Q153 / 200 · WBJEE 2013 · Physics
The number of atoms of a radioactive substance of half-life $T$ is $N_{0}$ at $t=0 .$ The time necessary to decay from $N_{0} / 2$ atoms to $N_{0} / 10$ atoms will be
- A$\frac{5}{2} T$
- B$\operatorname{tlog} 5$
- Correct answerC$T \log \left[\frac{5}{2}\right]$
- D$\frac{T}{2} \log 5$
Solution
We have, $\frac{N(t)}{N_{0}}=e^{-\lambda t}$ or $N(t)=N_{0} e^{-\lambda t}$
$\therefore$ For the given condition. $\frac{N_{0}}{2}=N_{0} e^{-\lambda t_{1}}$ and $\frac{N_{0}}{10}=N_{0} e^{-\lambda t_{2}}$
$\Rightarrow \quad \frac{1}{2}=e^{-\lambda t_{1}}$ and $\frac{1}{10}=e^{-\lambda t_{2}}$
or $\quad e^{\lambda t_{1}}=2$ and $e^{\lambda t_{2}}=10$
Taking log on both sides, $\lambda t_{1}=\log 2 \Rightarrow t_{1}=\frac{\log 2}{\lambda}$
or $\quad t_{1}=\frac{\log 2 \times \mathrm{T}}{\log 2}$
and $\quad \lambda t_{2}=\log 10$
$\Rightarrow \quad t_{2}=\frac{\log 10}{\lambda}$
of $\quad t_{2}=\frac{\log 10 \times \mathrm{T}}{\log 2}$
$\therefore \quad\left(t_{2}-t_{1}\right)=T\left[\frac{\log 10}{\log 2}-1\right]=T\left[\frac{\log 10-\log 2}{\log 2}\right]$
$=T\left[\frac{\log 5}{\log 2}\right]$
$\Rightarrow\left(t_{2}-t_{1}\right)=T \left[\frac{\log 5}{\log 2}\right]$
Hence, the correct answer is option: 'C'.