If $\mathrm{M}_\mathrm{o}$ is the mass of isotope ${}_{5}^{12} \mathrm{~B}$, $\mathrm{M}_\mathrm{p}$ and $\mathrm{M}_\mathrm{n}$ are the masses of proton and neutron, then nuclear binding energy of isotope is :

(A) $\left(5 \mathrm{M}_\mathrm{p}+7 \mathrm{M}_\mathrm{n}-\mathrm{M}_\mathrm{o}\right) \mathrm{C}^{2}$

(B) $\left(\mathrm{M}_\mathrm{o}-5 \mathrm{M}_\mathrm{p}\right) \mathrm{C}^{2}$

(C) $\left(M_{o}-12 M_{n}\right) C^{2}$

(D) $\left(\mathrm{M}_\mathrm{o}-5 \mathrm{M}_\mathrm{p}-7 \mathrm{M}_\mathrm{n}\right) \mathrm{C}^{2}$

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Modern Physics

M_{o}

is the mass of isotope

_{5}^{12} B

M_{p}

and

M_{n}

are the masses of proton and neutron, then nuclear binding energy of isotope is :

(A) $(5 M_{p} + 7 M_{n} - M_{o}) C^{2}$ ✓ Correct
(B) $(M_{o} - 5 M_{p}) C^{2}$
(C) $(M_{o} - 12 M_{n}) C^{2}$
(D) $(M_{o} - 5 M_{p} - 7 M_{n}) C^{2}$

Answer: (A)

Solution

For

_{5}^{12} B

: 5 protons and 7 neutrons make up the nucleus. The mass defect is

Δ m = 5 M_{p} + 7 M_{n} - M_{o}

Binding energy

= Δ m C^{2} = (5 M_{p} + 7 M_{n} - M_{o}) C^{2}

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Single correct option

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