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A freshly prepared sample of a radio iso tope of half-life 1386 \mathrm{sec} has activity 10^3disintegration per second. Given that \ln_2=0.693, the friction of the initial number of nuclei that will decay in the first 80 \mathrm{~sec} after preparation of the sample is:-

Option: 1

3%


Option: 2

4%


Option: 3

5%


Option: 4

6%


Answers (1)

best_answer

Number of nuclei decayed in time t,
%\begin{aligned} \mathrm{N_d=N_0 }&\mathrm{ \left(1-e^{-\lambda t}\right) }\\ \therefore \quad \text { decayed } &\mathrm{ =\left(\frac{N_d}{N_0}\right) \times 100 }\\ &\mathrm{ =\left(1-e^{-\lambda t} d\right) \times 100}\\ &\mathrm{Here, \lambda=\frac{0.693}{1386}=5 \times 10^{-4} \mathrm{~s}^{-1}}\\ \therefore \% \text { decayed } &\mathrm{ =(\lambda t) \times 100 }\\ &\mathrm{ =\left(5 \times 10^{-4}\right) \times 80 \times 100 }\\ &\mathrm{ =4 }\\ \end{aligned}

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Suraj Bhandari

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