A radioactive substance has a half-life of 10 | vedclass

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(C) The half-life of the radioactive substance is $T_{1/2} = 10^8 	ext{ years}$.Converting this into seconds: $T_{1/2} = 10^8 	imes 365 	imes 24

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$4.5 	imes 10^{19}$

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(C) The half-life of the radioactive substance is $T_{1/2} = 10^8 	ext{ years}$.Converting this into seconds: $T_{1/2} = 10^8 	imes 365 	imes 24 	imes 60 	imes 60 \approx 3.15 	imes 10^{15} \,s$.The activity is given as $R = 10^4 \,Bq$.The relationship between activity $R$, decay constant $\lambda$, and number of atoms $N$ is $R = \lambda N$.The decay constant is $\lambda = \frac{0.693}{T_{1/2}}$.Substituting $\lambda$ in the activity equation: $R = \frac{0.693}{T_{1/2}} 	imes N$.Rearranging for $N$: $N = \frac{R 	imes T_{1/2}}{0.693}$.Substituting the values: $N = \frac{10^4 	imes 3.15 	imes 10^{15}}{0.693} \approx 4.54 	imes 10^{19}$.Thus, the number of atoms present is approximately $4.5 	imes 10^{19}$.

$A$ radioactive substance has a half-life of $10^8$ years and an activity of $10^4 \,Bq$. The number of atoms of this substance present is

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