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
0
0
1
(92 ) (2 )
1
(184) (1.60
10
C)
5.54
10
m.
4
4
(4.78
10 J C)
e
e
r
K
π
π
−
−
×
=
=
=
×
×
P
P
(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
6
19
14
(4.78
10 eV) (1.6
10
J ev)
13.8 N.
(5.54
10
m)
K
F
r
−
−
×
×
=
=
=
×
38.21.
(a) I
DENTIFY
:
If the particles are treated as point charges,
1
2
0
1
.
4
q q
U
r
π
=
P
S
ET
U
P
:
1
2
q
e
=
(alpha particle);
2
82
q
e
=
(gold nucleus);
r
is given so we can solve for
U
.
E
XECUTE
:
19
2
9
2
2
13
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) I
DENTIFY
:
Apply conservation of energy:
1
1
2
2
.
K
U
K
U
+
=
+
S
ET
U
P
:
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
=
E
XECUTE
:
Conservation of energy thus says
13
1
2
5.82
10
J
3.63 MeV.
K
U
−
=
=
×
=
(c)
2
1
2
K
mv
=
so
13
7
27
2
2(5.82
10
J)
1.32
10
m/s
6.64
10
kg
K
v
m
−
−
×
=
=
=
×
×
E
VALUATE
:
/
0.044,
v c
=
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
π
=
34
2
1.05
10
kg m
s.
−
×
⋅
38.23.
I
DENTIFY
and
S
ET
U
P
:
Use the energy to calculate
n
for this state. Then use the Bohr equation, Eq.(38.10), to
calculate
L
.
E
XECUTE
:
2
(13.6 eV)/
,
n
E
n
= −
so this state has
13.6/1.51
3.
n
=
=
In the Bohr model.
L
n
=
U
so for this state
34
2
3
3.16
10
kg m /s.
L
−
=
×
⋅
U =
E
VALUATE
:
We will find in Section 41.1 that the modern quantum mechanical description gives a different
result.
38.24.
I
DENTIFY
and
S
ET
U
P
:
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
λ
is the wavelength of the photon that is absorbed or emitted.
E
XECUTE
:
4
1
2
2
1
1
(13.6 eV)
12.75 eV
4
1
E
E
E
⎛
⎞
Δ
=
−
= −
−
= +
⎜
⎟
⎝
⎠
.
15
8
(4.136
10
eV
s)(3.00
10
m/s)
97.3 nm
12.75 eV
hc
E
λ
−
×
⋅
×
=
=
=
Δ
.
15
3.08
10
Hz
c
f
λ
=
=
×
.
38.25.
I
DENTIFY
:
The force between the electron and the nucleus in
3+
Be
is
2
2
0
1
,
4
Ze
F
r
π
=
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) S
ET
U
P
:
Modify Eq.(38.18).
E
XECUTE
:
4
2
2
0
1
8
n
me
E
n h
= −
P
(hydrogen) becomes
2
2
4
2
2
3+
2
2
2
2
2
0
0
1
(
)
1
13.60 eV
(for Be
)
8
8
n
m Ze
me
E
Z
Z
n h
n h
n
⎛
⎞
⎛
⎞
= −
=
−
=
−
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
P
P
The groundlevel energy of
3+
Be
is
1
2
13.60 eV
16
218 eV.
1
E
⎛
⎞
=
−
= −
⎜
⎟
⎝
⎠
E
VALUATE
:
The groundlevel energy of
3+
Be
is
2
16
Z
=
times the groundlevel energy of H.
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 Spring '06
 Buchler
 Physics, Energy, Kinetic Energy, Potential Energy, Photon, ev, ΔE

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