Question

# A small amount of solution containing $$Na^{24}$$ radionuclide with activity $$A= 2.0.10^3$$ disintegration per second was injected into the bloodstream of a man. The activity of 1 $$cm^3$$ of the blood sample taken t= 5.0 hours later turned out to be A'= 16 disintegration per minute per $$cm^3$$ . The half-life of the radionuclide is T= 15 hours. Find the volume of the man's blood.

Solution

## Let $$V$$ = volume of blood in the body of the human being. Then the total activity of the blood is $$A'V$$. Assuming all this activity is due to the injected $$Na^{24}$$ and taking account of the decay of this radionuclide, we get                          $$V A' = A\ e^{-\lambda\ t}$$ Now                $$\lambda = \dfrac{ln\ 2}{15}$$ per hour ,                                                 $$t = 15$$ hourThus             $$V = \dfrac{A}{A'} e^{-ln\ 2/3} = \dfrac{2.0 \times 10^{3}}{(16 / 60)} e^{-ln\ 2/3} cc = 5.95$$ litre PhysicsNCERTStandard XII

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