**A Model of the Helium Atom
**

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

For
brevity we write 1/4pe_{0}
as k:

k=1/4pe_{0}.
(1)

Then
the net Coulomb force on each electron has the magnitude

F_{coulomb}=k2e^{2}/r_{e}^{2}-ke^{2}/4r_{e}^{2}
(2)

=7ke^{2}/4r_{e}^{2}.

There
is also a Newtonian force on each electron:

F_{newton}=mv_{e}^{2}/r_{e}.
(3)

Equating
the Coulombic and Newtonian forces produces

mv_{e}^{2}=7ke^{2}/4r_{e}
(4)

or

v_{e}^{2}r_{e}=7ke^{2}/4m.
(5)

By
deBroglie each electron is associated with a wavelength l:

l=h/mv_{e}.
(6)

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

l=2pr_{e}. (7)

Thus

h/mv_{e}=2pr_{e}, (8)

and

v_{e}r_{e}=h/2pm. (9)

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

v_{e}=7pke^{2}/2h. (10)

Solving
for r_{e} and v_{e}:

r_{e}=.302386792e-10,

v_{e}=3828451.

In
the model of Hydrogen we found:

r_{e}=.52916885e-19

v_{e}=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*.)