JEE Main 26-Aug-2021 Shift 2 Solved Paper

Given, at time t=0,NA(0)=2NB(0)
Decay constant is same for both radioactive atoms as λ.
For A → B ,

dNB(t)

dt

=λNA(t)−λNB(t)
Substituting NA(0)e−λt for NA(t) in above expression, we get

dNB(t)

dt

=λNA(0)e−λt−λNB(t)
=2λNB(0)e−λt−λNB(t)
⇒

dNB(t)

dt

+λNB(t)=2λNB(0)e−λt
Multiplying both sides by eλt, we get
eλt[

dNB(t)

dt

+λNB(t)]=2λNB(0)e−λt×eλt
⇒

d

dt

[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
⇒

NB(t)

NB(0)

=[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

dNB(t)

dt

=0
From Eq. (ii),

dNB(t)

dt

=−λC[1+2λt]e−λt+2Cλe−λt=0
Solving the above expression,
t=

1

2λ

s
Thus, maximum value of function will be at t=

1

2λ

s.
Hence, graph (c) is correct option.