Bohr's planetary atom modified 20120218
Neils Bohr unified Rutherford's nuclear atom with Balmer's formula for the spectrum of the hydrogen atom in 1913 leaving us with the indelible image of the planetary atom. The planetary atom is the prototype of tiny machines which illuminate matter. Planetary systems rotate in a plane while atomic systems precess from a plane into ellipsoids which we see in scanning tunneling microscope pictures but this does not detract from Bohr's work. We address the source of this in precession and electrostatic gravity.
  1. Balmer's formula
  2. Bohr's original atom
  3. Bohr's atom as a binary system
  4. Balmer series hydrogen spectral lines
  5. Ionization
Balmer's formula
Balmer's 1885 formula which fits many of the lines in the hydrogen spectrum can be written,
wavelength = 4*pi*rc/alpha3 *base2 *n2/(n2- base2), where rc is the classical radius of the atom and alpha is a constant 0.00729735 or 1/137.035989.
wavelength = 91.126705_nm *base2*n2/(n2- base2), for wavelengths greater than 91.1E-9_m.
The base=2 in the visible Balmer series of spectral lines and n is greater than two. When n=3 the Balmer series returns a wavelength of 656.3E-9_m, the red of hydrogen-alpha. The base=1 in the ultraviolet Lyman series and n is greater than one. When n=2 the Lyman series returns a wavelength of 121.5E-9 _m, called hydrogen Lyman-alpha.

Bohr's original atom
First - Bohr used Rydberg who used Balmer so Bohr also fits the spectral data. Bohr linked Rydberg's formula with the kinetic energy changes, due to the transition or jumping of an electron between various orbits, in electrons orbiting the nucleus of atoms. These kinetic energy changes emit and absorb photons visible in the hydrogen spectrum.

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) = 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 at the n orbit. While n and the orbits increase the velocity and angular momentum decrease.

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.
(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.

ke = electron kinetic energy
ke = hp *frequency = .5*me *ve2, substitute for hp, frequency and ve2
ke = 2*pi*me*c*rc/alpha *c/wavelength = .5*me*c2*alpha2/n2
ke = .5*me*c2*alpha2 /n2 = 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.
ke/base2 - ke/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.

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?

Bohr's atom as a binary system
Binary systems can generalize Bohr's planetary atom with an unmoving center proton to a system where both the electron and proton move producings red-electron and blue-proton currents and green magnetic fields which are forces due to moving charge. We see two atoms when their protons and electrons line-up in their in-phase orbits. The binary systems contains the proper amount of energy to agree with the Balmer series hydrogen spectral lines while also agreeing with the energy of ionization. When the hydrogen atom is ionized the electron-proton pair separate and absorb energy. When the electron-proton pair merge to become a hydrogen atom they give off energy.

A binary system with orbiting charged particles has wavelike orbits
  • The red dots are electrons and the blue dots are protons orbiting on their circular paths.
  • The red and blue sine waves are an edge view of the orbital plane and currents traced out by the electron and proton pair as they move across the page on a helical path like a spring on a string.
  • Both orbit each other on opposite sides of the center of mass of the system.
  • There are both wave and particle descriptions of atoms. We will be looking at both.
  • The orbital period of the proton and electron pair is the same. They are a dipole.
  • Looking at a point, in the orbital plane as they orbit, would show alternating charges and dipole forces at the frequency that the electron and proton orbit and pass in front of each other. Blue-red-blue-red or plus-minus-plus-minus at 6.6E15_hertz as the dipoles reverse direction at a wavelength of 45.5E-9_m in the extreme ultraviolet.
  • A distant static charge would only see the oscillating high frequency plus-minus-plus-minus merged to neutrality which excludes gravity from being caused by the interaction of dipoles and static charges.
  • A distant in-phase rotating dipole would however experience Coulomb, magnetic and gravitational forces.
The stationary unmoving center proton and orbiting electron view of the planetary atom conceals their binary wavelike behavior which the red and blue sine waves emphasize. Matrix mechanics is equivalent to wave equations for future reference.

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.

