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(D) The radioactive decay law states that the activity $A$ at time $t$ is given by $A = A_0 (1/2)^n$,where $n = t/T_{1/2}$ is the number of half-lives

Vedclass · Accepted Answer

$200$

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(D) The radioactive decay law states that the activity $A$ at time $t$ is given by $A = A_0 (1/2)^n$,where $n = t/T_{1/2}$ is the number of half-lives.Given $A_0 = 1600 \ counts \ s^{-1}$ at $t = 0$ and $A = 100 \ counts \ s^{-1}$ at $t = 8 \ s$.Substituting these values: $100 = 1600 	imes (1/2)^{8/T_{1/2}}$.$1/16 = (1/2)^{8/T_{1/2}} \Rightarrow (1/2)^4 = (1/2)^{8/T_{1/2}}$.Equating the exponents: $4 = 8/T_{1/2} \Rightarrow T_{1/2} = 2 \ s$.Now,for $t = 6 \ s$,the number of half-lives $n = 6/2 = 3$.The activity at $t = 6 \ s$ is $A = A_0 	imes (1/2)^n = 1600 	imes (1/2)^3$.$A = 1600 	imes (1/8) = 200 \ counts \ s^{-1}$.

The counting rate observed from a radioactive source at $t = 0$ was $1600 \ counts \ s^{-1}$,and at $t = 8 \ s$,it was $100 \ counts \ s^{-1}$. The counting rate observed as $counts \ s^{-1}$ at $t = 6 \ s$ will be

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