Question

Half-life period of a radiactive substance is $10\min$, then amount of substance decayed in $40\min$ will be

• A. $25\%$
• B. $50\%$
• C. $75\%$
• D. None of these

Answer 1

Answer: (D)

Number of half-lives in $40\min$
$n=\frac{40}{T_{1/2}}=\frac{40}{10}=4$
Amount of the substance remaining after $n$ half- lives
$N=N_0{\left(\frac{1}{2}\right)}^n=N_0{\left(\frac{1}{2}\right)}^4=\frac{N_0}{16}$
where, $N_0$ is original amount of the substance.
Amount of the substance decayed in $40\min$.
$N^{\prime }=N_0-N=N_0-\frac{N_0}{16}=\frac{15}{16}{N}_0$
$=\left(\frac{15}{16}\times 100\%\right)$ of $N_0$
$=15\times 6.25\%$ of $N_0$
$=93.75\%$ of $N_0$