The rate of radioactive disintegration at an | vedclass

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(D) The rate of disintegration is given by $R = N \lambda$,where $\lambda$ is the decay constant.The decay constant $\lambda$ is related to the half-l

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$3.17 	imes 10^{19}$

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(D) The rate of disintegration is given by $R = N \lambda$,where $\lambda$ is the decay constant.The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by the formula $\lambda = \frac{0.693}{T_{1/2}}$.Substituting $\lambda$ into the rate equation: $R = N 	imes \frac{0.693}{T_{1/2}}$.Rearranging to solve for the number of atoms $N$: $N = \frac{R 	imes T_{1/2}}{0.693}$.Given $R = 10^{10} \ s^{-1}$ and $T_{1/2} = 2.2 	imes 10^{9} \ s$:$N = \frac{10^{10} 	imes 2.2 	imes 10^{9}}{0.693} = \frac{2.2 	imes 10^{19}}{0.693} \approx 3.17 	imes 10^{19} \ atoms$.

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

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