Bohr's H Atom Revisited

Abstract. A reason is suggested why the electron in a Bohr Hydrogen atom doesn't radiate.

In1913 Niels Bohr introduced his model of the Hydrogen atom. He suggested that the atom consists of a central, resting proton (nucleus) with an electron orbiting in a circle. (In the lowest energy state the radius of the orbit is relec = .529177E-10 meters.) In every orbit the electron's speed is such that its angular momentum, melecvelecrelec, equals the constant h/2p. This restriction dictates that only certain discrete angular momenta are allowed. (For the lowest energy, velec=.2187542E7 meters/second.)

The orbiting electron in his model does not radiate ... a clear contradiction to the time-honored proposition that a charge going in a circle must radiate. Bohr's response to this objection was (reportedly) simply that, in the case of an atom, there is no radiation in any stable state (including the "ground" or lowest energy state).

In this article we take a more detailed look at Bohr's model. The objective is to ascertain if there is a physical explanation for why his model, suitably modified, does not radiate. Our strategy will be to demonstrate that the total tangential force on the electron ... its radiation reaction force and the tangential component of the Lorentz force from the proton ... sum to zero. If such is the case, then the total work done on the electron in any orbit would be zero, and the atom should neither emit nor absorb radiant energy.

Now a legitimate objection is that in Bohr's model (with its resting proton) the Lorentz force has no tangential component, and thus the electron's radiation reaction force would be unopposed. The atom should radiate (quite as many complained when Bohr introduced his model). In meeting this objection we shall suppose that, looking down from the positive z-axis, the electron and proton both circle the Origin in a counterclockwise direction in the xy plane, and at a common angular speed, w. The radius of the more massive proton's circular orbit, rprot, is of course smaller than that of the electron (relec). 

At any given instant the two particles are diametrically opposed. At time t=0, the electron is at (-relec,0) and the proton is at (rprot,0). The electron is moving at the speed velec, and the proton at the lesser speed vprot.  

Bohr, as we know, suggested that the nucleus of the atom was at rest. But if the linear momentum of the entire atom is zero at all times, then at any moment the proton must be moving in the opposite direction as the electron. Owing to this motion there are delayed time and distance factors that should be considered in computing the interactive force on the electron. And when these factors are appropriately sized, the y-component of the Lorentz  force on the electron will not be zero!  There is a small component that points in the same direction as the electron's velocity. And that direction is opposite to the  radiation reaction force that the electron experiences as a consequence of its circular motion.

The orbiting proton of course engenders a magnetic field at the electron. But the magnetic force on the electron is always perpendicular to its velocity, and plays no role in canceling the radiation reaction force. 

A schematic illustration of the situation at time t=0 is provided in Appendix A. We know from Bohr's angular momentum rule what the electron's angular frequency is:

w=velec/relec=4.133856155E16/second.    (1)

And in order for the total momentum to be zero, we presumably have

mprotwrprot=melecwrelec.    (2)

But we also have

rprot=vprot/w.    (3)

Dividing (2) by (3) and solving for vprot produces

vprot=velecmelec/mprot = 1193 meters/second.    (4)


rprot=vprot/w=2.8820E-14 meters.    (5)

At the delayed time, td, the proton is at an angle q below the x-axis, where

q=w(-td).    (6) 

We are now prepared to compute the Lorentz force to find the y-component of the interactive force experienced by the electron. In order to compute this, we shall invoke the formula for the proton's E field. This formula is relativistically rigorous, and takes into account the retarded effects. (Griffiths' text, Introduction To Electrodynamics, provides an example of the formula.) 

The y-component of this force is

Fy = -qEy.    (7)

For the radiation reaction force we have

FRR=q2w3relec/(6pe0c3).    (8)

 Using the computer we can "creep up" to just the delayed time needed to yield the result

Fy+FRR ~ 0.    (9)

Postscript: The modeling program was modified to compute Fy and FRR for the electron's force on the proton. Again, the objective was to see if the y-component of the interactive force on the proton is equal and opposite to the radiation reaction force on that  particle. It was found that Fy=1.16286192e-17, and FRR=-1.16286189e-17 ... again practically an equality. Thus together the electron and proton, and hence the atom, do not radiate.

