 # A radioactive nucleus decays by two different processes. Question:

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

1. (1) $9 \mathrm{sec}$.

2. (2) $6 \mathrm{sec}$.

3. (3) $55 \mathrm{sec}$

4. (4) $12 \mathrm{sec} .$

Correct Option: 1

Solution:

(1) Let $\lambda_{1}$ and $\lambda_{2}$ be the decay constants of two process. $N$ be the number of nuclei left undecayed after two process. From the law of radioactive decay we have]

$-\frac{d N}{d t}=\lambda_{1} N+\lambda_{2} N \quad\left[\because-\frac{d N}{d t}=\lambda N\right]$

$\Rightarrow-\frac{d N}{d t}=\left(\lambda_{1}+\lambda_{2}\right) N$

$\Rightarrow \lambda_{\text {eq. }}=\left(\lambda_{1}+\lambda_{2}\right)$

$\Rightarrow \frac{\ln 2}{T}=\frac{\ln 2}{T_{1}}+\frac{\ln 2}{T_{2}}$                $\left(\because \lambda=\frac{\ln 2}{T}\right)$

$\Rightarrow \frac{1}{T}=\frac{1}{T_{1}}+\frac{1}{T_{2}}$

$\Rightarrow \frac{1}{T}=\frac{1}{10}+\frac{1}{100}=\frac{11}{100} \quad$ [Given: $T_{1}=10 \mathrm{~s} \& T_{2}=100 \mathrm{~s}$ ]

$\Rightarrow T=\frac{100}{11}=9 \mathrm{sec}$