They Are All Photons

This is another interpretation of the set of data from the post "Sizing Them Up" dated 3 Dec 2014,

where $a_{\psi}=16.32\,nm$ is the value of $a_{\psi\,ne}$.

But if,

$(n_2=2n_e)\rightarrow (n_1=n_e)+n_e$

gives one photon $a_{\psi\,ne}$, then

$(n_3=3n_e)\rightarrow (n_2=2n_e)+n_e$

also gives one photon $a_{\psi\,ne}$.  And both,

$(n_4=4n_e)\rightarrow (n_2=2n_e)+2n_e$

$(n_2=3n_e)\rightarrow (n_1=2n_e)+n_e$

give photon $a_{\psi\,2ne}$.

What if they are all photons liberated, and we have,

where the smallest photon $a_{\psi\,ne}$ has the highest energy.

$a_{\psi}$ remains undetermined...