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 $\left(1/e^2\right)$ after a time:

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

Answer 1

The correct option is C $\frac{1}{2\lambda }$

Given : $(N_o{)}_A=(N_o{)}_B=N_o$

Using : $Kt=ln\frac{N}{N_o}$ where, $K$ is the decay constant

$⟹$ $Kt=lnN-ln{N}_o$

For substance A : $\lambda t=ln{N}_A-ln(N_o{)}_A$

$\therefore \lambda t=ln{N}_A-ln{N}_o$ .............(1)

For substance B : $5\lambda t=ln{N}_B-ln(N_o{)}_B$

$\therefore 5\lambda t=ln{N}_B-ln{N}_o$ .............(2)

Subtracting (2) from (1) we get:

$\lambda t-5\lambda t=ln\frac{N_A}{N_B}$

$-4\lambda t=ln\frac{1}{e^2}$

$-4\lambda t=-2$

$⟹t=\frac{1}{2\lambda }$