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(C) The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by the formula $\lambda = \frac{\ln 2}{T_{1/2}}$.For two simultaneous decay pro

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$300$

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(C) The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by the formula $\lambda = \frac{\ln 2}{T_{1/2}}$.For two simultaneous decay processes, the effective decay constant is $\lambda_{	ext{eff}} = \lambda_1 + \lambda_2$.Given $T_1 = 5\, 	ext{min} = 300\, s$ and $T_2 = 30\, s$.Thus, $\lambda_1 = \frac{\ln 2}{300}$ and $\lambda_2 = \frac{\ln 2}{30}$.$\lambda_{	ext{eff}} = \frac{\ln 2}{T_{	ext{eff}}} = \frac{\ln 2}{300} + \frac{\ln 2}{30}$.Dividing by $\ln 2$, we get $\frac{1}{T_{	ext{eff}}} = \frac{1}{300} + \frac{1}{30} = \frac{1 + 10}{300} = \frac{11}{300}$.Therefore, $T_{	ext{eff}} = \frac{300}{11}\, s$.Comparing this with $\frac{\alpha}{11}\, s$, we find $\alpha = 300$.

$A$ radioactive nucleus decays by two different processes. The half-life of the first process is $5$ minutes and that of the second process is $30\,s$. The effective half-life of the nucleus is calculated to be $\frac{\alpha}{11}\,s$. The value of $\alpha$ is $..............$

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