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Assuming that energy released by the fission of a single X92235X2922235U nucleus is 200MeV, calculate the number of fissions per second required to produce 1 watt power. - Physics

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प्रश्न

Assuming that energy released by the fission of a single \[\ce{^235_92U}\] nucleus is 200MeV, calculate the number of fissions per second required to produce 1-watt power.

संख्यात्मक

उत्तर

The fission of a single \[\ce{^235_92U}\] nucleus releases 200 MeV of energy

Energy released in the fission is given by the formula,

E = `"Pt"/"n" => "n"/"t" = "P"/"E"`

E = 200 MeV = 200 x 106 x 1.6 x 10-19

E = 3.2 x 10-11 J

`"n"/"t" = "P"/"E" = 1/(3.2 xx 10^-11)`

= 0.3125 x 1011 = 3.125 x 1010 

`"n"/"t"` = 3.125 x 1010 

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पाठ 9: Atomic and Nuclear physics - Evaluation [पृष्ठ १९२]

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सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 9 Atomic and Nuclear physics
Evaluation | Q 4.09. | पृष्ठ १९२
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