Modern Physics Ques 125
- ${ }^{131} I$ 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 $(Bq)$. It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After $11.5$ $ h, 2.5$ $ ml$ 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 $e^{2} \approx 1+x$ for $|x|<1$ and $\ln 2 \approx 0.7$ ).
(2017 Adv.)
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Answer:
Correct Answer: 125.$(5)$
Solution:
Formula:
- $I^{131}$ $\quad T _{1 / 2}=8$ Days $\quad Xe^{131}+\beta$
$A _0=2.4 \times 10^{5} Bq=\lambda N _0$
Let the volume is $V$,
$ \begin{aligned} t & =0 \quad A _0=\lambda N _0 \\ t & =11.5 h \quad A=\lambda N \\ 115 & =\lambda \quad (\frac{N}{V} \times 2.5) \\ 115 & =\frac{\lambda}{V} \times 2.5 \times\left(N _0 e^{-\lambda t}\right) \\ 115 & =\frac{\left(N _0 \lambda\right)}{V} \times(2.5) \times e^{-\frac{\ln 2}{8 \text { day }}(11.5 h)} \\ 115 & =\frac{\left(2.4 \times 10^{5}\right)}{V} \times(2.5) \times e^{-1 / 24} \\ V & =\frac{2.4 \times 10^{5}}{115} \times 2.5 \left[1-\frac{1}{24}\right] \\ & =\frac{2.4 \times 10^{5}}{115} \times 2.5 \left[\frac{23}{24}\right] \\ & =\frac{10^{5} \times 23 \times 25}{115 \times 10^{2}}=5 \times 10^{3} ml=5 L \end{aligned} $
BETA
