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In the process of nuclear fission of 1 g...

In the process of nuclear fission of `1 g` uranium, the mass lost is `0.92 mg`. The efficiency of power house run by the fission reactor is `10%`.To obtain `400` megawatt power from the power house, how much uranium will be required per hour? `(c=3xx10^(8) m s^(-1))`.

Text Solution

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Power to be obtained from power house =
400 mega watt
`:.` Energy obtained per hour = 400 mega watt `xx` 1hour =
`(400xx10^(6) "watt") xx 3600 ` second `=144xx10^(10)` joule
Here only 10% of input is utilized . In order to obtain `144xx10^10` joule of useful energy, the output energy from the power house `(10E)/(100)=144xx10^(10)J`
`E = 144 xx10^(11)` joule
Let , this energy is obtained from a mass - loss of `Deltam` kg Then `(Deltam)c^2 = 144xx10^(11) ` joule
`(Deltam) = (144xx10^(11))/(3xx10^(18))^2 = 16xx10^(-5)kg ` = 0.16 gm
Since 0.92 milli gram `(=0.92 xx10^(-3)" gm")` mass is lost in 1 gm uranium , hence for a mass loss of 0.16 gm , the uranium required is `=(1xx0.16)/(0.92xx10^(-3))=174 gm`
Thus to run the power house , 174 gm uranium is required per hour.
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