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Bohr's H Atom Revisited (Part 1)

In 1913 Niels Bohr introduced his model of the Hydrogen atom. No doubt influenced by Rutherford’s work, he suggested that the atom consists of a central, resting proton (nucleus) with an electron orbiting in a circle. According to the latest information on the Internet, in the atom’s lowest energy state, re1, the radius of the electron’s orbit is re1= 5.291772 × 10−11 meters. And the value for the electron’s mass is me= 9.10938356 × 10-31 kg. Bohr postulated that the electron's angular momentum, L=(me)(ve)(re), equals the constant Nh/2p, where N=1, 2, 3... (N=1 is the lowest "ground" state of the atom.)  Thus according to Bohr’s postulate the most probable value for the ground state ve is 2187691 m/sec.

Now according to Coulomb F, the force exerted on the electron by the proton, is generally F=q2/4peo(re2). And according to Newton 2 F=(me)(ve2)/(re). This yields ve=q2/(4peo(re)(me))1/2=2187691 m/sec.

Let us now consider Bohr’s assumption that the proton is at rest at the origin of an inertial rectangular coordinate system. Looking down from the positive z-axis, we imagine that the electron is at (-re,0,0) and is orbiting CCW around the origin. This would mean that at any given instant the entire atom’s momentum would simply equal the electron’s momentum. But the electron’s momentum varies in direction from moment to moment. Yet in Bohr’s model (and presumably by observation) the atom’s total momentum is zero at any given moment. The only way this condition can be satisfied is if the proton is not actually at rest. Assuming the atom’s center is at rest at the origin, the proton must rotate CCW around the atom’s center quite like the electron does.  And the electron and proton must be diametrically opposed at every instant. (Of course the proton orbital radius would be less than that of the electron, owing to the proton’s mass being greater than the electron’s.) Together the two should orbit around the origin at a common angular frequency, w.

But our critique of Bohr’s model is not yet done for the following reason: the electron should radiate energy, since any charged particle going around a circular path radiates. Indeed Lorentz had found that a charge, going in a circle of radius r and at speed v, should experience a “radiation reaction” force opposite at any moment to the direction of v. The formula for this force is FRadReact = q2w3(re)/(6peoc3). If this force is not counteracted then the charge should radiate and spiral into an increasingly smaller-radius quasi-circle. However, if it is counteracted by a component of the electric force from another charge, then the charge should not radiate. It should continue to travel at a constant speed in a constant-radius circle.

What might be the explanation for the needed counteraction to FRadReact acting on the orbiting electron? It cannot be a magnetic force, since such a force would act perpendicular to v. It must be a component of an acting electric force engendered by the rotating proton.

Now unknown to Bohr, later theorists worked out the formula for the electric field of a charge whose kinetic history is known. This formula entails terms such as the source charge’s retarded velocity and acceleration, which pertain to a time prior to their electric force on some distant charge. The key to this phenomenon is that electromagnetic fields are not instantly felt at distant points. Their influence travels at the speed of light from source charge to subject charge. The field felt by a subject charge at time t is a function of kinematical variables at the source charge at some earlier time tretarded = tpresent–L/c, where L is the distance between the object charge and the source charge’s position at the time tretarded. (Note in this case that L is a displacement, and is not angular momentum.) As mentioned earlier, the formula for the radiation reaction force, acting on the Bohr model electron at time t=0, is FRadReact=q2w3(re)/6peoc3. And the formula for the proton-engendered electric force acting on the electron is the negative of FRadReact. This force provides the needed counterforce to FRadReact. If the sum of these two forces is zero, then the net radiation emitted by the electron is zero.

We may determine where the proton was by “inching up”  to values for L and q using a computer (See Appendix A for a schematic, and Appendix B for a typical “inching up” program written in Python). Note that the radius of the proton’s orbit has been greatly exaggerated in the schematic for clarity.

It is worth mentioning that Bohr’s postulate, L=Nh/2p, does not apply to H atom protons. The formula that works in the proton case is Lp=N(hp)/2p, where the constant hp is a companion to h. hp has the value hp = 2p(Lp)/N = 3.608669e-37 = h(me/mp).

