In time t,7/8 of a radioactive sample disinte | vedclass

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(B) The fraction of the sample remaining after time $t$ is $N/N_0 = 1 - 7/8 = 1/8$.Using the radioactive decay law $N = N_0(1/2)^n$,we have $1/8 = (1/

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$\frac{4}{3}t$

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(B) The fraction of the sample remaining after time $t$ is $N/N_0 = 1 - 7/8 = 1/8$.Using the radioactive decay law $N = N_0(1/2)^n$,we have $1/8 = (1/2)^n$,which implies $(1/2)^3 = (1/2)^n$,so $n = 3$.Since $n = t/T_{1/2}$,the half-life is $T_{1/2} = t/3$.Now,for $15/16$ of the sample to disintegrate,the remaining fraction is $N/N_0 = 1 - 15/16 = 1/16$.Using $N/N_0 = (1/2)^n$,we have $1/16 = (1/2)^n$,which implies $(1/2)^4 = (1/2)^n$,so $n = 4$.The total time required is $t' = n 	imes T_{1/2} = 4 	imes (t/3) = \frac{4}{3}t$.

In time $t$,$7/8$ of a radioactive sample disintegrates. How much time will it take for $15/16$ of the sample to disintegrate?

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