A radioactive nuclide A of decay…

A radioactive nuclide A of decay constant 2λ decays into a nuclide B of decay constant λ, and B decays into a stable nuclide. At t=0 there are N0…

One or more options correct

A radioactive nuclide

\mathrm{A}

of decay constant

2\lambda

decays into a nuclide

\mathrm{B}

of decay constant

\lambda

, and

\mathrm{B}

decays into a stable nuclide. At

t=0

there are

N_0

nuclei of

\mathrm{A}

and none of

\mathrm{B}

. Choose the correct statement(s).

A. The number of $\mathrm{B}$ nuclei is maximum at $t=\dfrac{\ln 2}{\lambda}$ .
B. The maximum number of $\mathrm{B}$ nuclei is $N_0/4$ .
C. At the instant of maximum $\mathrm{B}$ , the numbers of $\mathrm{A}$ and $\mathrm{B}$ nuclei are equal.
D. For all times after the instant of maximum $\mathrm{B}$ , the activity of $\mathrm{B}$ exceeds the activity of $\mathrm{A}$ .

Show answer and solution

Answer: A, D

Here

N_A=N_0e^{-2\lambda t}

and the Bateman solution gives

N_B=\dfrac{2\lambda N_0}{\lambda-2\lambda}\left(e^{-2\lambda t}-e^{-\lambda t}\right)=2N_0\left(e^{-\lambda t}-e^{-2\lambda t}\right)

. Setting

dN_B/dt=0

means the production rate

2\lambda N_A

equals the decay rate

\lambda N_B

, i.e.

e^{-\lambda t}=1/2

, so

t=\ln2/\lambda

; A is true. At this time

N_B=2N_0(1/2-1/4)=N_0/2

, not

N_0/4

; B is false. There

N_A=N_0/4

, which is half of

N_B

, so C is false. At the maximum, the activities are equal (

2\lambda N_A=\lambda N_B

); afterwards

e^{-\lambda t}

decays more slowly than

e^{-2\lambda t}

, so the activity of B stays larger; D is true.

Original question written in the JEE Advanced 2016 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake