A solution containing active cobalt `60/27` Co having activity of `0.8 muCi` and decay constant `lambda` is injected in an animal's body. If `1cm^(3)` of blood is drawn from the animal's body after 10hrs of injection , the activity found was 300 decays per minute. What is the volume of blood that is flowing in the body ? (`1 Ci = 3.7 xx 10^(10)` decays per second and at `t = 10` hrs `e^(-lambdat) = 0.84`)

Options

A. 4 litres
B. 6 litres
C. 5 litres
D. 7 litres

Answer

C

Explanation

Let V be the total volume of blood and corresponding activity is A which is given by: `A=A_(0)e^(-lambdat)` where`A_(0)=0.8mu Ci = 0.8 xx 10^(-6) xx 3.7 xx 10^(10)` decays per second.
Let v be the volume of blood taken from the animal’s body. Activity for the v volume can be written as follows:
`A. =A/Vv rArr A. =(A_(0)e^(-lambdat))/V v`
`V=v A_(0)/A^(.)e^(-lambdat)`.....(1)
It is given that at t = 10 hrs e u = 0.84, hence on substituting the values we get the following:
`rArr V=(1 cm^(3))(0.8 xx 10^(-6) xx 3.7 xx 10^(10))/(300//60)(0.84)`
`=4.97 xx 10^(3) cm^(3)=4.97` lit s