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JEE Advanced 2017 Paper 1

Section: Physics

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Question : 10 of 54

Marks: +1, -0

^{131}l

is an isotope of Iodine that β decays to an isotope of Xenon with a half-life of 8 days. A small amount of a serum labelled with

^{131}l

is injected into the blood of a person. The activity of the amount of

131^{I}

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 the 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

\lvert x \rvert < < 1

2 \approx 0.7

[JEE Adv 2017 P1]

Your Answer:

Solution:

t_{\frac{1}{2}} = 8 \; \text{days} \; = 8 \times 24 \text{ hr}

R_{0} = 2.4 \times 10^{5} \text{ Bq}

\; \text{Using} \; \; \frac{0.691}{t_{\frac{1}{2}}} \times t = \ln \frac{R_0}{R}

\frac{0.692}{8 \times 24} \times 11.5 = \ln \frac{2.4 \times 10^{5}}{R}

\frac{2.4 \times 10^{5}}{R} = e^{0.041} = 1 + 0.041

R = \frac{2.4 \times 10^{5}}{1.041} = 2.3 \times 10^{5}

115 \text{ Bq} \; \text{is in volume} \; = 2.5 \text{ ml}

2.3 \times 10^{5} \text{ Bq} \; \text{is in volume} \; = \frac{2.5}{115} \times 2.3 \times 10^{5}

= 0.05 \times 10^{5}

= 5 \times 10^{3} \text{ ml}

= 5 \; \text{litres} \;

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