Question

A ${}^{32}P$ radionuclide with half-life $T=14.3$ days is produced in a reactor at a constant rate $q=2.7\times 10^9$ nuclei per second. How soon after the beginning of production of that nuclide will its activity be equal to $A=1.0\times 10^9$ disintegrations/s ?

• A. 9.5 days
• B. 8 days
• C. 7.5 days
• D. 6 days

Answer 1

Answer: (A)

According to the problem, the radio nuclide is being produced at a constant rate as well as it decay with rate $-\frac{dN}{dt}=\lambda N$ . Hence net rate of accumulation of radio nuclide is given by
$\frac{dN}{dt}=q-\lambda N\dots \left(1\right)$
or $\underset{0}{\overset{N}{\int }}\frac{dN}{q-\lambda N}=\underset{0}{\overset{t}{\int }}dt$
or $-\frac{1}{\lambda }{\left(\left[\log \left(q-\lambda N\right)\right]\right)}_0^N=t$
or $-\frac{1}{\lambda }\ln \left[\frac{\left(q-\lambda N\right)}{q}\right]=t$
or $t=\frac{{\left(T\right)}_{1/2}}{\ln 2}\ln \left(1-\frac{A}{q}\right)$
where $\lambda =\frac{\ln 2}{T_{1/2}}$ and $A=\lambda N$
On substituting numerical values, we get
$t=9.5$ days