A small amount of solution containing $N a^{24}$ radionuclide with activity $A = 2.0.1 0^{3}$ disintegration per second was injected into the bloodstream of a man. The activity of 1 $c m^{3}$ of the blood sample taken t= 5.0 hours later turned out to be A'= 16 disintegration per minute per $c m^{3}$ . The half-life of the radionuclide is T= 15 hours. Find the volume of the man's blood.

Question

Verified by Toppr

Answer 1

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 $N a^{24}$ and taking account of the decay of this radionuclide, we get

$V A^{'} = A e^{- λ t}$

Now $λ = \frac{l n 2}{1 5}$ per hour ,

$t = 15$ hour

Thus $V = \frac{A}{A ^{'}} e^{- l n 2 / 3} = \frac{2 . 0 \times 1 0 ^{3}}{( 1 6 / 6 0 )} e^{- l n 2 / 3} c c = 5.95$ litre

Answer 2

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

N a^{24}

and taking account of the decay of this radionuclide, we get

V A^{'} = A e^{- λ t}

Now

λ = \frac{l n 2}{1 5}

per hour ,

t = 15

hour

Thus

V = \frac{A}{A ^{'}} e^{- l n 2 / 3} = \frac{2 . 0 \times 1 0 ^{3}}{( 1 6 / 6 0 )} e^{- l n 2 / 3} c c = 5.95

litre