The smaller the particle the higher \(E_n\) is. Big particle absorb energy to break into smaller particles. When small particles coalesce energy is released as a photon that we observe as the emission spectrum.
Big particles are more susceptible to collision impacts and break into smaller particles, absorbing part of the energy of impact on breaking up. Energy is absorbed as a photon, from which we obtain the absorption spectrum.
But \(E_n\) is not \(\psi\). \(E_n\) is the result of \(\psi\) on a circular path at light speed \(c\).
So,
\(\lambda_\psi\ne\lambda_n\)
and the photon packets emitted or absorbed,
\(h.f_{vis}=h.\left(f_{n_1}-f_{n_2}\right)\)
\(\lambda_{vis}=\cfrac{c}{f_{vis}}=\cfrac{c}{f_{n_1}-f_{n_2}}\)
where \(f_{vis}\) and \(\lambda_{vis}\) are obtained from the experimental spectrum(s) observed.
Obviously,
\(\lambda_{vis}\ne\lambda_n\) and
\(\lambda_{vis}\ne \lambda_{\psi}\)
But,
\(\lambda_{\psi\,n}=\lambda_{n}\), \(m=1\)
only when \(m=1\) that there is one wavelength around the circular path of radius \(a_{\psi}\). The factor \(2\pi\) appears as the wavelength, \(\lambda_{\psi}\) is wrapped around a circular path.
And since the energy transitions as particles coalesce and disintegrate is the reverse of electron energy level transitions, we do not have an issue with negative energy.
And in the rest state of no collisions, we have \(a_{\psi\,n=1}\) where \(a_{\psi}\) is the smallest at the highest energy possible,
\(E_{n=1}=E_{max}\)
\(a_{\psi}=a_{\psi\,1}=a_{\psi\,c}\)
So, paradoxically \(\psi\) with \(a_{\psi\,c}\) has the highest energy but the smallest size.
