Question

Two radioactive substances $A$ and $B$ have decay constants $5\, \lambda$ and $\lambda$ respectively. At $t=0$ they have the same number of nuclei. The ratio of number of nuclei of $A$ to those of $B$ will be $\left(\frac{1}{e}\right)^{2}$ after a time interval

• A. $ \frac{1}{4\lambda } $
• B. $ 4\lambda $
• C. $ 2\lambda $
• D. $ \frac{1}{2\lambda } $

Answer 1

The Correct Option is D

Number of nuclei remained after time $t$ can be written as
$N=N_{0} e^{-\lambda t}$
where $N_{0}$ is initial number of nuclei of both the substances.
$N_{1}=N_{0} e^{-5 \lambda t}$ ...(i)
$N_{2}=N_{0} e^{-\lambda t}$ ...(ii)
Dividing E (i) by E (ii), we obtain
$\frac{N_{1}}{N_{2}}=e^{(-5 \lambda+\lambda) t}=e^{-4 \lambda t}=\frac{1}{e^{4 \lambda t}}$
But, we have given
$\frac{N_{1}}{N_{2}}=\left(\frac{1}{e}\right)^{2}=\frac{1}{e^{2}}$
Hence, $\frac{1}{e^{2}}=\frac{1}{e^{4 \lambda t}}$
Comparing the powers, we get
$2=4 \lambda t$
or $t=\frac{2}{4 \lambda}=\frac{1}{2 \lambda}$