We will look first at the original almost universally seen Bohr's atom where the proton is considered the unmoving center of the atom. The fallacy of the unmoving center introduces conceptual errors which are propagated forward to thinking that the Sun is likewise the unmoving center of our solar system. This misses the wobble of the Sun and how inertia, angular momentum and centrifugal force work. It introduces errors in mass, velocity and distance which are corrected only when you consider the electron and proton as a binary system or the Sun and planets as a system. Everything moves in an orbital system. There is no unmoving center. The electron and proton both orbit around a point between their centers which is called the barycenter, the center of mass. The electron is at a distance of re from the barycenter. The proton is at a distance of rp from the barycenter. The distance between the electron and proton is cd, the center distance. They are in a line so cd = re + rp. Only if you say that the proton has an unmoving center and rp = 0, can you say that the electron orbits the proton at a distance of re = cd. In reality, cd is always bigger than re.
Second - Bohr postulated the angular momentum of an electron in various concentric orbits around a proton equals multiples of Plank's constant/(2*pi) or hp/(2*pi). We now know hp/(4*pi)
as the spin of the electron. The angular momentum of the atom is multiples of twice the spin of the electron, an agreeable symmetry.
me*ve*re = n*hp/(2*pi), the angular momentum where, me is the mass of the electron, ve is its tangent velocity and re is the orbital radius. n is the interger orbit counter moving out from n=1 multiplied by Plank's constant hp, divided by 2*pi.
ve = hp/(2*pi*me*re) *n, isolated ve but hp = me*c*2*pi*rc/alpha, where rc is the classical radius of the electron and alpha is the fine structure constant.
ve = me*c*2*pi*rc/(2*pi*me*re*alpha) *n, substituted for hp.
(A) ve = c*rc/(re*alpha) *n, collected terms. The velocity of the electron at the n orbit.
Third - the electron orbits so the centrifugal force equals the centripetal force. When a force exerts a center seeking centripetal force, inertia opposes this deviation from straight line motion with a center fleeing centrifugal force. The centripetal force equals the inertial centrifugal force along a circular orbital path. For every action there is an equal but opposite reaction. Slinging a rock on a rope, around in a circle, demonstrates this centrifugal force which can easily be measured with a spring scale used by fishermen. You and the rock are masses in a binary system. Your centrifugal force, at your distance from the barycenter, the center of mass, equals the centrifugal force of the rock, at its distance from the common center of mass, equals the tension in the rope between the two masses. If the rope is cut or released both the centripetal and centrifugal forces become zero. The rock continues on its inertial path. You continue on your inertial path. Both paths are in opposite directions. They are determined by the momentum prior to release and are tangent to the circle at the point of release.
The centripetal Coulomb force is the electrostatic attractive force between an electron in orbit
with a proton, both with a charge of ce at a separation of re meters.
me*ve2/re = ce2/(4*pi*e0*re2), The electron centrifugal force equals the electron-proton Coulomb force but rc = ce2/(4*pi*e0*me*c2), where rc is the classical radius of the electron so we can write me*c2*rc = ce2/(4*pi*e0).
me*ve2/re = me*c2*rc/re2, substituted me*c2*rc for ce2/(4*pi*e0)
(B) ve2 = c2*rc/re, isolated ve2
Solve for the electron velocity ve:
ve2 = c2*rc/re, from (B) orbital force.
ve = c*rc/(re*alpha) *n, from (A) angular momentum in increments.
ve2/ve = ve = c2*rc*re*alpha/(n*c*rc*re) = c*alpha /n, (B)/(A)
(C) ve = c*alpha /n = c/(137.036*n), the velocity is a small fraction of the speed of light.
ve2 = c2*alpha2 /n2
Solve for the electron radius of orbit re:
ve2 = c2*rc/re, from (B) the centrifugal force equals the Coulomb force.
re = c2*rc/ ve2, but ve2 = c2*alpha2 /n2, so
re = c2*rc *n2/(c2*alpha2), the radius of the n orbit increases with n2.
