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A Model of the Helium Atom

Let us model the Helium atom, at time t=0, as (a) two electrons at x=-re and x=re, (b) two protons at x=-rp and x=rp (with rp<<re), and (c) two neutrons at z=rn and z=-rn. The diametrically opposed electrons are orbiting in a common circular orbit of radius re, at a common speed ve. re and ve are to be determined.

For brevity we write 1/4pe0 as k:

k=1/4pe0.    (1)

Then the net Coulomb force on each electron has the magnitude

Fcoulomb=k2e2/re2-ke2/4re2    (2)

            =7ke2/4re2.

There is also a Newtonian force on each electron:

Fnewton=mve2/re.    (3)

Equating the Coulombic and Newtonian forces produces

mve2=7ke2/4re    (4)

or

ve2re=7ke2/4m.    (5)

By deBroglie each electron is associated with a wavelength l:

l=h/mve.    (6)

In the ground state we assume that just one of these wavelengths fits into the electron orbital:

l=2pre.    (7)

Thus

h/mve=2pre,    (8)

and

vere=h/2pm.    (9)

From Eqs. (5) and (9) we thence find that

ve=7pke2/2h.    (10)

Solving for re and ve:

re=.302386792e-10,

ve=3828451.

In the model of Hydrogen we found:

re=.52916885e-19

ve=2187691

Thus the radius of the heavier Helium atom is less than that

of the Hydrogen atom (and the speed of the orbital electrons is

greater.)