﻿ 4. Bohr

# 4. Showing the results for Bohr atom hydrogen 1s simulation.

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