There is no SHM about the nucleus. The electron will simply be in circular motion around the nucleus at a radius \(r_o\) and a velocity of \(c\), as given by.
\(r_o=\cfrac{q^2}{4\pi \varepsilon_o{m_ec^2}}\)
However if the electron does not approach close enough ie \(> r_o\) then it may be possible that the electron over shoot the nucleus only to reverse direction at some distance \(x\) then,
\(\cfrac { 1 }{ 2 } m_{ e }c^{ 2 }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }x } =\cfrac { 1 }{ 2 } kx^{ 2 }\)
As a first approximation, the resonance is liken to be from Hooke's spring formulation. Speed in the perpendicular direction does not change and so does not concern us here.
\( \cfrac { k }{ { m }_{ e } } =\cfrac { c^{ 2 } }{ x^{ 2 } } =c^{ 2 }\left\{ \cfrac { 2\pi \varepsilon _{ o } }{ q^{ 2 } } m_{ e }c^{ 2 } \right\} ^{ 2 }\)
\(f=\cfrac { 1 }{ 2\pi } \sqrt { \cfrac { k }{ { m }_{ e } } }=\cfrac { 1 }{ 2\pi } c\left\{ \cfrac { 2\pi \varepsilon _{ o } }{ q^{ 2 } } m_{ e }c^{ 2 } \right\} \)
\(f=\cfrac{m_{ e }c^{ 3 }\varepsilon _{ o }}{q^2}\)
\(f\)=9.10938291e-31*(299792458)^3*8.854187817e-12/(1.602176565e-19)^2=8.4660e21 Hz
Was this resonance from hydrogen ever detected? The sun might have this resonance.
