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Q.
The half life of $_{92} U ^{238}$ against $\alpha$ -decay is $4.5 \times 10^{9}$ years. How many disintegrations occur per second in one gram of $_{92} U^{238}$ ?
Nuclei
Solution:
$T =4.5 \times 10^{9}$ years
$T=4.5 \times 10^{9} \times 365 \times 24 \times 60 \times 60 $
$\lambda=\frac{0.693}{ T } s ^{-1}$
Number of atoms in one gram of $_{92} U ^{238}$ is
$ N =\frac{6.03 \times 10^{23}}{238} $
$\frac{ dN }{ dt }=\lambda N$
$ \Rightarrow \frac{ dN }{ dt }=\frac{0.693 \times 6.03 \times 10^{23}}{238 \times 4.5 \times 10^{9} \times 365 \times 24 \times 60 \times 60} $
$\frac{ dN }{ dt }=2.34 \times 10^{4} s ^{-1}$