JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Modern Physics

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 nucleons.
The difference of binding energy of B and A will be ______ MeV.

Answer: 6

Solution

Nucleus A has

Z = 17

with an equal number of neutrons, so its mass number is

34

and its total binding energy is

1.2 \times 34 = 40.8

MeV. Nucleus B has

26

nucleons with binding energy

1.8

MeV per nucleon, giving

1.8 \times 26 = 46.8

MeV. The difference of the binding energies of B and A is

46.8 - 40.8 = 6

MeV.

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Integer answer

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