Q125 / 180 · NEET 2019 (Odisha) · Physics
The rate of radioactive disintegration at an instant for a radioactive sample of half life $2.2 \times 10^9 \mathrm{~s}$ is $10^{10} \mathrm{~s}^{-1}$. The number of radioactive atoms in that sample at that instant is
- A$3.17 \times 10^{20}$
- B$3.17 \times 10^{17}$
- C$3.17 \times 10^{18}$
- Correct answerD$3.17 \times 10^{19}$
Solution
Given, half life $T_{1 / 2}=2.2 \times 10^9 \mathrm{~s}$
Rate of disintegration, $R=10^{10} \mathrm{~s}^{-1}$
If $\mathrm{N}$ be the number of nuclei present, then the rate of disintegration is
$$
\begin{aligned}
\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N} \quad(\lambda=\text { decay constant }) \\
\Rightarrow \quad \mathrm{R}=\lambda \mathrm{N} \quad \text { or } \mathrm{N}=\frac{\mathrm{R}}{\lambda} \quad \ldots \text { (i) }
\end{aligned}
$$
Also, the half life is given by,
$$
\mathrm{T}_{1 / 2}=\frac{0.693}{\lambda} \Rightarrow \lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}
$$
From Eq. (i) and (ii), we get
$$
\mathrm{N}=\frac{\mathrm{R}}{0.693} \times \mathrm{T}_{1 / 2}=\frac{10^{10} \times 2.2 \times 10^9}{0.693}=3.17 \times 10^{19}
$$
Hence, the correct answer is option: 'D'.