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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

6.4 MeV

b

119 MeV

c

238 MeV

d

476 MeV

answer is C.

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Detailed Solution

B.EA  for  U238  is 7.5 MeVFor half of UraniumB.EA  is 8.5 MeVU238→X119+X119Energy released is=238[8.5−7.5]=238MeV
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