1000 Solved Problems in Modern Physics

4.3 Solutions 287

4.67 Power radiated,P=σAT^4 = 4 πR^2 σT^4

P 2

P 1

=

R^22

R^21

.

T 24

T 14

=

(4R 1 )^2

R^21

.

(2T 1 )^4

T 14

= 256

Furthermore,

P 2

P 1

=

dQ 2 /dt

dQ 1 /dt

=

m 2 s(dT/dt) 2

m 1 s(dT/dt) 1

wheresis the specific heat

Butm 2 ∝R 23 andm 1 ∝R^31

∴

(dT/dt) 2

(dT/dt) 1

=

P 2

P 1

.

R^31

R^32

=

256

43

= 4

4.68 (a)λm.T=b

T=

b

λm

=

2. 897 × 10 −^3

1 × 10 −^6

= 2 ,897 K

P 2

P 1

=

T 24

T 14

= 2

New temperature,T 2 =T 1 × 21 /^4 = 2 , 897 × 1. 189 = 3 ,445K

(b) The wavelength at which the radiation has maximum intensity

λm=

2. 897 × 10 −^3

3445

= 0. 84 × 10 −^6 m= 0. 84 μm

4.69 The mean value∈is determined from;

∈=

Σ∞n= 0 n∈e−βn∈

Σ∞n= 0 e−βn∈

=−

d

dβ

ln

∑∞

n= 0

e−βn∈

=−

d

dβ

ln

(

1 +e−β∈+e−^2 β∈+···

)

=−

d

dβ

ln

1

1 −e−β∈

where we have used the formula for the sum of terms of an infinite geometric

series.

∈=

∈e−β∈

1 −e−β∈

=

∈

eβ∈− 1

(β= 1 /kT)

4.70 (a)

uλdλ=

8 πhc

λ^5

.

1

ehc/λkT− 1

dλ (Planck’s formula) (1)

For long wavelengths (low frequencies) and high temperatures the ratio

hc

λkT1 so that we can expand the exponential in (1) and retain only the

first two terms

uλdλ=

8 πhc

λ^5 [(1+hc/λkT+...)−1]

=

8 πkT

λ^4

dλ

writingλ=cυ;dλ=−υc 2 dν