Q.

The binding energy per nucleon for U238  is about 7.5 MeV where as it is about 8.5 MeV for a nucleus having a mass half of Uranium. If U238  splits into two exact halves the energy released would be

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a

119 MeV

b

238 MeV

c

476 MeV

d

6.4 MeV          

answer is C.

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Detailed Solution

B.EA  for  U238  is 7.5 MeV

For half of Uranium

B.EA  is 8.5 MeV

U238X119+X119

Energy released is=238[8.57.5]=238MeV

 

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