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The disintegration rate of a certain radioactive sample at any instant is $4250$ disintegrations per minute. $10$ minutes later, the rate becomes $2250$ disintegrations per minute. The approximate decay constant is : $\mathrm{\text { (Take } \left.\log _{10} 1.88=0.274\right)}$
Option: 1
$0.02 \mathrm{~min}^{-1}$

Option: 2
$2.7 \mathrm{~min}^{-1}$

Option: 3
$0.063 \mathrm{~min}^{-1}$

Option: 4
$6.3 \mathrm{~min}^{-1}$

Answers (1)

best_answer

$\mathrm{A_0=4250}$

$\mathrm{A_t=2250 \text { disintegration per min}}$

$\mathrm{A_t=A_0 2^{-t / T_{v_2}}}$

$\mathrm{2250=42502^{-10 / T_{1 / 2}}}$

$\mathrm{\frac{9}{17}=2^{-10 / T_{1 / 2}}}$

$\mathrm{-\log _{10}\left(\frac{9}{17}\right)=\frac{10}{T_{1 / 2}} \log _{10} 2}$

$\mathrm{\because \frac{\ln (2)}{T_{1 / 2}}=\lambda}$

$\mathrm{\left(\log _e 10\right) \frac{\log _{10}(1.88)}{10}=\lambda}$

$\mathrm{\lambda=0.0274 \times 2.303}$

$\mathrm{=0.063 \mathrm{~min}^{-1}}$

Hence (3) is correct option.

Posted by

HARSH KANKARIA

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