Question

The half-life of that radioactive substance, which reduces to $1/64$ of its initial value in $15$ hours, will be

• A. $5hours$
• B. $2hours$
• C. $2.5hours$
• D. $4hours$

Answer 1

The correct option is C $2.5hours$

For radioactive decay
$\frac{N}{N_a}={\left(\frac{1}{2}\right)}^n$ where $n$ is the number of half lives
$\therefore \frac{1}{64}={\left(\frac{1}{2}\right)}^n$
or $n=6$
$T_{1/2}=\frac{t}{n}=\frac{15}{6}=2.5hrs$.