A radioactive nucleus decays by two different process 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 _______
To find the effective half-life of a radioactive nucleus decaying by two processes, we use the formula for the combined half-life when two decay processes are occurring simultaneously. The formula for the effective half-life \( T_{\text{eff}} \) is given by: \[ \frac{1}{T_{\text{eff}}} = \frac{1}{T_1} + \frac{1}{T_2} \] where \( T_1 \) and \( T_2 \) are the half-lives of the two processes. Given \( T_1 = 300\,s \) (5 minutes) and \( T_2 = 30\,s \), we have: \[ \frac{1}{T_{\text{eff}}} = \frac{1}{300} + \frac{1}{30} \] Calculate each term: \[ \frac{1}{300} = 0.00333 \] \[ \frac{1}{30} = 0.03333 \] Adding these, we get: \[ \frac{1}{T_{\text{eff}}} = 0.00333 + 0.03333 = 0.03666 \] Thus, \[ T_{\text{eff}} = \frac{1}{0.03666} \approx 27.27\,s \] The effective half-life \( T_{\text{eff}} \) is given as \( \frac{\alpha}{11} \) seconds, so: \[ \frac{\alpha}{11} = 27.27 \] Solving for \( \alpha \), we get: \[ \alpha = 27.27 \times 11 = 299.97 \approx 300 \] So, the value of \( \alpha \) is 300, which falls within the expected range of 300 to 300.
