A radioactive sample decays by 10% in one mon | vedclass

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(A) Let the initial amount of the radioactive sample be $N_0$. After $t = 1$ month,the amount remaining is $N = N_0(1 - 0.10) = 0.90N_0$.Using the rad

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$34.39$

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(A) Let the initial amount of the radioactive sample be $N_0$. After $t = 1$ month,the amount remaining is $N = N_0(1 - 0.10) = 0.90N_0$.Using the radioactive decay law $N = N_0 e^{-\lambda t}$,for $t = 1$ month:$0.90N_0 = N_0 e^{-\lambda(1)} \Rightarrow e^{-\lambda} = 0.90$.After $t = 4$ months,the amount remaining is $N' = N_0 e^{-\lambda(4)} = N_0 (e^{-\lambda})^4$.Substituting the value of $e^{-\lambda}$:$N' = N_0 (0.90)^4 = N_0 (0.6561)$.The fraction of the sample that has decayed is $1 - \frac{N'}{N_0} = 1 - 0.6561 = 0.3439$.Converting to percentage,the decayed fraction is $34.39\%$.

$A$ radioactive sample decays by $10\%$ in one month. What percentage of the sample will decay in four months (in $\%$)?

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