Question

Half life of a radioactive substance $A$ is two times the half life of another radioactive substance $B$. Initially the number of nuclei of $A$ and $B$ are $N_A$ and $N_B$ respectively. After three half lives of '$A$', number of nuclei of both become equal. The ratio of $\frac{N_A}{N_B}$ will be

• A. $\frac{1}{2}$
• B. $\frac{1}{8}$
• C. $\frac{1}{3}$
• D. $\frac{1}{6}$

Answer 1

The correct option is B $\frac{1}{8}$

The amount remaining after $n$ half lives can be calculated as:
$N=N_0{\left(\frac{1}{2}\right)}^n$
Remaining amount of $A=N_A{\left(\frac{1}{2}\right)}^3$
Remaining amount of $B=N_B{\left(\frac{1}{2}\right)}^6$
$N_A{\left(\frac{1}{2}\right)}^3=N_B{\left(\frac{1}{2}\right)}^6$
$\frac{N_A}{N_B}=\frac{8}{64}=\frac{1}{8}$