Modern Physics Ques 127
- A freshly prepared sample of a radioisotope of half-life $1386$ $ s$ has activity $10^{3}$ disintegrations per second. Given that $\ln 2=0.693$, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first $80 $ $s$ after preparation of the sample is
(2013 Adv.)
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Answer:
Correct Answer: 127.$(4)$
Solution:
Formula:
- Number of nuclei decayed in time $t$ is given by $N(t) = N_0 e^{-\lambda t}$
$ \begin{alignedat} N _d & =N _0\left(1-e^{-\lambda t}\right) \\ \therefore \quad \% \text { decayed } & =\left(1 - \frac{N _d}{N _0}\right) \times 100 \` & =\left(1-e^{-\lambda t}\right) \times 100 \end{aligned} $
Here, $\lambda=\frac{0.693}{1386}=5 \times 10^{-4} s^{-1}$
$\therefore \quad %$ decayed $\approx(\lambda t) \times 100$
$ =(5 × 10^{-4})(80)(100)=0.4 $
BETA
