Radioactivity of a sample (Z=22) decreases by | vedclass

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(C) The radioactive decay follows first-order kinetics. The decay constant $\lambda$ is given by: $\lambda = \frac{2.303}{t} \log \frac{N_0}{N_t}$.Giv

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(C) The radioactive decay follows first-order kinetics. The decay constant $\lambda$ is given by: $\lambda = \frac{2.303}{t} \log \frac{N_0}{N_t}$.Given that radioactivity decreases by $90 \%$, the remaining amount $N_t = N_0 - 0.90 N_0 = 0.10 N_0$.Substituting the values: $\lambda = \frac{2.303}{10} \log \frac{N_0}{0.10 N_0} = \frac{2.303}{10} \log(10) = \frac{2.303}{10} 	imes 1 = 0.2303 \ year^{-1}$.The half-life $t_{1/2}$ is given by: $t_{1/2} = \frac{0.693}{\lambda}$.$t_{1/2} = \frac{0.693}{0.2303} \approx 3.01 \ years$.Therefore, the half-life is approximately $3 \ years$.

Radioactivity of a sample $(Z=22)$ decreases by $90 \%$ after $10 \ years$. What will be the half-life of the sample (in $years$)?

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