Question

${}^{131}I\;$is an isotope of Iodine that $\beta$ decays to an isotope of Xenon with a half-life of $8$ days. A small amount of a serum labelled with ${}^{131}I\;$ is injected into the blood of a person. The activity of the amount of ${}^{131}I\;$ injected was $2.4 \times {10^5}\;Becquerel{\rm{ }}\left( {Bq} \right)$. It is known that the injected serum will get distributed uniformly in the bloodstream in less than half an hour. After $11.5$ hours, $2.5ml\;$ of blood is drawn from the person's body, and gives an activity of $115\;Bq$. The total volume of blood in the person's body, in litres is approximately (you may use ${e^x} \approx 1 + x\;for\;\mid x\mid < < 1\;and\;ln2 \approx 0.7$).

Answer 1

Here, total volume we need to find out here the volume can be obtained if we know the final concentration of the isotope. If N is the number of nuclei in the sample and $\Delta N$ undergo decay in time $\Delta t$ then $\dfrac{{\Delta N}}{{\Delta t}} \propto N$ or $\dfrac{{\Delta N}}{{\Delta t}} = - \lambda N$, where, $\lambda$ is called the radioactive decay constant or disintegration constant.it is used to derive law of radioactive decay.

Complete Step by step answer: So We can use the following relation to find out the required value.
$\dfrac{{\Delta N}}{{\Delta t}} = - \lambda N$
$\dfrac{{dN}}{N} = - \lambda t$
Integrating both sides,
$\int {_{{N_0}}^N} \dfrac{{dN}}{N} = - \lambda \int {_{{t_0}}^t} t$
$\Rightarrow \ln \left( {\dfrac{{N}}{{{N_0}}}} \right) = - \lambda t$
$\Rightarrow N = {N_0}{e^{ - \lambda t}}$
This is the law of radioactive decay.
Hence, on the basis of the above equation we can derive activity of a radioactive substance: The total decay rate R of a sample of one or more radionuclides is called the activity of that sample.
$R = \dfrac{{ - dN}}{{dt}}\; = \lambda N{e^{ - \lambda t}} = {R_0}{e^{ - \lambda t}} = \lambda N$
Here, ${R_0}$ is the radioactive decay rate at time$t{\rm{ }} = {\rm{ }}0$, and R is the rate at any subsequent time t.
Given data:

$${t_{\dfrac{1}{2}}} = 8days\; = 8 \times \;24\;hrs\\\ {R_0} = \;2.4 \times \;{10^5}Bq$$

Using,

$$\Rightarrow \dfrac{{0.692}}{{{t_{\dfrac{1}{2}}}}} \times t = \ln \dfrac{{{R_0}}}{R}\\\ \Rightarrow \dfrac{{0.692}}{{8 \times 24}} \times 11.5 = \ln \dfrac{{2.4 \times {{10}^5}}}{R}\\\ \Rightarrow \dfrac{{2.4 \times {{10}^5}}}{R} = {e^{0.041}} = 1 + 0.041\\\ \Rightarrow R = \dfrac{{2.4 \times {{10}^5}}}{{1.041}} = 2.3 \times {10^5}$$

Now, doing further calculation,

$$115\;Bq\;i{\mathop{\rm s}\nolimits} \;in\;volume = 2.5ml\\\ 2.3 \times {10^5}\;Bq\;i{\mathop{\rm s}\nolimits} \;in\;volume = \dfrac{{2.5}}{{115}} \times 2.3 \times {10^5}\\\ \Rightarrow = 0.05 \times {10^5}\\\ \Rightarrow = 5 \times {10^3}ml\;or\;5\;litres$$

By knowing the relationship between different units of radioactivity we can get to the appropriate answer,

$$1\;becquerel = 1\;Bq = 1\;radioactive\;deca\;yper\;second\\\ 1\;curie = 1\;Ci = {3.710^{10}}\;Bq(decays\;per\;second)$$

Hence, $5\;litres$ is the total volume of blood in the person's body.

Note: In the same way ageing of the tree is estimated by using carbon dating technique. The carbon isotope carbon-$14$is used as a radioactive isotope. For mean value of a radioactive substance: Mean life $\left( \tau \right)$ = sum of life of all atoms / total no of atoms present. $\tau = \dfrac{1}{\lambda }$ , where, $\lambda$ is the decay constant.