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(B) The decay constant for the first process is $\lambda_1 = \frac{\ln 2}{T_1}$ and for the second process is $\lambda_2 = \frac{\ln 2}{T_2}$.Since th

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$4762 	ext{ yr}$

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(B) The decay constant for the first process is $\lambda_1 = \frac{\ln 2}{T_1}$ and for the second process is $\lambda_2 = \frac{\ln 2}{T_2}$.Since the nucleus decays by two processes, the effective decay constant is $\lambda = \lambda_1 + \lambda_2$.Substituting the expressions for decay constants, we get $\frac{\ln 2}{T} = \frac{\ln 2}{T_1} + \frac{\ln 2}{T_2}$, which simplifies to $\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2}$.Thus, the effective half-life $T$ is given by $T = \frac{T_1 T_2}{T_1 + T_2}$.Given $T_1 = 5 	imes 10^3 	ext{ yr}$ and $T_2 = 10^5 	ext{ yr} = 100 	imes 10^3 	ext{ yr}$.$T = \frac{(5 	imes 10^3) 	imes (100 	imes 10^3)}{5 	imes 10^3 + 100 	imes 10^3} = \frac{500 	imes 10^6}{105 	imes 10^3} = \frac{500000}{105} \approx 4761.9 	ext{ yr}$.Rounding to the nearest integer, we get $4762 	ext{ yr}$.

$A$ radioactive nucleus can decay by two different processes. The half-lives of the first and second decay processes are $5 \times 10^3$ years and $10^5$ years respectively. Then, the effective half-life of the nucleus is

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