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(B) Let the initial number of nuclei be $N_0$.After $t = 1$ day,the number of decayed nuclei is $10 \%$ of $N_0$,so the number of nuclei remaining is

Vedclass · Accepted Answer

$19$

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(B) Let the initial number of nuclei be $N_0$.After $t = 1$ day,the number of decayed nuclei is $10 \%$ of $N_0$,so the number of nuclei remaining is $N(1) = N_0 - 0.10 N_0 = 0.90 N_0$.Using the radioactive decay law $N(t) = N_0 e^{-\lambda t}$,we have $0.90 N_0 = N_0 e^{-\lambda(1)}$,which gives $e^{-\lambda} = 0.9$.After $t = 2$ days,the number of nuclei remaining is $N(2) = N_0 e^{-\lambda(2)} = N_0 (e^{-\lambda})^2$.Substituting $e^{-\lambda} = 0.9$,we get $N(2) = N_0 (0.9)^2 = 0.81 N_0$.The number of decayed nuclei after $2$ days is $N_{decayed} = N_0 - N(2) = N_0 - 0.81 N_0 = 0.19 N_0$.Thus,the decayed percentage is $19 \%$.

Ten percent of a radioactive sample has decayed in $1$ day. After $2$ days,the decayed percentage of nuclei will be ...... $\%$

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