Obtain the binding energy (in MeV) of a nitrogen nucleus
$({}_{7}^{14}N), given m ({}_{7}^{14}N) = 14.00307 u .$

Solution

{}_{7}^{14}N

nucleus is made up of 7 protons and 7 neutrons.

Mass of nucleons forming nucleus = 7m_p + 7m_n = Mass of 7 protons + Mass of 7 neutrons

= 7 × 1.00783 + 7 × 1.00867 u

= 7.05431 + 7.06069 u

= 14.11550.u

Mass of nucleus, m_N = 14.00307 u

Therefore,

Mass defect = 14.11550 - 14.00307 = 0.11243 a.m.u.

Energy equivalent to mass defect = 0.11243 × 931 = 104.67 MeV

∴ Binding energy = 104.67 MeV.

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Wave Optics

Obtain the binding energy (in MeV) of a nitrogen nucleus
$({}_{7}^{14}N), given m ({}_{7}^{14}N) = 14.00307 u .$

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Obtain the binding energy (in MeV) of a nitrogen nucleus N714, given m N714 = 14.00307 u.

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Obtain the binding energy (in MeV) of a nitrogen nucleus
$({}_{7}^{14}N), given m ({}_{7}^{14}N) = 14.00307 u .$