4.3 Solutions
289
=
2
15
(3.14159)
5 (1.38065)
4
× 10
−92
(6.626068 × 10 −34 ) 3 (2.99792 × 10 8 ) 2
= 5.67 × 10
−8 W-m
−2 -K
−4
a value which is in excellent agreement with the experiment.
4.73 Number of modes per m
3 in the frequency interval dν is
N =
8πν
2 dν
c 3
But,
ν =
c
λ
; dν = −
dλ
λ 2 ; λ =
4,990 + 5,010
2
= 5,000 A
0
dλ = 5,010 − 4,990 = 20 A
0
∴ N =
8π dλ
λ 4 =
8π × 20 × 10
−10
(5 × 10 −7 ) 4
= 8.038 × 10
17
/m
3
4.74 (a)
(1)
P = AE λ dλ =
8π hc Adλ
λ 5 (e hc/λkT − 1)
(2)
Mean wavelength λ = 0.55 μm = 5.5 × 10
−7 m.
dλ = (0.7 − 0.4) μm = 3 × 10
−7 m
A = πr
2
= π(2.5 × 10
−3 )
2
= 1.96 × 10
−5 m
2
hc
λkT
=
(6.63 × 10
−34 )(3 × 10
8 )
(5.5 × 10 −7 )(1.38 × 10 −23 )(4,000)
= 6.55
Using the above values in (2) we find
P = AE λ dλ = 0.84 × 10
−6 W = 0.84 μW.
(b) hν =
hc
λ
=
6.63 × 10
−34
× 3 × 10
8
5.5 × 10 −7
= 3.616 × 10
−19
Number of photons emitted per second
n =
P
hν
= 0.84 × 10
−6
/3.616 × 10
−19
= 2.32 × 10
12
/s
4.75 u λ dλ =
8π hc
λ 5
1
e hc/λkT − 1
dλ
(1)
Put λ = c/ν
(2)
and dλ = −
c
ν 2
dν
(3)
in the RHS of (1) and simplify
u ν dν =
8π hν
3
c 3 (e hν/kT − 1)
dν
(4)