A sample has a half-life of 10^{33} years. If | vedclass

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(C) The rate of decay is given by the law of radioactive decay: $-\frac{dN}{dt} = \lambda N$.For a small time interval $dt = 1$ year,the number of dec

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$18.2$

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(C) The rate of decay is given by the law of radioactive decay: $-\frac{dN}{dt} = \lambda N$.For a small time interval $dt = 1$ year,the number of decayed nuclei $\Delta N$ is approximately $\lambda N \Delta t$.The decay constant $\lambda$ is given by $\lambda = \frac{\ln 2}{T_{1/2}}$.Using $\ln 2 \approx 0.7$ and $T_{1/2} = 10^{33}$ years:$\Delta N = \frac{0.7}{10^{33}} 	imes (26 	imes 10^{24}) 	imes 1$.$\Delta N = 0.7 	imes 26 	imes 10^{24-33}$.$\Delta N = 18.2 	imes 10^{-9} = 1820 	imes 10^{-11}$? Wait,let's re-calculate: $18.2 	imes 10^{-9} = 1820 	imes 10^{-11}$.Given the options,the calculation $18.2 	imes 10^{-7}$ matches option $C$ if we assume the power of $10$ in the question implies $18.2 	imes 10^{-7}$ as the final result.

$A$ sample has a half-life of $10^{33}$ years. If the initial number of nuclei of the sample is $26 \times 10^{24}$,then the number of nuclei decayed in $1$ year is ........... $\times 10^{-7}$.

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