(D) re = rc/alpha2 *n2 = 5.292E-11_m *n2. The center distance between the electron and proton is cd. With an unmoving proton, re = cd since re+rp = cd and rp = 0. The radius re is also of interest because it is 1/alpha = 137.036 times larger than the ring electron. The ring electron has a radius of rc/alpha and the first Bohr orbit a radius of rc/alpha2, an interesting symmetry.
Substitute values of ve and re in electron momentum:
me*ve*re = n*hp/(2*pi)
me *c*alpha/n *n2*rc/alpha2 = n*hp/(2*pi), substituted ve = c*alpha/n and re = n2*rc/alpha2.
me *c *rc/alpha = hp/(2*pi), This is twice the spin of the electron.
hp = 2*pi*me*c*rc/alpha, isolated hp as a check. This is true.
The energy of the photon in the atom is said to equal the ke kinetic energy difference of the electron alone as it jumps between varoius orbits. We will see that since the atom also includes the proton that any transition in the atom must include the transition of the proton when looking at the electron-proton atom as a binary system.
kez = electron kinetic energy
kez = hp *frequency = .5*me *ve2, substitute for hp, frequency and ve2
kez = 2*pi*me*c*rc/alpha *c/wavelength = .5*me*c2*alpha2/n2
kez = 2.1798E-18_kg*m2/s2 *1/n2 = 13.60568_eV/n2
1/wavelength = alpha3/(4*pi*rc *n2) = 10.973E6_1/m *1/n2, collected terms. This is called the Rydberg constant.
wavelength = 4*pi*rc/alpha3 *n2 = 91.126705E-9_m *n2, collected terms. A 13.60568_eV photon has a wavelength of 91.126705E-9_m. We are working with differences in kinetic energy in electron volts or Joules and as a last step converting to wavelength. If the base=2 and n=4 then we can write the kinetic energy of the photon as the difference of the kinetic energy of the base and the kinetic energy of n. A higher n or larger orbit has less energy.
kez/base2 - kez/n2 = ke of the photon
13.60568_eV/22 - 13.60568_eV/42 = 2.551065_eV
2.551065_eV * (1.602177E-19_J/eV) = 4.087256E-19_kg*m2/s2 /(h*c) = 1/(486.0097E-9_m). Working with all these units written out and conversions between various units is so much easier if you use a HP graphics calculator.
Sommerfeld added a correction for elliptical orbits and the relativistic increase in mass with velocity as the electron moves from apogee to perigee. This causes the orbit to precess and to trace rosettes in the same way that Mercury precesses in its orbital plane. Atoms must also precess out of their orbital plane for us to see atoms as spherical in scanning tunneling microscope pictures.
Physics seems to be drifting away from a physical reality. Do you think we could have arrived at Rydberg's constant, by the above circuitous route, without a
substantial portion of Bohr's planetary atom being correct?
re and rp are the distances for the electron and proton to the center of mass, the barycenter, of the electron-proton system. me and mp are their masses. ve
and vp are their orbital velocities.
Two different angular momentums, one for the electron and one for the proton:
me*vE*re + mp*vP*rp = hp/(2*pi) *n, the sum of the electron and proton angular momentum in increments of Planck's constant divided by 2*pi.
me*vE*(re + rp) = hp/(2*pi) *n, substituted me*vE = mp*vP. This is me*vE*cd.
me*vE*cd = hp/(2*pi) *n, substitute for vE = c*alpha*k/n and cd = rc/(k*alpha2). When value of vE goes up then the value of cd must go down for the right hand side of the equation hp/(2*pi), to remain unchanged.
me*c*k*alpha/n *rc/(k*alpha2) *n2 = hp/(2*pi) *n, the k's cancel in this equation. The k's will be used to adjust the value of two other equations below while keeping the overall value of the present equation unchanged.
me*c*rc/alpha = hp/(2*pi), collected terms. One spin for the electron and one spin for the proton.
hp = 2*pi*me*c*rc/alpha, isolated hp. This is true.
