JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Modern Physics

At time

t = 0

, a material is composed of two radioactive atoms

A

and

B

, where

N_{A} (0) = 2 N_{B} (0)

. The decay constant of both kind of radioactive atoms is

λ

. However, A disintegrates to B and B disintegrates to

C

. Which of the following figures represents the evolution of

N_{B} (t) / N_{B} (0)

with respect to time

t

[N_{A} (0) = No. of A atomsat t = 0 N_{B} (0) = No. of Batomsat t = 0]

Answer: (A)

Solution

A only decays, so

N_{A} (t) = N_{A} (0) e^{- λ t} = 2 N_{B} (0) e^{- λ t}

. B is fed by A and decays with the same constant:

\frac{d N _{B}}{d t} = λ N_{A} - λ N_{B} = 2 λ N_{B} (0) e^{- λ t} - λ N_{B}

Putting

u = N_{B} e^{λ t}

reduces this to

\frac{d u}{d t} = 2 λ N_{B} (0)

, so

u = N_{B} (0) (1 + 2 λ t)

and

\frac{N _{B} ( t )}{N _{B} ( 0 )} = (1 + 2 λ t) e^{- λ t}

Its derivative is

λ e^{- λ t} (1 - 2 λ t)

, which vanishes at

t = \frac{1}{2 λ}

.
The required curve therefore starts at the value

1

, rises to a maximum of

2 e^{- 1/2} \approx 1.21

t = \frac{1}{2 λ}

, and then falls away to zero.

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option