1000 Solved Problems in Modern Physics

420 7 Nuclear Physics – I

Now at timet

NE=NE^0 exp(−λEt) (3)

Solution of (2) is

NF=Aexp(−λEt)+Bexp(−λFt)(4)

whereAandBare constants which can be determined by the use of the

initial conditions. Att=0, the initial number ofFis zero, that isNF^0 =0.

Using this condition in (4) givesB=−Aand (4) becomes

NF=A[exp(−λEt)−exp(−λFt)] (5)

Further, dNF/dt=−λEAexp(−λEt)+λFAexp(−λFt)

Att= 0

dNF^0 /dt=λENE^0 =−λEA+λFA

orA=λENA^0 /(λB−λA)(6)

Using (6) in (5)

NF=

λENE^0

λF−λE

[exp(−λEt)−exp(−λFt)] (7)

The time at which the greatest number of RaF atoms is obtained by differ-

entiatingNFwith respect totin (7) and setting dNF/dt= 0

We findtmax=

1

λF−λE

ln

(

λF

λE

)

(8)

λE= 0. 693 / 5 = 0 .1386 day−^1 ,λF= 0. 693 / 138 = 0 .0052 day−^1

Number of bismuth atoms= 6. 02 × 1023 × 5 × 10 −^10 / 210 = 1. 43 × 1012

Using these values in (6) and (5) we find

T= 24 .6 days

NF(max)= 1. 37 × 1010

7.93 In the series decay, A→B→C, ifλA<< λB, that isτA>> τB, secular

equilibrium is reached andNB/NA=λA/λB=T 1 / 2 (B)/T 1 / 2 (A)

Here A=Radium and B=Radon

T 1 / 2 (Radium)=N(radium).T 1 / 2 (Radon)/N(radon)

=

(

1

6. 4 × 10 −^6

)(

3. 825

365

)

= 1 ,637 years

7.94 Activity|dN 1 /dt|=λN 1

|dN 2 /dt|=λN 2 =λN 1 exp(−λt)

Fractional decrease of activity

=[λN 1 −λN 1 exp(−λt)]/λN 1 = 1 −exp(−λt)= 4 / 100

Or exp(−λt)= 24 / 25

λ=

1

t

ln

(

25

24

)

Putt=1h,λ= 0. 0408

τ= 1 /λ= 24 .5h