A radio nuclide with disintegration constant `lambda` is produced in a reactor at a constant rate `alpha` nuclei per second. During each decay energy `E_0` is released. `20%` of this energy is utilized in increasing the temperature of water. Find the increase in temperature of `m` mass of water in time `t`. Specific heat of water is `s`. Assume that there is no loss of energy through water surface.

Answer

`772K`

Explanation

Rate of formation of nuclei at time t is
`(dN)/(dt) =alpha-lambdaN`
Where `N=`number of nuclei at that time.
`N=(alpha)/(lambda)(1-e^(- lambda t))`
Number of nuclei distintegrarted in time `t`,
`N_(d)=alphat-(alpha)/(lambda)(1-e^(-lambdat))`
Therefore, energy released till time `t`,
`E=e_(0)E_(0)N_(d)[alphat-(alpha)/(lambda)(1-e^(-lambdat))]`

`20%` of this energy is used for heating water.Hence,
`0.2E_(0)[alphat=(alpha)/(lambda)(1-e^(-lambdat))]`
`rArr Delta theta= (0.2E_(0)[alphat-(alpha)/(lambda)(1-(1)/(2))])/(ms)`
At `t=T_(1//2)`,
`Delta theta =(0.2E_(0)[2 lambdaT_(1//2)-(2lambda)/(lambda)(1-(1)/(2))])/(ms)`
`=0.2E_0[0.386]/ms=772 K` .