No Nucleus Needed

$\psi$ wraps around a sphere.  A certain amount of energy is associated with $\psi$ motion around the sphere.  If $h$ amount of energy is in the system, going once per second around the sphere, then travelling at light speed $c$ around a radius of $a_{\psi\,n}$, the amount of potential energy associated with this motion is,

$E_n=h.\cfrac{c}{2\pi a_{\psi\,n}}$

but,

$2\pi a_{\psi\,n}=m\lambda_m$

where $m$ is the number of wavelength, $\lambda_m$ of $\psi$ around the circle.

With $m=1$,

$E_n=h.\cfrac{c}{\lambda_m}=h.f$

which is Planck's relation given $m=1$.  When $\psi$ is a half wave in time and a half wave in space space,

$m=\cfrac{1}{2}$

$E_n=h.\cfrac{2c}{\lambda_m}=2h.f=h^{'}.f$

with a mysterious factor of $2$.

When $m=2, \,3...$

$E_n=h.\cfrac{c}{m\lambda_m}=\cfrac{h}{m}.f=h_m.f$

$h_m$ changes with $m$ because $c=f\lambda$.

And,

$h_{m_1}.f\gt h_{m_2}.f$

when $m_2\gt m_1$.

When there is more wavelengths around the circular path, stored energy decreases, which suggest another way $\psi$ can absorb and release packets of energy; by changing $m$.  Also, given $a_{\psi\,n}$ at fixed value, $f$ increases with $m$ but $h_m$ decreases, so, $E_n=h_m.f$ remains a constant.  $f$ increases with decreasing $a_{\psi}$, with increasing $E_n$.  Catastrophe when the number of particles with small $a_{\psi}$ is also large.

$E_n$ is independent of $m$.  $h$ is defined as one wavelength going around once per second along the circular path, then a corresponding increase in the number of wavelengths, $m$ increases the frequency by $m$ but this increase cancels with the decrease in $h$ by the reciprocal of $m$.

With these issues with $m$ aside, when two particles coalesce,

$a_{\psi\,n_1}\rightarrow a_{\psi\,n_2}$ 

the bigger particle has less potential energy associated with its $\psi$ in circular motion,  The amount of energy released due to this drop in potential is.

$\Delta E=E_{n_1}-E_{n_2}=h\cfrac{c}{2\pi a_{\psi\,n_1}}-h\cfrac{c}{2\pi a_{\psi\,n_2}}$

$\Delta E=hf_{n_1}-hf_{n_2}=h\left(f_{n_1}-f_{n_2}\right)$

$f_{n_1}\gt f_{n_2}$

using $h$ to denote all cases of $m$, as discussed above, we have an expression similar to the discrete energy level transitions expression for electrons around a nucleus,  In this case, however it is just $\psi$ alone going around a circular path and energy level decreases with increasing $a_{\psi\,n}$

No nucleus needed.