Question

A certain substance decays to ${\frac{1}{64}}^{th}$ of its initial activity in $24$ days. The half-life of the substance will be

• A. $6$ days
• B. $4$ days
• C. $3$ days
• D. $2$ days

Answer 1

The correct option is B $4$ days

Given $N=\frac{N_0}{64}=\frac{N_0}{2^6}$.
After $n$ half lives number of parent nuclei is
$N=\frac{N_0}{2^n}$.

Thus $24$ days is $6$ half-lives
$T_{\frac{1}{2}}=\frac{24}{6}=4days$