Q.

What is the total energy emitted for given fission reaction 01n+92235U⟶92236U⟶4098Zr+52136Te+201nThe daughter nuclei are unstable therefore they decay into stable end products M 4298o and X 54136e by successive emission of β-particles. Let mass of   01n=1.0087 amu, mass of  92235U=236.0526 amu, mass of  54136Xe=135.9170 amu, mass of M 4298o = 97.9054

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a

198 MeV

b

220 MeV

c

185 MeV

d

230 MeV

answer is A.

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Detailed Solution

01n+92235U⟶4298Mo+54136Xe+201n(1.0087+235.0439)=(97.9054+135.917+2.0174)Δm=0.2128∴  Total energy released during a fission reaction=0.2128×931MeV=198MeV
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