JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Modern Physics

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^{131}

I 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} I

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

^{131} I

injected was

2.4 \times 1 0^{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

for

∣ x ∣ ≪ 1

and

ln 2 \approx 0.7

Answer: 5

Solution

The decay constant is

λ = \frac{l n 2}{T _{1/2}} = \frac{0.7}{8 \times 24} = \frac{0.7}{192}

per hour. After time

t = 11.5

hours, the activity decreases to

A (t) = A_{0} e^{- λ t}

. Since

λ t = \frac{0.7 \times 11.5}{192} = \frac{8.05}{192} \approx 0.042 ≪ 1

, we use

e^{- λ t} \approx 1 - λ t \approx 1 - 0.042 \approx 0.958

. The total activity in the body after 11.5 hours is approximately

A (t) \approx 2.4 \times 1 0^{5} \times 0.958 \approx 2.3 \times 1 0^{5}

Bq. If 2.5 ml of blood has activity 115 Bq, then the total volume

V

V = \frac{2.5 \times 2.3 \times 1 0 ^{5}}{115} = \frac{2.5 \times 2.3 \times 1 0 ^{5}}{115} = \frac{5.75 \times 1 0 ^{5}}{115} = 5000

ml = 5 liters.

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Integer answer

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JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Modern Physics

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