Question 428: Atoms & Nuclei

Question

Two radioactive materials $A$ and $B$ having decay constant ' $7 \lambda$ ' and ' $\lambda$ ' respectively, initially have same number of nuclei. The time taken to have the ratio of number of nuclei of material B to that of A as ' e ' is

Accepted Answer

Let the initial number of nuclei of both $A$ and $B$ be $N_0$.
The number of nuclei left after time $t$ for any radioactive material is given by:
$$ N = N_0 e^{-\lambda t} $$
For material $A$ (decay constant $= 7\lambda$):
$$ N_A = N_0 e^{-7\lambda t} $$
For material $B$ (decay constant $= \lambda$):
$$ N_B = N_0 e^{-\lambda t} $$
We are told that the ratio $\frac{N_B}{N_A} = e$ at time $t$.
So,
$ \frac{N_B}{N_A} = \frac{N_0 e^{-\lambda t}}{N_0 e^{-7\lambda t}} = \frac{e^{-\lambda t}}{e^{-7\lambda t}} $
$ \frac{N_B}{N_A} = e^{- \lambda t + 7\lambda t} = e^{6\lambda t} $
Given:
$ e^{6\lambda t} = e $
Take natural logarithm (log base $e$) on both sides:
$ 6\lambda t = 1 $
$ t = \frac{1}{6\lambda} $
Correct answer:
Option B: $\boxed{\frac{1}{6\lambda}}$

Answer

$\frac{1}{\lambda}$

Answer

$\frac{1}{6 \lambda}$

Answer

$\frac{1}{7 \lambda}$

Answer

$\frac{1}{8 \lambda}$

Atoms & Nuclei

Two radioactive materials A and B having decay constant ' 7 \lambda ' and ' \lambda ' respectively, initially have same number of nuclei. The time taken to have the ratio of number of nuclei of material B to that of A as ' e ' is

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