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JEE Main 2020 (4 Sep, Shift 2) · Mass Defect & Binding Energy

Mass Defect & Binding Energy question from JEE Main 2020 (4 Sep, Shift 2)

Q18JEE Main 2020 · 4 SEP · SHIFT 2Easy

Mass Defect & Binding Energy

Find the Binding energy per nucleon for ${}_{50}^{120}Sn$ . Mass of proton mp = 1.00783 U, mass of neutron mn = 1.00867 U and mass of tin nucleus mSn = 119.902199 U. (take 1U = 931 MeV)

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Answer: (b)

$Z = 50$ , $N = 120 - 50 = 70$
$50(1.00783) + 70(1.00867) = 50.3915 + 70.6069 = 120.9984$ u
$\Delta m = 120.9984 - 119.902199 = 1.0962$ u, $BE = 1.0962 \times 931 = 1020.6$ MeV
$\dfrac{BE}{A} = \dfrac{1020.6}{120} \approx 8.5$ MeV, so the answer is (b).

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