1489
Chapter 39
1. According to Eq. 394,
E
n
L
– 2
. As a consequence, the new energy level
E'
n
satisfies
F
H
G
I
K
J
F
H
G
I
K
J
E
E
L
L
L
L
n
n
2
2
1
2
,
which gives
L
L
2 . Thus, the ratio is
/
2
1.41.
L L
2. (a) The groundstate energy is
E
1
h
2
8
m
e
L
2
n
2
6.63
10
34
J
s
2
8(9.11
10
31
kg) 300
10
12
m
2
1
2
6.70
10
19
J
4.19eV.
(b) With
m
p
= 1.67
10
– 27
kg, we obtain
2
34
2
2
2
22
1
2
2
27
12
3
6.63 10
J s
1
8.225 10
J
8
8(1.67 10
kg) 200 10
m
5.13 10 eV.
p
h
E
n
m L
3. Since
E
n
L
– 2
in Eq. 394, we see that if
L
is doubled, then
E
1
becomes (2.6 eV)(2)
– 2
= 0.65 eV.
4. We first note that since
h
= 6.626
10
–34
J·s and
c
= 2.998
10
8
m/s,
hc
6.626
10
34
J
s
2.998
10
8
m/s
1.602
10
19
J/eV
10
9
m/nm
1240eV
nm.
Using the
mc
2
value for an electron from Table 373 (511
10
3
eV), Eq. 394 can be
rewritten as
E
n h
mL
n
hc
mc
L
n
2
2
2
2
2
2
2
8
8
bg
c h
.
The energy to be absorbed is therefore
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View Full DocumentCHAPTER 39
1490
E
E
4
E
1
5
2
1
2
h
2
8
m
e
L
2
24
hc
2
8
m
e
c
2
L
2
24 1240eV
nm
2
8 511
10
3
eV
0.250nm
2
144eV.
5. We can use the
mc
2
value for an electron from Table 373 (511
10
3
eV) and
hc
=
1240 eV · nm by writing Eq. 394 as
E
n h
mL
n
hc
mc
L
n
2
2
2
2
2
2
2
8
8
bg
c h
.
For
n
= 3, we set this expression equal to 4.7 eV and solve for
L
:
L
n hc
mc
E
n
bg
c h
b
g
c
hb
g
8
3 1240
8 511
10
4 7
085
2
3
eV nm
eV
eV
nm.
.
.
6. With
m
=
m
p
= 1.67
10
– 27
kg, we obtain
E
1
h
2
8
mL
2
n
2
6.63
10
34
J.s
2
8(1.67
10
27
kg) 120
10
12
m
2
1
2
2.28
10
21
J
0.0143eV.
Alternatively, we can use the
mc
2
value for a proton from Table 373 (938
10
6
eV) and
hc
= 1240 eV · nm by writing Eq. 394 as
E
n
n
2
h
2
8
mL
2
n
2
hc
2
8
m
p
c
2
L
2
.
This alternative approach is perhaps easier to plug into, but it is recommended that both
approaches be tried to find which is most convenient.
7. To estimate the energy, we use Eq. 394, with
n
= 1,
L
equal to the atomic diameter,
and
m
equal to the mass of an electron:
2
2
34
2
2
10
2
2
31
14
1
6.63 10
J s
3.07 10
J=1920MeV
1.9 GeV.
8
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 Spring '11
 Atom, Energy, NZ, ev

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