Q23 CQ
Question
To obtain the most precise value of \({\rm{BE}}\) from the equation \({\rm{BE = (ZM}}{{\rm{(}}^{\rm{1}}}{\rm{H) + N}}{{\rm{m}}_{\rm{n}}}{\rm{)}}{{\rm{c}}^{\rm{2}}}{\rm{ - m}}{{\rm{(}}^{\rm{A}}}{\rm{X)}}{{\rm{c}}^{\rm{2}}}\) , we should take into account the binding energy of the electrons in the neutral atoms. Will doing this produce a larger or smaller value for\({\rm{BE}}\)? Why is this effect usually negligible?
Step-by-Step Solution
VerifiedAs a result, the value will be lower. The contribution will be overlooked in response to the second question due to its insignificance.
The spontaneous emission of radiation in the form of particles or high-energy photons as a result of a nuclear process is known as radioactivity.
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.