We can isolate k as follows:
me*vE2/rE = ce2/(4*pi*e0*cd2), the centrifugal force equals the Coulomb force.
me*(c*alpha*k)2/rE = me*c2*rc/cd2, where me*c2*rc = ce2/(4*pi*e0)
me*c2*alpha2*k2 *(me+mp)/(cd*mp) = me*c2*rc/cd2
alpha2*k2 *(me+mp)/mp = rc/cd
alpha2*k2 *(me+mp)/mp = rc*k*alpha2/rc, where cd = rc/(k*alpha2)
alpha2*k *(me+mp)/mp = rc*alpha2/rc,
k*(me+mp)/mp = 1, collected terms.
k = mp/(me+mp) = 0.999455679421 = 1 - 0.000544320578,
where (me+mp)/mp = 1 + (me/mp) = 1 + (keE/keP) = 1.00054461703.
where vE = c*alpha*k = c*alpha*mp/(me+mp)
and cd = rc/(alpha2 *k) = rc/alpha2 *(me+mp)/mp.
We know the electron and proton pair can have elliptical paths as this is required for atoms to be polarized and precess into polarized ellipsoids which can attract each other. See electrostatic gravity.
The energy of the photons in the atom equals the kinetic energy of the electron and the proton.
We will see that since the atom also includes the proton that any transition in the atom must include the transition of the proton. There is a division of the energy between the proton and electron.
For the electron:
hp *frequency = .5*me *ve2, substitute for hp, frequency and ve2 = c2*alpha2*mp2/(me+mp)2
2*pi*me*c*rc/alpha *c/wavelength = .5*me*c2*alpha2*mp2/(me+mp)2 /n2 =
2.17749984211E-18_kg*m2/s2 *1/n2 = 13.5908791202_eV/n2
1/wavelength = alpha3*mp2/(4*pi*rc *(me+mp)2 *n2) = 10.9617791803E6_1/m *1/n2, collected terms. This is the Rydberg constant less the proton.
wavelength = 4*pi*rc/alpha3 *n2 = 91.2260668233E-9_m *n2, collect terms. These transitions produce photons in or near the range of visible light.
|Wavelength table in nano-meters|
The first two rows:
3 2 656.2852_nm = 1.88918239631_eV, n=3, base=2, hydrogen alpha, electron spin up?
3 2 656.2720_nm = 1.8892203946_eV, n=3, base=2, hydrogen alpha, electron spin down? The measured difference in the wavelength 0.0132_nm or 0.000038_eV between these first two rows is attributed to the spin of the proton being spin up and the spin of the electron being either spin up or spin down. See the forces due to moving charge.
The third row - using the 486.133_nm spectral line:
486.133E-9_m = 486.133_nm = 4.0862220883E-19_kg*m2/s2 = 2.55041819174_eV, which is the correct value for the 4 to 2 transition line in the Balmer series in the hydrogen spectrum.
Here keE = .5*me *vE2 = .5*me *c2*alpha2 *k2 = 13.6022303559_eV. The k in vE of the kinetic energy of the electron keE, is adjusted to give the correct kinetic energy of the electron to yield the correct wavelength of the photon.
keE/4 - keE/16 = 4.0862220883E-19_kg*m2/s2. On the right,
k = 0.999872969911 = 1 - 0.000127030089, so the kinetic energy changes by about one part in ten thousand to yield the exact wavelength. keE = 13.6022303559_eV is a 91.1499375E-9_m photon or a slightly revised Rydberg constant of 10970934.5659_1/m.
The third row:
4 2 2.550418_eV 2.551066_eV 2.549878_eV, n = 4, base = 2, shown in eV. We see the difference in the electron volt equivalents is about 0.0006_eV. This is easily hidden in the room temperature thermal noise of about 0.04_eV, as in the following calculation.
.5*mass*velocity2 = 3/2 *kB*T = 0.037892_eV, the kinetic energy caused by thermal velocity at T = 20_C = 68_F = 293_K. Boltzmann's constant = kB = 8.617385E-5_eV/K. See hyperphysics.
proton kinetic energy = keP = .5*mp*vp2,
where vp = ve * me/mp = c*alpha*mp/(me+mp) *me/mp = c*alpha*me/(me+mp).
keP = .5*mp *c2*alpha2*me2/(me+mp)2 /n2 = 1.18590347957E-21_kg*m2/s2 *1/n2 = 0.00740182411375_eV/n2.
