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A small amount of solution containing Na^(24) radionulide with activity A= 2.0.10^(3) distintegrations per second was injected in the bloodstream of man. The activity of 1 cm^(3) of blood sample taken t= 5.0 hours later turned out to be A'= 16 disintegrations per miunute per cm^(3). The half life of of the radionuclide is T=15 hours. Find the volume of the man's blood. |
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Answer» Solution :LET `V=` volume of blood in the body of the human beign. Then the total activity of the blood is `A'V`. Assuming all this activity is due to the inhected `Na^(24)` and taking ACCOUNT of the decay of this radionuclide, we get `VA'=Ae^(-lambdat)` Now `lambda=(IN2)/(15) per hour, t= 5 hour` Thus `V=(A)/(A')e^(In 2//3)=(2.0xx10^(3))/((16//60))e^(-In2//3)C c= 5.95 litre` |
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