Given, at time
t=0,NA(0)=2NB(0) Decay constant is same for both radioactive atoms as λ.
For A → B ,
=λNA(t)−λNB(t) Substituting
NA(0)e−λt for
NA(t) in above expression, we get
=λNA(0)e−λt−λNB(t) =2λNB(0)e−λt−λNB(t) ⇒+λNB(t)=2λNB(0)e−λt Multiplying both sides by
eλt, we get
eλt[+λNB(t)]=2λNB(0)e−λt×eλt ⇒[NB(t)eλt]=2λNB(0) Integrating both sides,
[NB(t)eλt]=2λNB(0)t+C ...(i)
Putting t = 0 in above expression,
[NB(0)eλ×0]=2λNB(0)×0+C ⇒C=NB(0) Putting value of C in Eq. (i)
[NB(t)eλt]=2λNB(0)t+NB(0) ...(ii)
⇒NB(t)=NB(0)[1+2λt]e−λt ⇒=[1+2λt]e−λt As,
NB(t)=NB(0)[1+2λt)e−λt ⇒NB(t)=C[1+2λt]e−λt ...(iii)
The maximum value of
NB(t) is obtained at
=0 From Eq. (ii),
=−λC[1+2λt]e−λt+2Cλe−λt=0 Solving the above expression,
t=s Thus, maximum value of function will be at
t=s.
Hence, graph (c) is correct option.