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JEE Main levelSingle correct optionDifficulty 2 of 3as021

The nucleus of $X_{2}^{4} X_{22}^{24} He$ has mass $4.00151 u$ . Take proton mass $1.00728 u$ , neutron mass $1.00866 u$ and $1 u = 931.5 MeV / c^{2}$ . The binding energy per nucleon is closest to

A28.3 MeV
B3.54 MeV
C14.1 MeV
D7.07 MeV

Answer: D

Mass defect = 2(1.00728) + 2(1.00866) - 4.00151 = 0.03037 u. Binding energy = 0.03037 x 931.5 = 28.29 MeV. Per nucleon: 28.29/4 = 7.07 MeV.

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