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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 megawatt x 1 hour `= (400 x 10 ^(6)` watt) x 3600 second ` = 144 x 10 ^(10)` joule
Here only `10 %` of input is utilised. In order to obtain `144 x 10 ^(10) ` joule of useful energy, the output energy from the power house `(10 E)/(100) = 144 x 10 ^(10) JE = 144x 10 ^(11)` joule
Let, this energy is obtained from a mass-loss of `Delta m` kg. Then
` (Delta m)c ^(2) = 144 xx 10 ^(11) ` joule
`Delta m = (144 xx 10^(11))/(( 3xx 10 ^(8))^(2))= 16 xx 10 ^(-5) kg= 0.16 g`
Since `0.92` milli gram `(=0.92 x 10 ^(-3)g)` mass is lost in 1 g uranium, hence for a mass loss of `0.16 g,` the uranium required is `= (1 xx 0.16)/(0.92 xx 10 ^(-3)) = 174 g`
Thus to run the power house, 174 gm uranium is required per hour.
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