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A nuclear reactor generates power at 50%...

A nuclear reactor generates power at 50% efficiency by fission of `._(92)^(235)U` into two equal fragments of `._(46)^(116)U` into two equal fragments of `._(46)^(116)Pd` with the emission of two gamma rays of 5.2 MeV each and three neutrons. The average binding energies per particle of `._(92)^(235)U` and `._(46)^(116)Pd` are 7.2 MeV and 8.2MeV respectiveley. Calculate the energy released in one fission event. Also-estimate the amount to `.^(235)U` consumed per hour to produce 1600 megawatt power.

A

128 gm

B

`1.4 kg`

C

`140.5 gm`

D

`281 gm

Text Solution

Verified by Experts

The correct Answer is:
A
`(50P)/(100) = 1600 MW, P =([BE_("product")- Be _("reactant")])/(t)`
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