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Q. 13.3  Obtain the binding energy( in MeV ) of a nitrogen nucleus (_{7}^{14}\textrm{N}), given m (_{7}^{14}\textrm{N})=14.00307\; \; u

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mn = 1.00866 u

mp= 1.00727 u

Atomic mass of Nitrogen m= 14.00307 u

Mass defect \Deltam=7\timesmn+7\timesm- m

\Deltam=7\times1.00866+7\times1.00727 - 14.00307

\Deltam=0.10844

Now 1u is equivalent to 931.5 MeV

Eb=0.10844\times931.5

Eb=101.01186 MeV

Therefore binding energy of a Nitrogen nucleus is 101.01186 MeV.

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