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 $\frac{1}{2\lambda }$

Decay equation is given by $N=N_o{e}^{-\lambda t}$

Where, $\lambda$ is Decay constant; $N$ is number of atom at time $t$

So, number of atom of $A$ is given by, $N_A=N_o{e}^{-5\lambda t}$

And that of $B$ are given by, $N_B=N_o{e}^{-\lambda t}$

Taking ration we get: $\frac{N_A}{N_B}=e^{-4\lambda t}$

Given that: $\frac{N_A}{N_B}=e^{-2}$

So, this will happen when, $t=\frac{1}{2\lambda }$