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Radioactivity of a sample ( $\mathrm{z}=22$ ) decreases $90 \%$ after 10 years. What will be the half life of the sample?
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The correct answer is:
3 years
Hints: $t=10$ yrs $\quad t_{\frac{1}{2}}=?$
$$
\lambda=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{N}_0}{\mathrm{~N}_{\mathrm{t}}}
$$
Since radioactivity decreases $90 \%$ in 10 yrs. $\Rightarrow N_0=100 \& N_t=10$
Thus $\lambda=\frac{2.303}{10} \log \frac{100}{10} \Rightarrow \lambda=\frac{2.303}{10}$
since $\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{\lambda}=\frac{2.303 \times \log 2}{\lambda} \Rightarrow \mathrm{t}_{\frac{1}{2}}=\frac{2.303 \times \log 2}{2.303 / 10}$
$\Rightarrow \mathrm{t}_{\frac{1}{2}}=(\log 2) \times 10 \simeq 3$ years
$$
\lambda=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{N}_0}{\mathrm{~N}_{\mathrm{t}}}
$$
Since radioactivity decreases $90 \%$ in 10 yrs. $\Rightarrow N_0=100 \& N_t=10$
Thus $\lambda=\frac{2.303}{10} \log \frac{100}{10} \Rightarrow \lambda=\frac{2.303}{10}$
since $\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{\lambda}=\frac{2.303 \times \log 2}{\lambda} \Rightarrow \mathrm{t}_{\frac{1}{2}}=\frac{2.303 \times \log 2}{2.303 / 10}$
$\Rightarrow \mathrm{t}_{\frac{1}{2}}=(\log 2) \times 10 \simeq 3$ years
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