Physics 6C HW#9 and #10 solution
June 5, 2008
11.34
a)
The energy of the photon is found as
E
=
E
f

E
i
=

13
.
606
eV
n
2
f


13
.
606
eV
n
2
i
here
n
f
= 2 and
n
i
= 3
= 13
.
606
eV
(
1
4

1
9
)
=
1.89 eV
b)
E
=
hc
λ
λ
=
6
.
626
×
10

34
J
·
s
(2
.
998
×
10
8
m/s
)
1
.
89
eV
(1
.
602
×
10

19
J/eV
)
=
656 nm
c)
f
=
c
λ
=
3
×
10
8
m/s
6
.
56
×
10

7
m
=
4
.
57
×
10
14
Hz
11.37
Δ
E
= (

13
.
6
eV
)(
1
n
2
f

1
n
2
i
)
Where for Δ
E
>
0 we have absorption and for
Δ
E <
0 we have emission.
1. for
n
i
=
2
and
n
f
=
5,
Δ
E
=
2
.
86
eV(absorption)
2. for
n
i
=
5
and
n
f
=
3,
Δ
E
=

0
.
967
eV(emission)
3. for
n
i
=
7
and
n
f
=
4,
Δ
E
=

0
.
572
eV(emission)
4. for
n
i
=
4
and
n
f
=
7,
Δ
E
=
0
.
572
eV(absorption)
a)
E
=
hc
λ
so the shortest wavelength is emitted in tran
sition
2
.
b)
The atom gains most energy in transition
1
.
c)
The atom loses energy in transition
2 and 3
.
1
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11.39
a)
r
n
=
a
0
n
2
where
a
0
= 0
.
0529 nm
r
2
= 4
a
0
=
0.212 nm
b)
p
=
m
e
v
2
=
s
m
e
k
e
e
2
r
2
=
9
.
95
×
10

25
Kg m/s
c)
l
2
=

r
2
×
p

=
m
e
v
2
r
2
=
2
.
11
×
10

34
kg
·
m
2
/s
d)
K
2
=
p
2
2
2
m
e
= 5
.
43
×
10

19
J
=
3.40 eV
e)
U
2
=

k
e
e
2
r
2
=

1
.
09
×
10

18
J
×
1
eV
1
.
60
×
10

19
=
6.80 eV
f)
E
2
=
K
2
+
U
2
= 3
.
40
eV

6
.
80
eV
=
3.40 eV
28.21
a)
For an electron,
λ
=
h
p
K
=
p
2
2
m
e
p
=
p
2
m
e
K
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 Spring '01
 staff
 Thermodynamics, Atom, Energy, Photon, Light, ev, tran sition

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