A radioactive with decay constant `lambda` is being produced in a nuclear ractor at a rate `q_0` per second, where `q_(0)` is a positive constant and `t` is the time. During each decay, `E_(0)` energy is released. The production of radionuclide starts at time`t=0`.
Average power developed in time t due to the decay of the radionuclide is

Options

A. `((a_(0)t)/(2)-(a_(0))/(lambda)+(a_(0))/(lambda^(2)t)-(a_(0))/(lambda^(2)t)e^(-lambdat))E_(0)`
B. `((a_(0)t)/(2)+(a_(0))/(lambda)+(a_(0))/(lambda^(2)t)-(a_(0))/(lambda^(2)t)e^(-lambdat))E_(0)`
C. `((a_(0)t)/(2)+(a_(0))/(lambda)+(a_(0))/(lambda^(2)t)+(a_(0))/(lambda^(2)t)e^(-lambdat))E_(0)`
D. `((a_(0)t)/(2)+(a_(0))/(lambda)+(a_(0))/(lambda^(2)t)+(a_(0))/(lambda^(2)t)e^(-lambdat))E_(0)`

Answer

A

Explanation

`P_(av)=([overset(t)underset(0)inta_(0)t-(a_(0))/(lambda)+(a_(0))/(lambda)e^(-lambda t)]E_(0)dt)/(underset(0)overset(t)intdt)`
`=[(a_(0)t)/(2)-(a_(0))/(lambda)+(a_(0))/(lambda^(2)t)-(a_(0))/(lambda^(2)t)e^(-lambda t)]E_(0)`
`(Delta m) C^(2)=E rArr Delta m=3.63xx10^(-8)m`