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 $- \frac{ d N ⁡}{ d ⁡ t} = \lambda N ⁡$ . Hence net rate of accumulation of radio nuclide is given by
$\frac{ d N ⁡}{ d ⁡ t} = q ⁡ - \lambda N ⁡ \ldots \left(1\right)$
or $\int\limits _{0}^{ N } \frac{ d ⁡ N ⁡}{ q ⁡ - \lambda N ⁡} = \int\limits _{0}^{t} d ⁡ t$
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