1000 Solved Problems in Modern Physics

4.3 Solutions 289

=

2

15

(3.14159)^5 (1.38065)^4 × 10 −^92

(6. 626068 × 10 −^34 )^3 (2. 99792 × 108 )^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 dνis

N=

8 πν^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=

8 πdλ

λ^4

=

8 π× 20 × 10 −^10

(5× 10 −^7 )^4

= 8. 038 × 1017 /m^3

4.74 (a)

(1)

P=AEλdλ=

8 πhcAdλ

λ^5 (ehc/λkT−1)

(2)

Mean wavelengthλ= 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× 108 )

(5. 5 × 10 −^7 )(1. 38 × 10 −^23 )(4,000)

= 6. 55

Using the above values in (2) we find

P=AEλdλ= 0. 84 × 10 −^6 W= 0. 84 μW.

(b)hν=

hc

λ

=

6. 63 × 10 −^34 × 3 × 108

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 × 1012 /s

4.75uλdλ=

8 πhc

λ^5

1

ehc/λkT− 1

dλ (1)

Putλ=c/ν (2)

and dλ=−

(c

ν^2

)

dν (3)

in the RHS of (1) and simplify

uνdν=

8 πhν^3

c^3 (ehν/kT−1)

dν (4)