Question

A radio nuclide of ${}^{32}\text{P}$ with half-life 14.3 day is produced in a nuclear reactor at a constant rate, $q=2.7\times {10}^9$ nuclei per second. Period soon after the beginning of production of that nuclide will its activity be equal to $1.7\times {10}^9$ dps is :

• A. $9.5day$
• B. $7.5day$
• C. $6day$
• D. None of these

Answer 1

The correct option is A $9.5day$

The radio nucleide is formed at a constant rate $q$.
$Thedecayrate-\frac{dN}{dt}=\lambda .N$
$Therateofaccumulation=\frac{dN}{dt}=(q-\lambda .N)$
${\int }_0^N\frac{dN}{q-\lambda N}={\int }_0^{t}dt$
$-\frac{1}{\lambda }\left[{\log }_e\right(q-\lambda N){]}_0^N=t$
$t=-\frac{1}{\lambda }\log \left[\frac{q-\lambda N}{q}\right]$
$=\frac{2.303}{\lambda }{\log }_{10}\left[\frac{q}{q-\lambda N}\right]$
$=\frac{2.303\times 14.3}{0.693}\log \frac{2.7\times {10}^9}{1.7\times {10}^9}$
$=9.5day$