Here is the most important first result from the three results that I will show. The results confirm that the classical Bohr Model falls out from QSA model which encompasses QM and QFT.

Please always refer to these wiki:

This is the result of simulating two particles with a width of 1823 which is close to 1822.8885 for electron compton wavelength (just simplification)interacting at a seperation of around Bohr radius which is:

1/(m*alpha)=1/(.00054858*.007297352569) = 249801.3

The raw data is below from the program with int=50. also make this
change in the program to get these results

for (mk = 2475; mk <= 500000000; mk++)

also

d0 =1823; // Particle 1 size

d1 = 1823; // Particle 2 size

long long kj =20000000000; // # of random throws (approx 30 min
for each distance)

but next I give the important data that we will discuss:

distance energy (P.E.) | charge^2(e^2) | Expectation value(Ex) |

249323 0.0000120326 | 3.000003457 | 2.219640631 |

249423 0.0000120278 | 2.999998876 | 2.219325817 |

249523 0.0000120229 | 3.000000804 | 2.217591031 |

249623 0.0000120181 | 3.000000809 | 2.217731633 |

249723 0.0000120133 | 2.999998829 | 2.215744702 |

249823 0.0000120085 | 3.000006682 | 2.215434488 |

249923 0.0000120037 | 2.999998356 | 2.214921159 |

because I have the 1/r law I interpret the energy as e^2/r ,
e=charge

so if you multiply distance *energy= e^2=3 as shown, the average of above e^2=
3.000001, but we will take 3 to simplify.

then because we know alpha I deduce that ( from alpha=e ^2/(h*c))

h*c=e^2/alpha= 3/ .007297352569= 411.108

from other arguments I have h=c= sqrt(411.108)= 20.2758

Now, the important part which Expectation value(Ex for short)

after inspection I find it to be related to the classical bohr model variables

Ex=v^2/(2*m*e^4) ---------- eq 1

solving for v^2=(2*m*e^4)*Ex --------------- eq 2

from above simulation the average of Ex= 2.2172 almost

hence v^2= (2*.0005485*9)*2.2172= 0.0218936

v= sqrt(0.0218936)= 0.14797

now we compute v/c=0.14797/20.2758= 0.0072976

v/c should be alpha we have a very good match with some error mostly because of
Ex which we can simulate with higher j thows to get more accuracy and also due
to the approxomation of 1823 and 1822.8885

**Great we proved that Ex is what it is,
and h=c**

next

from eq 1 we can compute the kinetic energy

K.E.= (m^2*e^4)*Ex=(1/2)*m*v^2=.5*.00054858*0.0218936

= 0.000006005195544

2*K.E.= 0.000012010

That is Bohr Model P.E.= 2*K.E.

**So the energy has the interpretation
of potential energy and Ex is related to K.E. , that makes perfect sense**

also if we take 1/(2*Ex)=1/(2*2.2172)=0.22551 almost m*c^2

m*c^2=.00054858*411.108= 0.225526

errors should be taken into account as mentioned earlier

Q.E.D.

2475 249323 1.2032598102434993e-005
5.9941919763095141e-006 3.0000034566933995 1.823 2.2196406306904919
-2.1562643940237649

2476 249423 1.2027755561853412e-005 5.9917972737823617e-006 2.9999988755041636
1.823 2.2193258174578432 -2.1564566940604664

2477 249523 1.202294299276971e-005 5.9894015680537819e-006 3.0000008043848765
1.823 2.2175910306876858 -2.1541575969249607

2478 249623 1.2018126569310113e-005 5.9870183764102081e-006 3.0000008086108982
1.823 2.2177316327685048 -2.1541184268552342

2479 249723 1.2013306059579073e-005 5.9846252913626688e-006 2.9999988291162647
1.823 2.2157447018220182 -2.1534291553698495

2480 249823 1.2008528765272037e-005 5.9822392351484673e-006 3.000006681726556
1.823 2.2154344880241297 -2.1517840085696207

2481 249923 1.2003690562073311e-005 5.9798542464554e-006 2.9999983563450483
1.823 2.2149211586316824 -2.1521569239379232