Search any question & find its solution
Question:
Answered & Verified by Expert
$15 / 16$ th of a radioactive sample disintegrates in $2 \mathrm{~h}$. Mean life of radioactive sample is approximately
Options:
Solution:
1081 Upvotes
Verified Answer
The correct answer is:
$43 \mathrm{~min}$
The undecayed nuclei in a radioactive sample are given by
Given, $\quad N=N_{0}-\frac{15}{16} N_{0}=\frac{1}{16} N_{0}, t=2 \mathrm{~h}$
$\Rightarrow \quad \frac{N_{0}}{16}=N_{0}\left(\frac{1}{2}\right)^{\frac{2}{t_{1 / 2}}}$
$\Rightarrow \quad\left(\frac{1}{2}\right)^{\frac{2}{t_{1 / 2}}}=\frac{1}{16}=\left(\frac{1}{2}\right)^{4}$
$\Rightarrow \quad \frac{2}{t_{t / 2}}=4 \Rightarrow t_{1 / 2}=\frac{1}{2} \mathrm{~h}=30 \mathrm{~min}$
$\therefore$ Mean life, $\quad \tau=1.44 t_{1 / 2}$ $=1.44 \times 30=43.2 \mathrm{~min} \simeq 43 \mathrm{~min}$
Given, $\quad N=N_{0}-\frac{15}{16} N_{0}=\frac{1}{16} N_{0}, t=2 \mathrm{~h}$
$\Rightarrow \quad \frac{N_{0}}{16}=N_{0}\left(\frac{1}{2}\right)^{\frac{2}{t_{1 / 2}}}$
$\Rightarrow \quad\left(\frac{1}{2}\right)^{\frac{2}{t_{1 / 2}}}=\frac{1}{16}=\left(\frac{1}{2}\right)^{4}$
$\Rightarrow \quad \frac{2}{t_{t / 2}}=4 \Rightarrow t_{1 / 2}=\frac{1}{2} \mathrm{~h}=30 \mathrm{~min}$
$\therefore$ Mean life, $\quad \tau=1.44 t_{1 / 2}$ $=1.44 \times 30=43.2 \mathrm{~min} \simeq 43 \mathrm{~min}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.