shusi (oos225) – HW 15 – turner – (58465)
1
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20
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001
10.0points
A certain laser outputs pure red light (photon
energy 1
.
8 eV) with power 575 mW.
How many photons per second does this
laser emit?
Consider just four of the energy levels in a
certain atom, as shown in the diagram below.
n
= 2
n
= 3
n
= 4
Convert eV to Joules:
003(part1of3)10.0points
Consider just four of the energy levels in a
certain atom, as shown in the diagram below.
n
= 2
n
= 3
n
= 4
(1
.
8 eV)
1
.
6
×
10

19
J
eV
≈
2
.
88
×
10

19
J
.
Then, number of photons emitted per sec
ond is simply
0
.
575 W
all possible transitions among these levels?
Which transition corresponds to the highest
frequency light emitted?
Which transition
corresponds to the lowestfrequency?
002
10.0points
Suppose we have reason to suspect that a cer
How many spectral lines will result from
all possible transitions among these levels?
Which transition corresponds to the highest
frequency light emitted?
Which transition
corresponds to the lowestfrequency?
tain quantum object has only three quantum
states:
E
0
, E
1
,
and
E
2
.
When we excite such
an object we observe that it emits electro
magnetic radiation of three different energies:
2
.
48 eV (green), 1
.
77 eV (orange), and 0
.
71 eV
(infrared).
If we now illuminate the (cooled) object
with continuous radiation from 0
.
4 eV to
2
.
7 eV, we observe absorption at 0
.
71 eV
and at 2
.
48 eV. From this data, what is the
difference in energy between
E
2
and
E
1
?
3.
three; level 2 to level 1 transition; level 4
to level 3 transition.
4.
three; level 4 to level 3 transition; level 2
to level 1 transition.
Explanation:
The energy levels can be taken as
E
0
= 0
E
1
= 0
.
71 eV
E
2
= 2
.
48 eV
.
If the levels were
E
0
= 0,
E
1
= 1
.
77 eV,
and
E
2
= 2
.
48 eV, then we would observe
absorption at 1
.
77 eV and 2
.
48 eV instead.
n
= 1
The highestfrequency transition is from
quantum level 4 to level 1.
The lowest
frequency transition is from quantum level
shusi (oos225) – HW 15 – turner – (58465)
2
4 to level 3.
004(part2of3)10.0points
An electron deexcites from the fourth quan
tum level to the third and then directly to the
ground state. Two photons are emitted.
How does the sum of their frequencies com
pare to the frequency of the single photon
that would be emitted by deexcitation from
the fourth level directly to the ground state?
4.
The sum is equal to the frequency of the
single photon.
correct
1.
Because energies are additive, so are the
frequencies.
However, the wavelength rela
tionship is not that simple: wavelength is in
versely proportional to the frequency, so the
reciprocals of the wavelengths are additive:
1
λ
4
→
3
+
1
λ
3
→
1
=
1
λ
4
→
1
005(part3of3)10.0points