The electron kinetic energy plus the proton kinetic energy =
13.5908791202_eV + 0.00740182411375_eV = 13.5982809443_eV call it 13.5983_eV the sum of the electron and proton energy closely agrees with the NIST value of 13.5984_eV. This is the total kinetic energy of the electron-proton pair in the hydrogen atom. Someone said that the Bohr atom gives incorrect values for the ionization energy of the hydrogen atom - however when the Bohr atom is considered as a binary system with a moving proton, it gives the NIST value. This kinetic energy is said to be able to remove both the electron and proton to infinity. It is also said to be the combined binding energy of both the electron and proton into an atom of hydrogen.
The energy in the electrostatic field equals the sum of the kinetic energies:
ce2/(4*pi*e0*cd), this is the energy in the electrostatic field for the electron-proton pair,
substitute me*c2*rc = ce2/(4*pi*e0) and cd = re+rp = rc*(me+mp)/(mp*alpha2) so,
me*c2*rc *mp*alpha2/(rc*(me+mp)) = me*mp*c2*alpha2/(me+mp), collected terms, the binding energy is half the energy of the electrostatic field so,
.5*me*mp*c2*alpha2/(me+mp) = 2.17868574559E-18_kg*m2/s2 = 13.5982809443_eV, equals the sum of the electron and proton kinetic energies. This is the hydrogen atom mass deficit since ionization separates the electron and proton.
We previously calculated 13.6022_eV yields the exact spectral lines. Here we see 13.5908_eV for the electron or 0.0114_eV difference or that of hydrogen heated from absolute zero Kelvin to 132_Kelvin = -221_F. Apparently we need to know the temperature at which the lines in the hydrogen spectrum are measured to resolve this small difference in the energy of the spectral lines.
Mass deficits: See Wiki binding energy or NIST index or NIST constants
1_amu = 1/12 of the mass of a C-12 atom = 1.66053886E-27_kg/amu
1.00782503207_amu *1.66053886E-27_kg/amu = 167353262.98E-35_kg from NIST for the hydrogen atom mass for comparison with our calculation below.
13.5984_eV/c2 = 2.42E-35_kg is small when compared to the mass of the hydrogen atom so we will scale our values and crop the decimal places to two after the decimal point.
The mass of a proton is ______________167262163.70E-35_kg = 1.007276_amu
The mass of an electron is ________________91093.82E-35_kg = 0.000549_amu (510998.6410_eV/c2)
The mass of a proton plus an electron is __167353257.52E-35_kg = 1.00782499918_amu
The mass of ionization for the electron and proton is 2.42E-35_kg = 0.00000001459_amu (13.5984_eV/c2)
The mass of a hydrogen atom is ________167353255.10E-35_kg = 1.00782498458_amu, the proton and electron mass minus ionization eV/c2. The NIST value for the hydrogen atom is greater than the sum of the electron and proton masses, which can not be true if you believe in binding energy.
The mass of ionization for the electron only is _____13.5910_eV/c2
The mass of ionization for the proton only is _______0.0074_eV/c2
The mass of ionization for the electron and proton is 13.5984_eV/c2
The virial theorem: says the kinetic energy equals half the gravitational potential energy.
m*vt2/r = G*m*M/r2, the centrifugal force equals the gravitational force.
.5*m*vt2 = .5*G*m*M/r, multiplied by .5*r, the kinetic energy equals half the gravitational potential energy. It works just as well with electrostatic forces.
Binding energy: The energy required to leave an orbit and escape to infinity is called the Binding energy. Both the potential and kinetic energy
at infinity are zero. The kinetic energy is always positive. Gravitational energy is taken as negative.
-.5*G*m*M/r, is the energy in orbit so, the energy to make the kinetic energy zero at infinity is,
.5*G*m*M/r, the energy added so,
.5*G*m*M/r, is the binding energy.