Question

A radioactive nucleus decays by two different processes. The half life for the first process is $10\text{s}$ and that for the second is $100\text{s}$. The effective half life of the nucleus is close to

• A. $\text{9 sec}$
• B. $\text{6 sec}$
• C. $\text{55 sec}$
• D. $\text{12 sec}$

Answer 1

The correct option is A $\text{9 sec}$

Let ${\lambda }_1$ and ${\lambda }_2$ be the decay constannts of two process. $N$ be the number of nuclei left undecayed after two process. From the law of radioactive decay we have

$-\frac{dN}{dt}={\lambda }_{1}N+{\lambda }_{2}N[\because -\frac{dN}{dt}=\lambda N]$
$\Rightarrow -\frac{dN}{dt}=({\lambda }_1+{\lambda }_2)N$
$\Rightarrow {\lambda }_{eq.}=({\lambda }_1+{\lambda }_2)$
$\Rightarrow \frac{\ln 2}{T}=\frac{\ln 2}{T_1}+\frac{\ln 2}{T_2}(\because \lambda =\frac{\ln 2}{T})$
$\Rightarrow \frac{1}{T}=\frac{1}{T_1}+\frac{1}{T_2}$
$\Rightarrow \frac{1}{T}=\frac{1}{10}+\frac{1}{100}=\frac{11}{100}$

[Given: $T_1=10\text{s}$ & $T_2=100\text{s}$]

$\Rightarrow T\approx 9\text{s}$