In the above problem if each decay produces $E_0$ energy, then find
(a) power produced at time t
(b) total energy produced upto time t

Answer: A::B::DSolution: (a) :' $N=\frac{\alpha }{\lambda }\left(1-e^{-\lambda t}\right)$
At time t, number of decays per second $=\lambda N=\alpha \left(1-e^{-\lambda t}\right)$
Each decay produces $E_0$ energy. Therefore, energy produced per second or power.
$=\left(\nu mberofdecaysper\sec ond\right)\left(e\ne rgy\prod uced\in eachdecay\right)$=(lambdaN)E_0=alphaE_0(1-e^(-lambdat))$\text{or}$P=alphaE_0(1-e^(-lambdat))$\left(b\right)Powerisafunctionoftime.Theref\text{or}e,\to tale\ne rgy\prod ucedup\to timetcanbeobta\in edby\int egrat\in gthispower\text{or}$E_(Total) = int_0^tPdt$A<ernateMethodE\ne rgyis\prod ucedonly\in decay.Up\to timet\to tal$alphat$\nu c\le iare\prod uced\text{and}N\nu c\le iare\le ft.So,\to tal\nu mberof\nu c\le idecayed.$N_d=alphat-N=alphat-alpha/lambda(1-e^(-lambdat))=alpha[t-1/lambda(1-e^(-lambdat))]$Eachdecay\prod uces$E_0$e\ne rgy.Theref\text{or}e,\to tale\ne rgy\prod ucedup\to timet,$E_(Total) = N_dE_0=alphaE_0[t-1/lambda(1-e^(-lambdat))]