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47. (a) Since
E
2
=
−
0
.
85 eV and
E
1
=
−
13
.
6eV+10
.
2eV =
−
3
.
4 eV, the photon energy is
E
photon
=
E
2
−
E
1
=
−
0
.
85 eV
−
(
−
3
.
4eV)=2
.
6eV
.
(b) From
E
2
−
E
1
=(
−
13
.
6eV)
µ
1
n
2
2
−
1
n
2
1
¶
=2
.
6eV
we obtain
1
n
2
2
−
1
n
2
1
=
−
2
.
6eV
13
.
6eV
≈−
3
16
=
1
4
2
−
1
2
2
.
Thus,
n
2
=4and
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Unformatted text preview: n 1 = 2. So the transition is from the n = 4 state to the n = 2 state. One can easily verify this by inspecting the energy level diagram of Fig. 4016....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Energy, Photon

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