Using the following data
Mass hydrogen atom = $1.00783 u$
Mass of neutron = $1.00867 u$
Mass of nitrogen atom ( $_{7} N^{14}$ )= $14.00307 u$
The calculated value of the binding energy of the nucleus of the nitrogen atom ( $_{7} N^{14}$ ) is close to

Question

TestHub · Accepted Answer

The binding energy of nucleus may be defined as the energy equivalent to the mass defect of the nucleus.

If $∆ m$ is mass defect then according to Einstein's mass-energy relation.

Binding Energy

$= ∆ m c^{2} = [\{Z m_{p} + (A - Z) m_{n}\} - M] c^{2}$

$= (7 \times 1.00783 + 7 \times 1.00867 - 14.00307) c^{2}$

$or BE = 0 .1124 \times 931 .5 MeV$

or $BE = 104.6 MeV$

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