A radionuclide with decay constant `lambda` is being produced in a nuclear reactor at a rate `a_(0) t` per second, where `a_(0)` is positive constant and `t` is the time. During each decay, `E_(0)` energy is released. The production of radionuclide starts at time `t=0`.
Instantaneous power developed at time `'t'` due to the decay of the radionuclide is

Options

A. `(a_(0)t+(a_(0))/(lambda)+(a_(0))/(lambda)e^(-lambda t))E_(0)`
B. `(a_(0)t+(a_(0))/(lambda)-(a_(0))/(lambda)e^(-lambda t))E_(0)`
C. `(a_(0)t-(a_(0))/(lambda)-(a_(0))/(lambda)e^(-lambda t))E_(0)`
D. `(A_(0)T-(a_(0))/(lambda)-(a_(0))/(lambda) e^(-lambdat)E_(0))`

Answer

C

Explanation

`N=(at_(0))/(lambda)-(a_(0))/(lambda_(2))+(a_(0))/(lambda^(2))rArr`
`P_("inst")= lambdaNE_(0)=[a_(0)t-(a_(0))/(lambda)+(a_(0))/(lambda)e^(-lambda t)]E_(0)`