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Q.

A solution containing active cobalt $_{27}^{60} Co$ having an activity of 0.8 $μ$ Ci and decay constant $λ$ is injected in an animal's body. When 1cm³ blood is drawn from the animal's body after 10 h of injection, the activity was found to be 300dpm (decays per minute). then the total volume of blood in the animal's body is close to [1 Ci = 3.7 x 10¹⁰ dps(decays per second) and at t = 10 h the value of $e^{- λt}$ = 0.84]

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a

4 L

b

8 L

c

5 L

d

7 L

answer is C.

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Detailed Solution

Let total volume of blood is V, initial activity $A_{0} = 0.8 μCi$ its activity at time t $A = A_{0} e^{- λt}$
activity of x volume $A^{'} = (\frac{A}{V}) X = X (\frac{A_{0}}{V}) e^{- λt} \Rightarrow V = X (\frac{A_{0}}{A^{1}}) e^{- λt}$
$V = (1 {cm}^{3}) (\frac{8 \times 10^{- 7} \times 3 .7 \times 10^{10}}{300 / 60}) (0 .84) = 4 .97 \times 10^{3} {cm}^{3} = 4 .97 litre$

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A solution containing active cobalt 2760Co having an activity of 0.8 μ Ci and decay constant λ is injected in an animal's body. When 1cm3 blood is drawn from the animal's body after 10 h of injection, the activity was found to be 300dpm (decays per minute). then the total volume of blood in the animal's body is close to [1 Ci = 3.7 x 1010 dps(decays per second) and at t = 10 h the value of e−λt = 0.84]