1000 Solved Problems in Modern Physics

418 7 Nuclear Physics – I

Fig. 7.18

7.86 The flow of heat in a material placed between the walls of a coaxial cylinder

is given by

dQ

dt

=

2 πL

ln

(

r 2

r 1

)(T 1 −T 2 )(1)

Number of decays of radon atoms per second

dN/dt= 100 × 10 −^3 × 3. 7 × 1010 = 3. 7 × 109 disintegration/second

Energy deposited byα′s=^3.^7 ×^109 ×^5 .5MeV/s

= 2. 035 × 1010 MeV/s= 3. 256 × 10 −^7 J= 0. 779 × 10 −^3 Cal/s

Using the values,k= 0 .025 Cal cm−^2 s−^1 C−^1 ,L=5cm,r 1 =2 mm and

r 2 =6 mm in (1), and solving for (T 1 −T 2 ) we find (T 1 −T 2 )= 1. 09 ◦C

7.87 In the series decay A→B→C, ifλA<λBthe transient equilibrium occurs

when

NB/NA=λA/(λB−λA)

Here A=Uranium and B=Radium

λA= 1 /τA= 0. 693 / 4. 5 × 109 year−^1 ,λB= 1 /τB= 0. 693 / 1 ,620 year−^1

N(Rad)/N(U)≈ 1 , 620 / 4. 5 × 109 = 1 / 2. 78 × 106

7.88 Use the law of radioactivity

NA=NA^0 exp(−λAt)(1)

NB=NB^0 exp(−λBt)(2)

Dividing (2) by (1)

NB/NA=exp(λA−λB)t (BecauseNA^0 =NB^0 )

Take logeon both sides

t=

1

(λA−λB)ln

(

NB

NA

)