At time t = 0, a material is composed of two radioactive atoms A and… | JEE Main 2021 (26 Aug, Shift 2)

Q71JEE Main 2021 · 26 AUG · SHIFT 2Difficult

Law of Radioactive Decay

At time t = 0, a material is composed of two radioactive atoms A and B, where NA(0) = 2NB(0). The decay constant of both kind of radioactive atoms is $\lambda$ . However, A disintegrates to B and B disintegrates to C. Which of the following figures represents the evolution of NB(t) / NB(0) with respect to time t?

[NA (0) = No. of A atoms at t = 0

NB (0) = No. of B atoms at t = 0]

Show answer and solution

Answer: (c)

$N_A = 2N_0 e^{-\lambda t}$ and $\dfrac{dN_B}{dt} = \lambda N_A - \lambda N_B$ , with $N_B(0) = N_0$ .
Solution: $N_B = N_0(1 + 2\lambda t)e^{-\lambda t}$ , so $\dfrac{N_B}{N_B(0)} = (1 + 2\lambda t)e^{-\lambda t}$ . It starts at 1 and first rises.
Maximum: $\dfrac{d}{dt}\left[(1+2\lambda t)e^{-\lambda t}\right] = \lambda e^{-\lambda t}(1 - 2\lambda t) = 0 \Rightarrow t = \dfrac{1}{2\lambda}$
After that it falls towards zero.
So the answer is (c): starts at 1, peaks at $t = \dfrac{1}{2\lambda}$ , then decays.

Report a mistake

Law of Radioactive Decay question from JEE Main 2021 (26 Aug, Shift 2)

More questions on Law of Radioactive Decay