
e
by
2
.
Ze
(a) SET UP:
Modify Eq.(38.18).
EXECUTE:
4
0
1
8
n
me
E
nh
P
(hydrogen) becomes
4
3
+
2
1( )
1
1
3
.
6
0
e
V
(for Be )
88
n
mZe
me
EZ
Z
n
=
−
=
−
The groundlevel energy of
3+
Be
is
1
2
13.60 eV
16
218 eV.
1
E
=
−
EVALUATE:
The groundlevel energy of
3+
Be
is
2
16
Z
=
times the groundlevel energy of H.
(b) SET UP:
The ionization energy is the energy difference between the
n
→∞
level energy and the
1
n
=
level
energy.
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Chapter 38
EXECUTE:
The
n
→∞
level energy is zero, so the ionization energy of
3+
Be
is 218 eV.
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 Spring '06
 Buchler
 Physics, Energy, Kinetic Energy, Potential Energy, Photon

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