Q 12-13-083JEE MainJEE Main 2023 (1 Feb, Shift 2)Easy
Nucleus $A$ having $Z=17$ and equal number of protons and neutrons has $1.2$ MeV binding energy per nucleon. Another nucleus $B$ of $Z=12$ has total $26$ nucleons and $1.8$ MeV binding energy per nucleon. The difference of binding energy of $B$ and $A$ will be ______ MeV.
Numerical value type. Enter your answer.
Answer: 6
$A$: $34\times1.2=40.8$ MeV. $B$: $26\times1.8=46.8$ MeV. Difference $=6$ MeV.
Solution by Sreeraj P, M.Sc Physics