Question

Radioactive material '$A$' has decay constant '$8\lambda$' and material '$B$' has decay constant '$\lambda$'. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A' will be $\frac{1}{e}$?

• A. $\frac{1}{9\lambda }$
• B. $\frac{1}{\lambda }$
• C. $\frac{1}{7\lambda }$
• D. $\frac{1}{8\lambda }$

Answer 1

The correct option is C $\frac{1}{7\lambda }$

As number of nuclei $N=N_0{e}^{-\lambda t}$
Where $\lambda$ is the decay constant and $N_o$ is the initial number of nuclei
Given, initialy the number of nuclei are same so, $N_{A_0}=N_{B_0}=N_0$ and$\frac{N_B}{N_A}=\frac{1}{e}$

For A, $\begin{array}{}{N}_A=N_0{e}^{(-8\lambda )t}& \text{(1)}\end{array}$
For B, $\begin{array}{}{N}_B=N_0{e}^{-\lambda t}& \text{(2)}\end{array}$

Dividing eq.(2) by eq.(1)
$\Rightarrow e^{-1}=\frac{e^{-\lambda t}}{e^{-8\lambda t}}$
$\Rightarrow e^{7\lambda t}=e^{-1}$
$7\lambda t=-1$
$t=-\frac{1}{7\lambda }$
So, the best option is $t=\frac{1}{7\lambda }$
There is slight error in the question, since the decay constent of material A is greater than material B. Hence the ratio of number of nuclie of material B to that of A is always greater than $1$, but it is given less then 1 $\left(\frac{1}{e}\right)$.