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. Numerical answer type. Work it out, then open the answer.

Answer: 6. Nucleus A: Z = N = 17, so A = 34 and BEA = 34 × 1.2 = 40.8 MeV Nucleus B: A = 26, so BEB = 26 × 1.8 = 46.8 MeV BEB - BEA = 46.8 - 40.8 = 6 MeV

Nucleus A having Z = 17 and equal number of protons and neutrons has… | JEE Main 2023 (1 Feb, Shift 2)

Q14JEE Main 2023 · 1 FEB · SHIFT 2Medium

Mass Defect & Binding Energy

Nucleus A having $Z = 17$ and equal number of protons and neutrons has $1.2\text{ MeV}$ binding energy per nucleon.

Another nucleus $\text{B}$ of $Z = 12$ has total 26 nucleons and $1.8\text{ MeV}$ binding energy per nucleons. The difference of binding energy of $\text{B}$ and $\text{A}$ will be _____________ $\text{MeV}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 6

Nucleus A: $Z = N = 17$ , so $A = 34$ and $BE_A = 34 \times 1.2 = 40.8$ MeV
Nucleus B: $A = 26$ , so $BE_B = 26 \times 1.8 = 46.8$ MeV
$BE_B - BE_A = 46.8 - 40.8 = 6$ MeV

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