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 $40\,\min$.
$N'=N_{0}-N=N_{0}-\frac{N_{0}}{16}=\frac{15}{16} N_{0}$
$=\left(\frac{15}{16} \times 100 \%\right)$ of $N_{0}$
$=15 \times 6.25 \%$ of $N_{0}$
$=93.75 \%$ of $N_{0}$