Q=Ef−Ei
=(12×7.86+4×7.07+4×7.07)−(20×8.03)
Q=−9.72MeV
JEE Main 2019 — Physics Modern Physics
Consider the nuclear fission, Ne20→2He4+C12. Given that the binding energy/nucleon of Ne20,He4andC12 are 8.03MeV,7.86MeV, respectively. Identify the correct statement:
Held on 10 Jan 2019 · Verified 6 Jul 2026.
Energy of 12.4MeV will be supplied.
Energy of 9.72MeV has to be supplied.
Energy of 3.6MeV will be released.
8.3MeV energy will be released.
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