The radionuclide `.^(56)Mn` is being produced in a cyclontron at a constant rate `P` by bombarding a manganese target with deutrons. `.^(56)Mn` has a half-life of `2.5 h` and the target contains large numbers of only the stable manganese isotopes `.^(56)Mn`. The reaction that produces `^(56)Mn` is
`.^(56)Mn` +`d` rarr `.^(56)Mn` +`p`
After being bombarded for a long time, the activity of `.^(56)Mn` becomes constant, equal to `13.86 xx 10^(10) s^(-1)`. (Use `l n2=0.693`, Avagadro number` =6 xx 10^(23)`, atomic weight of `.^(56)Mn`=`56 g mol^(-1))`.
After the activity of `.^(56)Mn` becomes constant, number of `.^(56)Mn` nuclei present in the target is equal to .

Options

A. `5xx10^(11)`
B. `20xx10^(11)`
C. `1.2xx10^(14)`
D. `1.8xx10^(15)`

Answer

D

Explanation

As Rate of decay=Rate of production
`P= lambda N rArr N=(P)/(lambda)=(Pt_(1//2))/(ln 2)= 1.8xx10^(15)`