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Photons, Electrons, and Atoms
3815
38.67.
IDENTIFY:
Assuming that Betelgeuse radiates like a perfect blackbody, Wien’s displacement and the Stefan
Boltzmann law apply to its radiation.
SET UP:
Wien’s displacement law is
3
peak
2.90 10 m K
T
λ
−
×⋅
=
, and the StefanBoltzmann law says that the
intensity of the radiation is
I
=
σ
T
4
, so the total radiated power is
P =
AT
4
.
EXECUTE: (a)
First use Wien’s law to find the peak wavelength:
m
=
(2.90
×
10
–3
mK
⋅
)/(3000 K) = 9.667
×
10
–7
m
Call
N
the number osf photons/second radiated.
N
×
(energy per photon) =
IA =
AT
4
.
N
(
hc/
m
) =
AT
4
.
4
m
AT
N
hc
λσ
=
.
78
2
4
8
2
4
34
8
(9.667 10 m)(5.67 10 W/m
K )(4 )(600 6.96 10 m) (3000 K)
(6.626 10
J s)(3.00 10 m/s)
N
π
−−
−
××
⋅×
×
=
×⋅ ×
.
N
= 5
×
10
49
photons/s.
(b)
2
4
42
4
B B
BB
S
4
SS
S
4
600
3000 K
4
5800 K
IA
AT
RT
R
R
σπ
⎛⎞
==
=
⎜⎟
⎝⎠
= 3
×
10
4
EVALUATE:
Betelgeuse radiates 30,000 times as much energy per second as does our sun!
38.68.
IDENTIFY:
The blackbody radiates heat into the water, but the water also radiates heat back into the blackbody.
The net heat entering the water causes evaporation. Wien’s law tells us the peak wavelength radiated, but a
thermophile in the water measures the wavelength and frequency of the light in the water.
SET UP:
By the StefanBoltzman law, the net power radiated by the blackbody is
()
44
sphere
water
dQ
T
dt
=−
. Since
this heat evaporates water, the rate at which water evaporates is
v
dQ
dm
L
dt
dt
=
. Wien’s displacement law is
3
m
2.90 10 m K
T
−
=
, and the wavelength in the water is
w
=
0
/
n
.
EXECUTE:
(a) The net radiated heat is
sphere
water
dQ
T
dt
and the evaporation rate is
v
dQ
dm
L
dt
dt
=
, where
dm
is the mass of water that evaporates in time
dt
. Equating these two rates gives
v
sphere
water
dm
LA
T
T
dt
.
(
)
24
4
sphere
water
v
4
T
dm
dt
L
−
=
.
82
4
2
4
4
4
3
5.67 10 W/m
K
4
(0.120 m)
(498 K)
(373 K)
1.92 10 kg/s
0.193 g/s
2256 10 J/Kg
dm
dt
−
−
⎡⎤
−
⎣⎦
×
=
×
(b)
(i) Wien’s law gives
m
= (0.00290 m K
⋅
)/(498 K) = 5.82
×
10
–6
m
But this would be the wavelength in vacuum. In the water the thermophile organism would measure
w
=
0
/
n
=
(5.82
×
10
–6
m)/1.333 = 4.37
×
10
–6
m = 4.37 μm
(ii) The frequency is the same as if the wave were in air, so
f = c/
0
= (3.00
×
10
8
m/s)/(5.82
×
10
–6
m) = 5.15
×
10
13
Hz
EVALUATE:
An alternative way is to use the quantities in the water:
0
/
/
cn
f
n
=
= c/
0
, which gives the same
answer for the frequency. An organism in the water would measure the light coming to it through the water, so the
wavelength it would measure would be reduced by a factor of 1/
n
.
38.69.
IDENTIFY:
The energy of the peakintensity photons must be equal to the energy difference between the
n
= 1
and the
n
= 4 states. Wien’s law allows us to calculate what the temperature of the blackbody must be for it to
radiate with its peak intensity at this wavelength.
SET UP:
In the Bohr model, the energy of an electron in shell
n
is
2
13.6 eV
n
E
n
, and Wien’s displacement law
is
3
m
2.90 10 m K
T
−
=
. The energy of a photon is
E = hf = hc/
.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Photon, Radiation

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