Q157 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Physics
A radioactive element which can decay by two processes, has half-life $t_1$ for first process and half-life $t_2$ for second process. Let $\langle t\rangle$ be the effective average-life of this element.
Which of the following is correct?
- A$\langle t\rangle < \frac{t_1 t_2}{t_1+t_2}$
- Correct answerB$\langle t\rangle=\frac{t_1 t_2}{t_1+t_2}$
- C$\langle t\rangle>\frac{t_1 t_2}{t_1+t_2}$
- D$\langle t\rangle=\ln 2\left(\frac{t_1+t_2}{t_1 t_2}\right)$
Solution
Radioactive decay rate is directly proportional to number of nucleus present at any instant,
$\frac{d N}{d t}=-\lambda N$ $\ldots(\mathrm{i})$
Here, $\lambda$ is a constant related to half-life $\left(\lambda=\frac{\ln 2}{T_{1 / 2}}\right)$,
where, $T_{1 / 2}$ is half-life time.
In given case, total rate of decay is
$\frac{d N}{d t}=-\left(\lambda_1+\lambda_2\right) N=-\lambda_{\text {effective }} \times N$ $\ldots(\mathrm{ii})$
Here, $\lambda_1=\frac{\ln 2}{t_1} \text { and } \lambda_2=\frac{\ln 2}{t_2}$
Similarly, $\lambda_{\text {effective }}=\frac{\ln 2}{t}$
Substituting these values in Eq. (ii), we have
$\left(\frac{\ln 2}{T_1}+\frac{\ln 2}{T_2}\right) N=\frac{\ln 2}{t} N$
$\Rightarrow \quad \frac{1}{t}=\frac{1}{t_1}+\frac{1}{t_2}=\frac{t_1+t_2}{t_1 t_2}$ or $\quad t=\frac{t_1 t_2}{t_1+t_2}$
Hence, the correct answer is option: 'B'.