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 $30 \mathrm{~s}$. The effective half-life of the nucleus is calculated to be $\frac{\alpha}{11} \mathrm{~s}$. The value of $\alpha$ is __________.
Answer (integer)
300
Solution
<p>$\Rightarrow {\lambda _{eff}} = {\lambda _1} + {\lambda _2}$</p>
<p>$$ \Rightarrow {{\ln 2} \over {{t_{1/2}}}} = {{\ln 2} \over {{{({t_{1/2}})}_1}}} + {{\ln 2} \over {{{({t_{1/2}})}_2}}}$$</p>
<p>$$ \Rightarrow {t_{1/2}} = {{{{({t_{1/2}})}_1} \times {{({t_{1/2}})}_2}} \over {{{({t_{1/2}})}_1} + {{({t_{1/2}})}_2}}} = {{300 \times 30} \over {300 + 30}}s = {{300} \over {11}}s$$</p>
<p>$\Rightarrow \alpha = 300$</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Radioactivity
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