SATHEE: Modern Physics 3 Question 26

Modern Physics 3 Question 26

27. $^{131} I$ 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 $2.4 \times 10^{5}$ Becquerel $(B q)$ . It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After $11.5 h, 2.5 m l$ of blood is drawn from the person’s body, and gives an activity of $115 B q$ . 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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Solution:

$I^{131} T_{1 / 2} = 8$ Days $\to X e^{131} + β$

$A_{0} = 2.4 \times 10^{5} B q = λ N_{0}$

Let the volume is $V$ ,

$\begin{aligned} t & = 0 A_{0} = λ N_{0} \\ t & = 11.5 h A = λ N \\ 115 & = λ \frac{N}{V} \times 2.5 \\ 115 & = \frac{λ}{V} \times 2.5 \times (N_{0} e^{- λ t}) \\ 115 & = \frac{(N_{0} λ)}{V} \times (2.5) \times e^{- \frac{\ln 2}{8 day} (11.5 h)} \\ 115 & = \frac{(2.4 \times 10^{5})}{V} \times (2.5) \times e^{- 1 / 24} \\ V & = \frac{2.4 \times 10^{5}}{115} \times 2.5 1 - \frac{1}{24} \\ = \frac{2.4 \times 10^{5}}{115} \times 2.5 \frac{23}{24} \\ = \frac{10^{5} \times 23 \times 25}{115 \times 10^{2}} = 5 \times 10^{3} m l = 5 L \end{aligned}$

Modern Physics 3 Question 26

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Modern Physics 3 Question 26

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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