Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is :

38Sulphur=2.48hhalf-life38Cl=0.62hhalf-life38Ar(stable)

Assume that we start with 1000 38S nuclei at time t = 0. The number of 38Cl is of count zero at t = 0 and will again be zero at t = . At what value of t, would the number of counts be a maximum?

Hint: The rate of disintegration depends on the active nuclei.
Step 1: Find the decay rate of two nuclei.
Consider the chain of two decays.
38S2.48h38Cl0.62h38Ar
At time t, let 38S has N1(t) active nuclei and 38Cl has N2(t) active nuclei.
dN1dt=-λ1N1=rateofformationofCl38.
Also,dN2dt=-λ2N2+λ1N1
But,N1=N0e-λ1t
dN2dt=λ1N0e-λ1t-λ2N2.....(i)
Step 2: Find the no. of counts of 38Cl.
Multiplying by eλ2tdt and rearranging
eλ2tdN2+λ2N2eλ2tdt=λ1N0e(λ2-λ1)tdt
Integrating both sides:
N2eλ2t=N0λ1λ2-λ1e(λ2-λ1)t+C
Sinceatt=0,N2=0,C=-N0λ1λ2-λ1
So,N2eλ2t=N0λ1λ2-λ1e(λ2-λ1)t-1..............ii
Step 3: Find the condition when no. of counts of 38Cl is maximum.
For maximum count, dN2dt=0
λ1N0e-λ1t-λ2N2=0[FromEq.(i)]
N0N2=λ2λ1eλ1t[FromEq.(ii)]
eλ2t-λ2λ1·λ1(λ1-λ1)eλ1t[e(λ2-λ1)t-1]=0
oreλ2t-λ2(λ2-λ1)eλ2t+λ2(λ2-λ1)eλ1t=0
1-λ2(λ2-λ1)+λ2(λ2-λ1)e(λ1-λ2)t=0
λ2(λ2-λ1)e(λ1-λ2)t=λ2(λ2-λ1)-1
e(λ1-λ2)t=λ1λ2
Step 4: Find the time when no. of counts of 38Cl is maximum.
t=(logeλ1λ2)(λ1-λ2)
=loge(2.480.62)2.48-0.62
=loge41.86=2.303×2×0.30101.86(λ=0.693T1/2)
=0.745S
Note Do not apply directly the formula of radioactive. Apply formulae related to chain decay.