From the previous post "Drag and A Sense of Lightness"
\(F_D=Av^2\)
If the helical path of the electron at terminal velocity is due only to the drag force.
\(\cfrac{m_ev^2}{r}=Av^2=(1-\cfrac{r_{ec}}{r_e})\cfrac{m_e}{r_{e}}v^2\)
\(\cfrac { 1 }{ r } =(1-\cfrac { r_{ ec } }{ r_{ e } } )\cfrac { 1 }{ r_{ e } } \)
\( r=r_{ e }\cfrac{1}{(1-\cfrac { r_{ ec } }{ r_{ e } } )}\)
\(r\) = 2.5e-11/(1-5.636e-5) = 2.50014e-11 m
Since,
\((1-\cfrac { r_{ ec } }{ r_{ e } } )<1\)
\(r>r_e\)
That means, under normal circumstances, the electron will never be in a helical radius smaller than the atomic radius. It will not crash into the positive nucleus. The presence of a positively charged nucleus adds to the centripetal force and the radius of circular motion is smaller.
Have a nice day.
