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(B) Let the initial amount of the radioactive sample be $N_0$.After time $t$,the amount remaining undecayed is $N(t) = 0.9 N_0$.According to the law o

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$19$

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(B) Let the initial amount of the radioactive sample be $N_0$.After time $t$,the amount remaining undecayed is $N(t) = 0.9 N_0$.According to the law of radioactive decay,$N(t) = N_0 e^{-\lambda t}$.Thus,$e^{-\lambda t} = 0.9$.After a total time $2t$,the amount remaining undecayed is $N(2t) = N_0 e^{-\lambda (2t)} = N_0 (e^{-\lambda t})^2$.Substituting the value of $e^{-\lambda t}$,we get $N(2t) = N_0 (0.9)^2 = 0.81 N_0$.The amount that has decayed in time $2t$ is $N_0 - N(2t) = N_0 - 0.81 N_0 = 0.19 N_0$.Therefore,the percentage of the initial sample that decays in time $2t$ is $19\%$.

$90\%$ of a radioactive sample is left undecayed after time $t$ has elapsed. What percentage of the initial sample will decay in a total time $2t$ : ..............$\%$

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