At t = 0 activity of radioactive substance is 1600 Bq and at t = 8 sec activity remains 100 Bq. The activity at t = 2 sec is :

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800 Bq

400 Bq

600 Bq

200 Bq

answer is D.

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Detailed Solution

$t = 0 N_{0} = 1600 B q$

$t = 8 s N = 100 B q$

$t = 2 s N = ?$

$\frac{N}{N_{0}} = {(\frac{1}{2})}^{\frac{t}{T}}$

$\Rightarrow \frac{100}{1600} = {(\frac{1}{2})}^{\frac{8}{T}}$

$\Rightarrow \frac{1}{2^{4}} = \frac{1}{2^{\frac{8}{T}}}$

$4 = \frac{8}{T}$

$T_{\frac{1}{2}} = T = 2 s$

$\frac{N}{1600} = {(\frac{1}{2})}^{\frac{2}{2}} = \frac{1}{2}$

$N = 800 B q$

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