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
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
a
198 MeV
b
220 MeV
c
185 MeV
d
230 MeV
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
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