A final note worth mentioning: The requirement that the atom's linear momentum equal zero leads to a proton angular momentum not equal to h/2p. Thus that restriction applies only to the electron in Bohr's atom.

Appendix A


The Bohr Atom at a Typical Delayed Time


Appendix B contains the Python code used to compute the values of FRR and Fy.

Appendix B

#For the articls, Bohr Revisited.
import math
c=2.99792458e8 #Speed of light
h=6.626068e-34 #Planck constant
eps0=8.85418782e-12 #Permittivity constant
q=1.60217662e-19 #Elementary charge
ElectronMass=9.10938356e-31 #Electron rest mass
ProtonMass=1.6726219e-27 #Proton rest mass
while N<=10:
N=int(input('level=?(>10 to exit) '))
print('Computing level ',N)
ElectronOrbitRadius=N**2*ElectronOrbitRadius1 #N'th level electron orbit radius
ElectronSpeed=h/(2*math.pi*ElectronMass*ElectronOrbitRadius1*N) #Electron orbital speed
ElectronKineticEnergy=ElectronMass*ElectronSpeed**2/2 #Electron kinetic energy
ElectronOrbitalPeriod=2*math.pi*ElectronOrbitRadius/ElectronSpeed #Electron & proton orbital period
ElectronOrbitalFrequency=1/ElectronOrbitalPeriod #Orbital fequency
omega=2*math.pi*ElectronOrbitalFrequency #Angular frequency
ProtonOrbitalSpeed=ElectronMass*ElectronSpeed/ProtonMass #Net momentum of atom is zero
ProtonOrbitalRadius=ProtonOrbitalSpeed/omega #Proton orbital radius
ProtonKineticEnergy=ProtonMass*ProtonOrbitalSpeed**2/2 #Proton kinetic energy
PE=-q**2/(4*math.pi*eps0*(ElectronOrbitRadius+ProtonOrbitalRadius)) #Potential energ of atom
TotalEnergy=ElectronKineticEnergy+ProtonKineticEnergy+PE #Total energy of atom
maxN=1e8 #Number of iterations in oebital computation
DelayedTime=(ElectronOrbitRadius1+ProtonOrbitalRadius1)/c #Time for signal to travel from prot to elec
dt=-DelayedTime/maxN #Time increment
loopN=-1 #Level no.
theta=0. #See diagram
RadiationReactionForce=q**2*omega**3*ElectronOrbitRadius/(6*math.pi*eps0*c**3) #Rad React force on elec
RetardedProtonSpeed=ProtonOrbitalSpeed #Proton travels at constant speed
ar=RetardedProtonSpeed**2/ProtonOrbitalRadius #Angular acceleration of proton
Fy=0. #Force of prot on elec
#Find parameters when forces on electron sum to zero
while RadiationReactionForce+Fy>=0:
#Print params and exit by clicking X in display
if loopN>maxN:
print('loopN,RadiationReactionForce,Fy',loopN,'; ',RadiationReactionForce,', ',Fy)
print('Infinite Loop; Aborting')
#If necessary, keep 'inching up'
RetardedVelocityX=ProtonOrbitalSpeed*math.sin(theta) #See diagram
L=math.sqrt((ElectronOrbitRadius+ProtonOrbitalRadius)**2-(ProtonOrbitalRadius*math.sin(theta))**2) #See diagram
#Compute electric field at electron (see Griffith's text)
Fy=-q*Ey #Interactive force on electron
#When Fy+RadiationReactionForce~0, display results.
print('RadiationReactionForce,Fy= ',RadiationReactionForce,', ',Fy)
#Do another level.