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JEE Advance - Physics (2017 - Paper 1 Offline - No. 12)

$${}^{131}{\rm I}$$ is an isotope of Iodine that $$B$$ decays to an isotope of Xenon with a half-life of $$8$$ days. A small amount of a serum labelled with $${}^{131}{\rm I}$$ is injected into the blood of a person. The activity of the amount of $${}^{131}{\rm I}$$ 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 $$11.5$$ hours, $$2.5$$ ml of blood is drawn from person's body, and gives an activity of $$115$$ $$Bq$$. The total volume of blood in the person's body, in liters is approximately (you may use $${e^x} \approx 1 + x\,\,$$ for $$\left| x \right| < < 1$$ and $$\ln 2 \approx 0.7).$$
Answer
5

Explanation

The decay rate is related to half life by $$\lambda = {{\ln 2} \over {{t_{1/2}}}}$$. A sample having N0 radioactive nuclei at time t = 0 will have $$N = {N_0}{e^{ - \lambda t}}$$ radioactive nuclei at time t. Thus, activity (A = $$\lambda$$N) of a sample reduces from its initial value A0 to a value $$A = {A_0}{e^{ - \lambda t}}$$ at time t.

Let V be the total volume of blood and N be the total number of radioactive nuclei at time t. The nuclei are distributed uniformly in blood. A sample of volume v will have n = (v/V)N radioactive nuclei in it. Thus, activity of this sample is

$$a = \lambda n = \lambda \left( {{v \over V}N} \right) = {v \over V}(\lambda N) = {v \over V}A = {v \over V}{A_0}{e^{ - \lambda t}}$$.

Substitute a = 115 Bq, v = 2.5 ml, A0 = 2.4 $$\times$$ 105 Bq, $$\lambda$$ = ln 2/t1/2 $$\approx$$ 0.7/(8 $$\times$$ 24) hr$$-$$1 and t = 11.5 hr to get

$$V = {{(2.5)(2.4 \times {{10}^5})} \over {115}}{e^{ - {{0.7 \times 11.5} \over {8 \times 24}}}}$$

$$ \approx {{(2.5)(2.4 \times {{10}^5})} \over {115}}\left( {1 - {{0.7 \times 11.5} \over {8 \times 24}}} \right)$$

= 4998 ml $$\approx$$ 5 litre.

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