A radioactive sample intended for irradiation of a hospital patient is prepared at a nearby laboratory. The sample has a half-life of 83.61h. What should its initial activity be if its activity is to be $\mathbf{7}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{Bq}$when it is used to irradiate the patient 24h later?

Half-life of the sample, ${\mathrm{T}}_{1/2}=83.61\mathrm{h}$

Activity of the sample after $24\mathrm{h},\mathrm{R}=7.4\times {10}^8\mathrm{Bq}$

Time of decay, t = 24 h

The average disintegration per second of a sample is inversely proportional to the exponential value of the decay constant at the given time. Thus, with time the value decreases exponentially from an initial quantity representing its decay per second. Using this concept, we can directly get the required value of the initial quantity that is directly proportional to the exponential.

Formula:

The rate of undecayed sample remaining after a given time, $R=R_0{e}^{-\lambda t}.....\left(1\right)$

where, $R_0$is the initial present amount of the sample.

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

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

Using the given data in the rearranged equation (1) after substituting value from equation (2), we can get the initial activity of the radioactive sample as follows:

$$\begin{gathered} {\mathrm{R}}_0={\mathrm{Re}}^{\mathrm{t}\mathrm{In}2/{\mathrm{T}}_{1/2}} \\ =\left(7.4\times {10}^8\mathrm{Bq}\right){\mathrm{e}}^{\left(24\mathrm{h}\right)\mathrm{In}2/83.61\mathrm{h}} \\ =9.0\times {10}^8 \end{gathered}$$

Hence, the value of initial activity is $9.0\times {10}^8\mathrm{Bq}$.