A dose of $\mathbf{8}\mathbf{.}\mathbf{60}\mathbf{}\mathbf{\mu Ci}$of a radioactive isotope is injected into a patient. The isotope has a half-life of 3.0h. How many of the isotope parents are injected?

The rate of decay of the radioactive isotope,$\mathrm{R}=8.60\mathrm{\mu Ci}\left(\frac{{10}^{-6}\mathrm{Ci}}{1\mathrm{\mu Ci}}\right)=8.60\times {10}^{-6}\mathrm{Ci}$

The half-life of the isotope,${\mathrm{T}}_{1/2}=3.0\mathrm{h}\left(\frac{86400\mathrm{s}}{1\mathrm{h}}\right)=1.08\times {10}^4\mathrm{s}$

The radioactive decay is due to the loss of the elementary particles from an unstable nucleus to convert them into a more stable one. Thus, a material with unstable nuclei can give two or more atoms production for the given decay process. Thus the required value of the number of nuclei is found using the number of nuclei relation in to the decay rate of the substance.

Formulae:

The rate of decay,$\mathrm{R}=\mathrm{\lambda N}$ (1)

where,$\lambda$is the disintegration constant, and Nis the number of undecayed nuclei.

The disintegration constant,$\mathrm{\lambda }=\frac{\mathrm{In}2}{{\mathrm{T}}_{1/2}}$ (2)

where,${\mathrm{T}}_{1/2}$is the half-life of the substance.

At first, we calculate the disintegration constant of the nuclide using the given half-life in equation (2) as follows:

$\begin{array}{c}\mathrm{\lambda }=\frac{\ln 2}{\text{1.08}\times {\text{10}}^4\text{\hspace{0.17em}s}}\\ =6.42\times {10}^{-5}/\text{s}\end{array}$

Thus, the number of isotope parents injected is given using the above value in equation (1) as follows:

$\begin{array}{c}\mathrm{N}=\frac{\mathrm{R}}{\mathrm{\lambda }}\\ =\frac{\text{8.60}\times {\text{10}}^{-6}\text{\hspace{0.17em}Ci}\times \text{3.7}\times {\text{10}}^{10}\text{\hspace{0.17em}Bq/Ci}}{6.42\times {10}^{-5}/\text{s}}\\ =4.96\times {10}^9\end{array}$

Hence, the value of the injected isotope parents is $4.96\times {10}^9$.