1.

In a nuclear reactor an element `X` decays to a radio active element `Y` at a constant rate `10^(15)` atoms per sec. Each decay release `100 MeV` energy. Half life of `Y` equals `T` and decays to a stable product `Z`. Each decay of `Y` releases `50 MeV`. All energy released inside the reactor is used to produce electricity at an afficiency of `25%`. Calculate the electrical power in `kw` generated in the reactor in steady state.

Answer» At steady state energy released per sec
`etaxx r(E_(1)+E_(2)), eta=25% , r= 10^(15)`
`E_(1)=100xx10^(6)xx1.6xx10^(-19)=1.6xx10^(-11)J`
`E_(2)=50xx10^(6)xx1.6xx10^(-19)=0.8xx10^(-11)J`


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