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Atoms & Nuclei

JEE Main level, numerical answer, medium

51 of 239

Question 51 of 239

Nucleus a having $Z=17$ and equal number of protons and neutrons has $1.2MeV$ binding energy per nucleon. Another nucleus $B$ of $Z=12$ has total $26$ nucleons and $1.8MeV$ binding energy per nucleons. The difference of binding energy of $B$ and $A$ will be ______ $MeV$ .

Show answer and solution

Answer: 6

A,Z=17

and have equal numbers of protons and neutrons i.e. nucleons

=2\times 17=34

So, for

A

,
Total binding energy

B{E}_{A}=1.2\times 34=40.8MeV

Another nucleus

B

mass number

=26

So, total binding energy of

B

B{E}_{B}=1.8\times 26MeV

=46.8MeV

Difference of

BE=B{E}_{B}-B{E}_{A}=6MeV

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