Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The rate of radioactive disintegration at an instant for a radioactive sample of half-life $2.2 \times 10^9 \text{ s}$ is $10^{10} \text{ s}^{-1}$ . The number of radioactive atoms in that sample at that instant is:

3.7 × 10^{20}

3.17 × 10^{17}

3.17 × 10^{18}

3.17 × 10^{19}

Step-by-Step Solution

Recall the Radioactive Decay Formulae: The rate of disintegration ( $R$ or Activity) is given by $R = \lambda N$ , where $\lambda$ is the decay constant and $N$ is the number of radioactive atoms. The decay constant is related to the half-life ( $T_{1/2}$ ) by the formula: $\lambda = \frac{0.693}{T_{1/2}}$ .
Combine Equations: Substituting $\lambda$ into the rate equation: $R = \left( \frac{0.693}{T_{1/2}} \right) N \implies N = \frac{R \cdot T_{1/2}}{0.693}$
Substitute Values: $R = 10^{10} \text{ s}^{-1}$ $T_{1/2} = 2.2 \times 10^9 \text{ s}$ $N = \frac{10^{10} \times 2.2 \times 10^9}{0.693}$
Calculate: $N = \frac{2.2 \times 10^{19}}{0.693} \approx 3.1746 \times 10^{19}$
Conclusion: The number of radioactive atoms is approximately $3.17 \times 10^{19}$ .

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