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13.Nuclei
medium
Let $N_{\beta}$ be the number of $\beta $ particles emitted by $1$ gram of $Na^{24}$ radioactive nucler (half life $= 15\, hrs$) in $7.5\, hours$, $N_{\beta}$ is close to (Avogadro number $= 6.023\times10^{23}\,/g.\, mole$)
A
$6.2\times10^{21}$
B
$7.5\times10^{21}$
C
$1.25\times10^{22}$
D
$1.75\times10^{22}$
(JEE MAIN-2015)
Solution
We know that $N_{\beta}=N_{0}\left(1-e^{-\lambda t}\right)$
$N_{\beta}=\frac{6.023 \times 10^{23}}{24} \cdot\left[1-e^{\left(-\frac{\ln ^{2} }{15} \times 75\right)}\right]$
on solving we get, $\mathrm{N}_{\beta}=7.4 \times 10^{21}$
Standard 12
Physics
