The rate of radioactive disintegration at an instant for a radioactive sample of half-life $2.2 imes 10^9

Question

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

Accepted Answer

1.  **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: $\lambda = \frac{0.693}{T_{1/2}}$ .
2.  **Combine Equations:** Substituting $\lambda$ into the rate equation:
    $$ R = \left( \frac{0.693}{T_{1/2}} ight) N \implies N = \frac{R \cdot T_{1/2}}{0.693} $$
3.  **Substitute Values:**
    *   $R = 10^{10} 	ext{ s}^{-1}$
    *   $T_{1/2} = 2.2 	imes 10^9 	ext{ s}$
    $$ N = \frac{10^{10} 	imes 2.2 	imes 10^9}{0.693} $$
4.  **Calculate:**
    $$ N = \frac{2.2 	imes 10^{19}}{0.693} \approx 3.1746 	imes 10^{19} $$
5.  **Conclusion:** The number of radioactive atoms is approximately $3.17 	imes 10^{19}$.

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