Question

Let $N_{\beta }$ be the number of $\beta$ particles emitted by $1$ gram of ${}^{24}Na$ radioactive nuclei (half life $=15hrs)$ in $7.5$ hours, $N_{\beta }$ is close to (Avogadro number $=6.023\times {10}^{23}/g$ mole)

• A. $1.75\times {10}^{22}$
• B. $6.2\times {10}^{21}$
• C. $7.5\times {10}^{21}$
• D. $1.25\times {10}^{22}$

Answer 1

The correct option is C $7.5\times {10}^{21}$

If there are N radioactive nuclei at some time $t$, then the number $dN$ which would decay in any given time interval $dt$ would be proportional to $N:$

$\frac{dN}{dt}=kN$

$[lnN{]}_{N_{\circ }}N=-kt$

$N=N_{\circ }{e}^{-kt}$

given $N=\frac{N}{2}$ when $t=t_{\frac{1}{2}}$

$K=\frac{ln2}{t_{\frac{1}{2}}}$

finding the remaining molecules left after $t=7\frac{1}{2}$ hrs

$N=N_{\circ }{e}^{-k7\frac{1}{2}}$

$N=\frac{N_{\circ }}{\sqrt{2}}$

Number of particles emitted = $N_{\circ }-\frac{N_{\circ }}{\sqrt{2}}$

$N_{\circ }-\frac{N_{\circ }}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}\frac{1}{24}\times 6.023\times {10}^{23}=7.34\times {10}^{21}$, hence correct answer is option C