1000 Solved Problems in Modern Physics

1.3 Solutions 79

Mx(t)=Eext=

∑∞

x= 0

extB(x)

=

∑∞

x= 0

(

N

r

)

(pet)r(1−p)N−r

=(pet+ 1 −p)N

∑(N

r

)

prqN−r

=(pet+ 1 −p)N

μ^0 n=

∂nMx(t)

∂tn

|t= 0

Thereforeμ^01 =∂∂Mt|t= 0 =Npet(q+pet)N−^1 |t= 0 =Np

Thus the mean=Np

(c)μ^02 =

∂^2 M

∂t^2

=[N(N−1)p^2 et(q+pet)N−^2 +Npet(q+pet)N−^1 ]t= 0

=N(N−1)p^2 +Np

Butμ 2 =μ^02 −(μ^01 )^2 =N(N−1)p^2 +Np−N^2 p^2

=Np−Np^2 =Np(1−p)=Npq

orσ=

√

Npq

1.95 Total counting rate/minute,m 1 = 245

Background rate/minute,m 2 = 49

Counting rate of source,m=m 1 −m 2 = 196

m 1 =

n 1

t 1

±

√

n 1

t 1

;m 2 =

n 2

t 2

±

√

n 2

t 2

; Net countm=m 1 −m 2

σ=(σ 12 +σ 22 )^1 /^2 =

(

n 1

t 12

+

n 2

t 22

) 1 / 2

=

(

m 1

t 1

+

m 2

t 2

) 1 / 2

σ=

(

m 1

t 1

+

m 2

t 2

) 1 / 2

=

(

49

100

+

245

20

) 1 / 2

= 3. 57

Percentage S.D.=mσ× 100 =^3196.^57 × 100 = 1 .8%

1.96 (a)B(x)=

N!

x!(N−x)!

pxqN−x

Using Sterling’s theorem