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

Hint: 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.