Question 187: Atoms & Nuclei

Question

At time t = 0, a material is composed of two radioactive atoms A and B, where N
A
(0) = 2N
B
(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 N
B
(t) / N
B
(0) with respect to time t?
[N
A
(0) = No. of A atoms at t = 0
N
B
(0) = No. of B atoms at t = 0]

Accepted Answer

A $$	o$$ B, B $$	o$$ C
$${{d{N_B}} \over {dt}} = \lambda {N_A} - \lambda {N_B}$$
$${{d{N_B}} \over {dt}} = 2\lambda {N_{{B_0}}}{e^{ - \lambda t}} - \lambda {N_B}$$
$${e^{ - \lambda t}}\left( {{{d{N_B}} \over {dt}} + \lambda {N_B}} ight) = 2\lambda {N_{{B_0}}}{e^{ - \lambda t}} 	imes {e^{\lambda t}}$$
$${d \over {dt}}({N_B}{e^{\lambda t}}) = 2\lambda {N_{{B_0}}}$$, on integrating
$${N_B}{e^{\lambda t}} = 2\lambda t{N_{{B_0}}} + {N_{{B_0}}}$$
$${N_B} = {N_{{B_0}}}[1 + 2\lambda t]{e^{ - \lambda t}}$$
$${{d{N_B}} \over {dt}} = 0$$ at $$ - \lambda [1 + 2\lambda t){e^{ - \lambda t}} + 2\lambda {e^{ - \lambda t}} = 0$$
$${N_{{B_{\max }}}}$$ at $$t = {1 \over {2\lambda }}$$

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Atoms & Nuclei

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