The binding energy per nucleon of ₈₃²⁰⁹Bi is _______ MeV. [Take m(₈₃²⁰⁹Bi) = 208.980388 u, mp = 1.007825 u, mn = 1.008665 u, 1 u = 931 MeV /c² ] (a) 7.48 (b) 7.84 (c) 8.79 (d) 6.94

Answer: (b). Z = 83, N = 209 - 83 = 126 83 × 1.007825 + 126 × 1.008665 = 83.649475 + 127.091790 = 210.741265 u Δ m = 210.741265 - 208.980388 = 1.760877 u, so BE = 1.760877 × 931 = 1639.4 MeV BE/A = (1639.4)/(209) = 7.84 MeV, so the answer is (b).

The binding energy per nucleon of ₈₃²⁰⁹Bi is _______ MeV. [Take… | JEE Main 2026 (2 Apr, Shift 2)

Q1JEE Main 2026 · 2 APR · SHIFT 2Easy

Mass Defect & Binding Energy

The binding energy per nucleon of ${}_{83}^{209}Bi$ is _______ MeV.

[Take $m({}_{83}^{209}Bi) = 208.980388$ u, $m_p = 1.007825$ u, $m_n = 1.008665$ u, $1$ u $= 931$ MeV $/c^{2}$ ]

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Answer: (b)

$Z = 83$ , $N = 209 - 83 = 126$
$83 \times 1.007825 + 126 \times 1.008665 = 83.649475 + 127.091790 = 210.741265$ u
$\Delta m = 210.741265 - 208.980388 = 1.760877$ u, so $BE = 1.760877 \times 931 = 1639.4$ MeV
$\dfrac{BE}{A} = \dfrac{1639.4}{209} = 7.84$ MeV, so the answer is (b).

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Mass Defect & Binding Energy question from JEE Main 2026 (2 Apr, Shift 2)

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