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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