\(\cfrac{{m}_{e}c^2}{{r}_{pe}}=\cfrac{q.q}{4\pi\varepsilon_o{r}^2_{pe}}\)
In general for a electrically neutral system of \(2n\) charges, \(n\) positive and \(n\) negative,
\(\cfrac{{m}_{e}c^2}{{r}_{n}}=\cfrac{n^2q^2}{4\pi\varepsilon_o{r}^2_{n}}\)
\({r}_{n}=\cfrac{n^2q^2}{4\pi\varepsilon_o{m}_{e}c^2}\)
\({ r }_{ n+1 }=\cfrac { (n+1)^{ 2 }q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \)
obviously \({ r }_{ n+1 } \gt { r }_{ n}\), but the corresponding potential energy term,
\(PE_{n}= \cfrac{n^2q^2}{4\pi\varepsilon_o{r}_{n}}={m}_{e}c^2\) which is a constant
Suppose a nucleus of \(n+1\) positive charge, emits a positive-negative charge pair, both charges leave the atom, then \(n\) negative charges will be caught at a higher orbit. This can be the case of disassociated hydrogen gas where a number of the ionized atoms come together to for a loose nucleus, under high temperature.
The potential energy difference is,
\(\Delta PE=\cfrac { n^{ 2 }q^{ 2 } }{ 4\pi \varepsilon _{ o }{ r }_{ n } }-\cfrac { n^{ 2 }q^{ 2 } }{ 4\pi \varepsilon _{ o }{ r }_{ n+1 } } \)
\(=\cfrac { n^{ 2 }q^{ 2 } }{ 4\pi \varepsilon _{ o } } \left\{ \cfrac { 1 }{ { r }_{ n } } -\cfrac { 1 }{ { r }_{ n+1 } } \right\} \)
\(=n^{ 2 }{ m }_{ e }c^{ 2 }\left\{ \cfrac { 1 }{ n^{ 2 } } -\cfrac { 1 }{ { (n+1)^{ 2 } } } \right\} \)
In general, for \(n_1\) remaining positive charges from a nucleus originally containing \(n_2\) positive charges,
\(\Delta PE_{n_1,n_2}= {n_1}^{ 2 }{ m }_{ e }c^{ 2 }\left\{ \cfrac { 1 }{ {n_1}^{ 2 } } -\cfrac { 1 }{ { {n_2}^{ 2 } } } \right\} \)
obviously \(n_1 \lt n_2 \). The final position of the negative charge has a lower potential than before the charge pair leaves the atom. The speed of the electrons at both positions is \(c\), unchanged. So, the energy difference \(\Delta PE\) is totally emitted, as with the charge pair. Where did this energy go? How did the energy went?
If similar process happens for other elements of higher atomic numbers, where bare nuclei come together to form loose positive charge centers, than it is expected that \({n}_{2}\) be a multiple of \({n}_{1}\). ie.
\(\Delta PE_{n}= { m }_{ e }c^{ 2 }\left\{1 -\cfrac { 1 }{ { {n}^{ 2 } } } \right\} \)
where \(n_2=n.n_1\) and \(n=2,3,4...\)
In this case, \(\Delta PE_n\) seems to have lost its atomic number signature. The last expression assume that the nucleus of the elements are not restricted by their geometry and come together freely. In actual cases, some \(n\) are not possible for some elements which gives that element its characteristic spectra lines.
