At time t = 0, a container has `N_(0)` radioactive atoms with a decay constant `lambda`. In addition, c numbers of atoms of the same type are being added to the container per unit time. How many atoms of this type are there at t = T ?

Options

A. `(c)/(lamda)` exp `(-lamda T)- N_(0)` exp `(-lamda T)`
B. `(c)/(lamda)` exp `(-lamda T) + N_(0)` exp `(-lamda T)`
C. `(c)/(lamda) {1- "exp" (-lamda T)} + N_(0)` exp `(-lamda T)`
D. `(c)/(lamda) {1+ "exp" (-lamda T)} + N_(0)` exp `(- lamda T)`

Answer

C

Explanation

Given, At t = 0 initial Nucleation `= N_(0)` Before Addition the number of atoms per unit time `(dN)/(dt)= - lamda N_(0) e^(-lamda t)`
After addition (atoms per unit time) `(dN)/(dt)= - lamda N_(0) e^(-lamda t) + Ce^(- lamda t)`
`rArr dN= - lamda N_(0) e^(- lamda t) dt + Ce^(-lamda t) dt`
Now, on integrating above equation
`N= (- lamda N_(0)e^(- lamda t))/(- lamda) + (C)/(- lamda) e^(- lamda t) + K`
`rArr N = N_(0) e^(-lamda t)- (C)/(lamda) e^(-lamda t) + K` ...(i)
where K is the constant of Integration
At t = 0 sec
`N_(0)= N_(0) e^(0) - (C)/(lamda) e^(0) + K`
`rArr N_(0) = N_(0) - (C)/(lamda) + K`
`rArr K= (C)/(lamda) t= T` (Given)
On putting value of K in eq. (i) we get,
`N= N_(0)e^(- lamda T) - (C)/(lamda) e^(- lamda T) + (C)/(lamda)`
`N = (C)/(lamda) (1- e^(-lamda T)) + N_(0) e^(-lamda T)`