4.3 Solutions 285
WhereC=constant which can be determined as follows.
ΣNi=N=CΣe−βEi (11)orC=N
Σe−βEi(12)
Equation (10) then becomesNi=Ne−βEi
Σe−βEi(13)
The denominator in (13)
Z=Σe−βEi (14)
Is known as the partition function. It can be shown that the quantityβ=1
kT(15)
wherekis the Boltzmann constant andTis the absolute temperature.
α=N
Z
(16)
4.3.4 Blackbody Radiation...............................
4.61 Electric power=power radiated
W=σT^4 A
A= 2 πrl= 2 π× 10 −^3 × 1. 0 = 6. 283 × 10 −^3 m^2T=[
W
σA] 1 / 4
=
[
1 , 000
5. 67 × 10 −^8 × 6. 283 × 10 −^3
] 1 / 4
= 1 , 294 K
4.62 The Solar constantSis the heat energy received by 1 m^2 of earth’s surface per
second. IfRis the radius of the sun andrthe earth-sun distance, then the total
intensity of radiation emitted from the sun will beσT^4 Wm−^2 and from the
sun’s surfaceσT^4. 4 πR^2. The radiation received per second per m^2 of earth’s
surface will be
S=σT^4.4 πR^2
4 πr^2
Solving,σT^4 =S.r^2
R^2= 1 , 400
(
1. 5 × 108
7 × 105
) 2
= 6. 43 × 107 Wm−^2T=
(
6. 43 × 107
σ) 1 / 4
=
(
6. 43 × 107
5. 67 × 10 −^8
) 1 / 4
= 5 ,800 K
4.63 Using the analogy between radiation (photon gas) and gas molecules, the pho-
tons move in a cavity at random in all directions, rebounding elastically from
the walls of the cavity. The pressure exerted by an ideal photon gas is