A radioactive element which can decay by two | vedclass

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Question

(C) The radioactive decay rate is given by $\frac{dN}{dt} = -\lambda N$. For two simultaneous decay processes, the total decay constant is $\lambda_{\

Vedclass · Accepted Answer

$\langle t \rangle  >  \frac{t_1 t_2}{t_1+t_2}$

Vedclass · Answer

(C) The radioactive decay rate is given by $\frac{dN}{dt} = -\lambda N$. For two simultaneous decay processes, the total decay constant is $\lambda_{	ext{eff}} = \lambda_1 + \lambda_2$. Since $\lambda = \frac{\ln 2}{T_{1/2}}$, we have $\frac{\ln 2}{T_{	ext{eff}}} = \frac{\ln 2}{t_1} + \frac{\ln 2}{t_2}$. This simplifies to $\frac{1}{T_{	ext{eff}}} = \frac{1}{t_1} + \frac{1}{t_2}$, so $T_{	ext{eff}} = \frac{t_1 t_2}{t_1+t_2}$. The average life $\langle t angle$ is related to the effective half-life by $\langle t angle = \frac{T_{	ext{eff}}}{\ln 2}$. Substituting $T_{	ext{eff}}$, we get $\langle t angle = \frac{1}{\ln 2} \left( \frac{t_1 t_2}{t_1+t_2} ight)$. Since $\ln 2 \approx 0.693    1$. Therefore, $\langle t angle  >  \frac{t_1 t_2}{t_1+t_2}$.

$A$ radioactive element which can decay by two processes has half-life $t_1$ for the first process and half-life $t_2$ for the second process. Let $\langle t \rangle$ be the effective average-life of this element. Which of the following is correct?

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