Appendix A The Bohr Atom at a Typical Delayed Time

Appendix B

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

me=ElectronMass

ProtonMass=1.6726219e-27    #Proton rest mass

mp=ProtonMass

ve1=q/math.sqrt(4*math.pi*eps0*me*re1)

L1=me*ve1*re1

print('L1=',L1)

L1=h/(2*math.pi)

print('L1=',L1)

input("OK?")

#Any of the lowest 10 energies can be computed.

N=0

while N<=10:

N=int(input('level=?(>10 to exit) '))

if(N<11):

print('Computing level ',N)

L=N*L1

ve=q**2/(4*math.pi*eps0*N*L1)

print('ve1=',ve1,', ve=',ve)

input('OK?')

ve=h/(2*math.pi*me*re1*N)

print('ve=',ve)

input('OK?')

#ElectronKineticEnergy1=ElectronMass*ElectronSpeed1**2/2           #Electron kinetic energy

#ElectronKineticEnergy=ElectronMass*ElectronSpeed**2/2

re=N*L1/(me*ve)

taue=2*math.pi*re/ve

nue=1/taue    #Orbital fequency

omega=2*math.pi*nue    #Angular frequency

vp1=me*ve1/mp #total momentum=0

vp=me*ve/mp

rp1=vp1/omega

rp=vp/omega

#ProtonKineticEnergy=ProtonMass*ProtonOrbitalSpeed**2/2

#TotalEnergy=ElectronKineticEnergy+ProtonKineticEnergy+PE    #Total energy of atom

maxN=1e8    #Maximun number of iterations in orbital computation

DelayedTime=(re+rp)/c

dt=-DelayedTime/maxN    #Time increment

loopN=-1

theta=0.    #See diagram

RetardedProtonSpeed=vp    #Proton travels at constant speed

ar=RetardedProtonSpeed**2/rp

Fy=0.    #Initial trial force of proton on electron

#Find parameters when forces on electron sum to zero

loopN=loopN+1

#If infinite loop, Print params and exit by clicking X in display

if loopN>maxN:

print('Possible infiite loop.')

print('Abort by clicking X in your display.')

input('OK?')

#If necessary, keep 'inching up’ to loop termination condition.

RetardedTime=-loopN*dt

theta=omega*RetardedTime

RetardedVelocityX=vp*math.sin(theta)    #See diagram

RetardedVelocityY=vp*math.cos(theta)

RetardedAccelerationX=-(vp**2)/rp*math.cos(theta)

RetardedAccelerationY=vp**2/rp*math.sin(theta)

L=math.sqrt((re+rp)**2-(rp*math.sin(theta))**2)    #See diagram

alpha=math.atan(rp*math.sin(theta)/L)

Lx=-L*math.cos(alpha)

Ly=L*math.sin(alpha)

#Compute electric field at electron (see Griffith's text)

ux=c*Lx/L-RetardedVelocityX

uy=(c*Ly/L-RetardedVelocityY)

u=math.sqrt(ux**2+uy**2)

ux=-u*math.cos(alpha)

uy=u*math.sin(alpha)

Ey=q/(4*math.pi*eps0)*L/(Lx*ux+Ly*uy)**3

Ey=Ey*(uy*(c**2-RetardedProtonSpeed**2)-(Lx*(ux*RetardedAccelerationY-uy*RetardedAccelerationX)))

Fy=-q*Ey    #Interactive force on electron

L=math.sqrt((re+rp)**2-(rp*math.sin(theta))**2)    #See diagram

alpha=math.atan(rp*math.sin(theta)/L)

Lx=-L*math.cos(alpha)

Ly=L*math.sin(alpha)

print("L=",L)

print("alpha=",alpha)

print("Lx=",Lx)

print("ve=",ve," vp=",vp," rp=",rp)

print(N*h/(2*math.pi*me*re*ve)) #should be 1.