1000 Solved Problems in Modern Physics

4.3 Solutions 283

orΔQ=TΔS=kTlnΔW

=(1. 38 × 10 −^23 )(300) ln 10^8

= 7. 626 × 10 −^20 J= 0 .477 eV

4.58 The Gaussian (normal) distribution is

f(x)=

1

σ

√

2 π

e−(x−μ)

(^2) / 2 σ 2
whereμis the mean andσis the standard deviation. The probability is found
from
(a)P(μ−σ<x<μ+σ)=
∫μ+σ
μ−σ
f(x)dx
Lettingz=x−σμ
P(− 1 <z<1)=

∫^1

− 1

φ(z)dz

= 2

∫ 1

0

φ(z)dz(from symmetry)

= 2 × 0. 3413 = 0 .6826 (from tables)

or 68.26%(shown shaded under the curve, Fig 4.4)

Fig. 4.4

(b) Similarly

P(μ− 2 σ)<x<μ+ 2 σ)= 0 .9544 or 95.44%

(c)P(μ− 3 σ)<x<(μ+ 3 σ)= 0 .9973 or 99.73%

4.59 P(n,T)=

e−

(n+^12 )ω

kT

Σ∞n= 0 e−

(n+^12 )ω

kT

=

e−(n+

(^12) )ω/kT
e−
(^12) ω/kT
Σ∞n= 1 enω/kT