A radioactive sample decays frac{7}{8} times | vedclass

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Question

(A) Let the initial quantity be $N_0$.Given that the sample decays by $\frac{7}{8}$ of its original quantity,the remaining quantity $N$ is:$N = N_0 -

Vedclass · Accepted Answer

$5$

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(A) Let the initial quantity be $N_0$.Given that the sample decays by $\frac{7}{8}$ of its original quantity,the remaining quantity $N$ is:$N = N_0 - \frac{7}{8}N_0 = \frac{1}{8}N_0$.We know the radioactive decay law is $N = N_0 (\frac{1}{2})^n$,where $n$ is the number of half-lives.$\frac{1}{8}N_0 = N_0 (\frac{1}{2})^n$$(\frac{1}{2})^3 = (\frac{1}{2})^n$Thus,$n = 3$.Since $n = \frac{t}{T_{1/2}}$,where $t = 15$ minutes and $T_{1/2}$ is the half-life:$3 = \frac{15}{T_{1/2}}$$T_{1/2} = \frac{15}{3} = 5$ minutes.

$A$ radioactive sample decays $\frac{7}{8}$ times its original quantity in $15$ minutes. The half-life of the sample is $......$ minutes.

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