Correct Answer - `772K`
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` .