Question

Obtain the binding energy of a nitrogen nucleus ${(}_7{N}^{14})$ from the following data: $m_H=1.00783$ a.m.u., $m_n=1.0087$ a.m.u.; $m_N=14.00307$ a.m.u.

Answer 1

$m_H$ = 1.00783 amu

$m_n$ = 1.0087 amu

$m_N$ = 14.0030u amu

In${}_7{N}^{14}$, these are 7 protons and 7 neutrons.

$\therefore$ mass defect,$\Delta =(7m_H+7m_n)-m_N$

=(7×1.00783)+(7×1.00807)-14.00307=0.11243 amu

Binding energy of nitrogen nucleus

= $\Delta$×931 MeV =0.11243×931

=104.67