Question

A `P^(32)` radionuclide with half-life `T= 14.3` days is prodiced in a reactor at a constant rate `q= 2.7.10^(9)` nuclei per second. How soon after the beiginning of production of that radionuclide will its activity be equal to `A= 1.0.10^(9) dis//s`?

Answer 1

`(dN)/(dt)= g(dN)/(dt)=underset("supply")underset (uarr) g-lambda underset(decay)underset(darr)N`
We see that `N` will approach a constant value `(g)/(lambda)`. This can also be proved directly. Multiply by `e^(lambda t)` and write
`(dN)/(dt)e^(lambda t)+lambdae^(lambda t)N="ge"^(lambda t)`
Then `(d)/(dt) (Ne^(lambda t))= "ge"^(lambda t)`
or `Ne^(lambda t)=(g)/(lambda)e^(lambda t)+ const`
At `t=0` when the production is started, `N=0`
`0= (g)/(lambda)+constant`
Hence `N=(g)/(lambda)(1-e^(-lambda t))`
Now the activity is `A= lambdaN= g(1-e^(-lambda t))`
From the problem
`(1)/(2.7)= 1-e^(-lambda t)`
This gives `lambda t= 0.463`
so `t=(4.463)/(lambda)=(0.463xxT)/(0.693)=9.5 days`
Algebraically `t=-(T)/(In 2) In(1-(A)/(g))`