The Binding Energy (\({\rm{BE}}\)) takes into account the electrons' binding energy in neutral atoms. It is provided by:
\(\begin{align}EB{\rm{ }} &= \left( {\left( {Z{m_p} + N{m_n} + Z{m_e}} \right) - m\left( {^AX} \right)} \right){c^2}\\ &= \left( {\left( {Z\left( {{m_p} + {m_e}} \right) + N{m_n}} \right) - m\left( {^AX} \right)} \right){c^2}..................(1)\end{align}\)
Considering the following masses:
\(\begin{align}{m_p} &= 1.007276{\rm{ }}u\\{m_n} &= 1.008665{\rm{ }}u\\{m_e} &= 0.00054858{\rm{ }}u\end{align}\)
In the equation one as:
\(\begin{align}EB{\rm{ }} &= \left( {(Z(1.007276 + 0.00054858)u + N(1.008665)u) - m\left( {^AX} \right)} \right){c^2}\\ &= \left( {(Z(1.00782458)u + N(1.008665)u) - m\left( {^AX} \right)} \right){c^2}...................(2)\end{align}\)
If comparing this result with the formula for\({\rm{EB}}\):
\(\begin{align}EB{\rm{ }} = \left( {\left( {ZM\left( {^1H} \right) + N{m_n}} \right) - m\left( {^AX} \right)} \right]{c^2}{\rm{ }}\\\left. { = [(Z(1.007825)u) + N(1.008665)u) - m\left( {^AX} \right)} \right){c^2}....................(3)\end{align}\).
We notice that the different between the second equation and the third equation is:
\(\begin{align}\Delta EB \approx Z\left( { - 4.2 \times {{10}^{ - 7}}} \right)\left( {931.5\,{\rm{MeV/}}{{\rm{c}}^{\rm{2}}}} \right){{\rm{c}}^{\rm{2}}}\\ = Z\left( { - 3.91 \times {{10}^{ - 4}}} \right)\,{\rm{MeV}}.................(4)\end{align}\)
For instance, the hydrogen atom\((Z = 1)\)we get:
\(\Delta EB \approx - 3.91 \times {10^{ - 4}}\,{\rm{MeV}}\)
Using this modification, we were able to obtain a lower figure, but the difference is too minor to be considered.
Therefore, It will result in a lower value. In response to the second question, the contribution will be ignored due to its insignificance.
