When Big Particle Collide

\(E=hf=h\cfrac{c}{\lambda}=h\cfrac{c}{2\pi a_{\psi}}\)

The bigger \(a_{\psi}\) has lesser energy, \(E\).

If the energy transition \(2n_e\rightarrow n_e + n_e\) results in the absorption line, then \(a_{\psi\,14.77}\) has also grow to \(2n_e\), even though unaffected by the high temperature high and voltage.  It has grow big because of the high collision rate.

How did the time particle grow bigger by collisions?

From the post "No Solution But Exit Velocity Anyway" dated 14 Jul 2015,

\(v^{ 2 }_{ { max } }=\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\}  e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }\)

When is \(v_{max}=0\)?  When

\(\psi_n=\psi_{max}\)

two big particles collide and coalesce.

To suggest that \(\psi_{max}\) changes with temperature would be wrong, as only temperature particles are affected by the temperature field force.  Big temperature particle are also more stable when the temperature field is high, for the difference in field potential that give raise to a pinch force, that pulls \(\psi\) away is less.  A high surrounding potential will result in a negative (inward) pinch force that compresses \(\psi\).

For,

\((e^{ 2\psi _{ max } }-1)=0\)

\(\psi_{max}=0\)

this is a null answer, just as \(\psi_n=0\).

Note:  The placid field remains.