  1. me*ve = mp*vp, the momentum of the electron and proton are equal.
  2. me*re = mp*rp, the mass times distance products are equal and balanced.
  3. ve/re = vp/rp, this is (1)/(2), the angular velocity and orbital periods are equal.
  4. me*ve2/re = mp*vp2/rp, this is (1)*(3), the centrifugal forces are equal to each other.
  5. cd = re + rp, where cd is the center distance between the electron and proton.
    cd = re + me*re/mp, since rp = me*re/mp.
    cd = re*(1+me/mp) = re*(mp+me)/mp = re*(1+1/1836.15) = re*1.0005446
    cd = mp*rp/me + rp, since re = mp*rp/me.
    cd = rp*(mp/me + 1) = rp*(mp+me)/me,
  6. cd = rc/alpha2 *n2 = 5.29178E-11_m *n2, from (D), but cd must vary in an elliptical orbit.
  7. re = cd *mp/(mp+me) *n2 = rc/alpha2 *mp/(mp+me)) *n2 = 5.28889E-11_m *n2, the distance between the electron and the center of mass.
  8. rp = cd *me/(mp+me) *n2 = rc/alpha2 *me/(mp+me) *n2 = 2.88042E-14_m *n2, the distance for orbit n between the proton and the center of mass.
  9. ve = c*alpha/n = c/(137.036*n) = 2187691.56_m/s *1/n, from (C), the velocity the distance for orbit n of the electron around the center of mass.
  10. vp = ve*me/mp = me/mp *c*alpha/n = me/mp *c/(137.036*n) = 1191.45_m/s *1/n = 2665.2 miles per hour, the velocity the distance for orbit n of the proton around the center of mass.
  11. ve/(2*pi*re) = frequency = 6.58327E15_1/s *1/n3
    12.
  12. c/frequency = wavelength = 45.54E-9_m. This is in the extreme ultraviolet, EUV. Here is a reference to the Solar Dynamics Observatory, (SDO) "EUV wavelengths range between 50 and 5 nanometers, which coincide with the characteristic absorption wavelengths of inner-shell electrons in the atoms that compose matter. As a result, EUV light directed onto a standard mirror or lens at normal incidence is absorbed rather than reflected, making it undetectable. For this reason, EUV light is also absorbed by Earth’s atmosphere, which is why telescopes must travel to space to study the light emitted from the Sun."

Two different angular momentums, one for the electron and one for the proton:
(Eq1) 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.

Centrifugal force of the electron = centrifugal force of the proton = Coulomb force:
(Eq3) me*vE2/rE = mp*vP2/rP = ce2/(4*pi*e0*cd2) = 8.2297492806E-8_kg*m/s2,
where k = mp/(me+mp) and vE = c*alpha*k
and cd = rc/(alpha2*k) and rE = cd*k = rc/alpha2. These are fantastically strong forces with respect to the tiny masses. The electron and proton are locked together. The electron acceleration, force/mass, is 9.034E22_m/s2. The Earth's g = 9.8_m/s2.

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. Bohr only included the energy of the 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
.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 the range of visible light.

For the proton:
hp *frequency = .5*mp *vp2, and vp2 = ve2*me2/mp2 = c2*alpha2*me2/(me+mp)2.
.5*mp*vp2 = .5*mp *c2*alpha2*me2/(me+mp)2
.5*mp *c2*alpha2*me2/(me+mp)2 = 1.18590347957E-21_kg*m2/s2 *1/n2 = 0.00740182411375_eV/n2
hp*c/wavelength = .5*mp *c2*alpha2*me2/(me+mp)2
2*pi*me*c*rc/alpha *c/wavelength = .5*mp *c2*alpha2*me2/(me+mp)2, substituted for hp
1 /wavelength = alpha3*me*mp /(4*pi*rc*(me+mp)2*n2) = 5969.97158832_1/m *1/n2. This is the Rydberg constant for a proton.
wavelength = 4*pi*rc*mp/(alpha3*me) *n2 = 1.67504984774E-4_m *n2, collect terms. In the electromagnetic spectrum these transitions produce photons in the range of microwave or infrared waves. When the electrons produce light the protons produce heat. Has anyone noticed these long wavelength proton transistions which seem to be required for ionization?
Balmer series hydrogen spectral lines
wavelength = 4*pi*rc/alpha3 *base2 *n2/(n2- base2).
wavelength = 91.126705_nm *base2*n2/(n2- base2), is Balmer's formula which can be rewritten,
1/wavelength = 10.973E6_1/m *(1/base2-1/n2), per Rydburg. Clearly, Balmer is better.
Wavelength table in nano-meters
nbasemeasuredBohrBohr*(1+(me/mp))
32656.2852_nm656.1128_nm656.4702_nm
32656.2720_nm656.1128_nm656.4702_nm
42486.1330_nm486.0095_nm486.2742_nm
52434.0470_nm433.9372_nm434.1734_nm
62410.1740_nm410.0705_nm410.2938_nm
72397.0072_nm396.9078_nm397.1239_nm
82388.9049_nm388.8076_nm389.0193_nm
92383.5384_nm383.4426_nm383.6514_nm
  • First three columns; n, base, measured wavelength from hyperphysics.
  • Fourth column, calculated wavelength from Bohr, Balmer or Rydberg.
    wavelength = 91.126705_nm *base2*n2/(n2- base2), these wavelengths are a little too short so the transitions show too much energy.
  • Fifth column, calculated wavelength from Bohr *(1+me/mp) considering the hydrogen atom as a binary system.
    wavelength = 91.176410_nm *base2*n2/(n2- base2), these wavelengths are a little too long so the transitions show not quite enough energy. The proton also has long wavelength and low energy transitions which occur at the same time as the electron transitions but they are not listed.

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 when necessary. 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 *k*T = 0.037892_eV, the kinetic energy caused by thermal velocity at T = 20_C = 68_F = 293_K. Boltzmann's constant = k = 8.617385E-5_eV/K. See hyperphysics.

Ionization
is the energy added to separate the electron and proton in a hydrogen atom. It can also be seen as the binding energy given off when the electron and proton become a hydrogen atom. It is the sum of the kinetic energy of the electron and proton. NIST gives 13.5984_eV for the ionization energy.

electron kinetic energy = keE = .5*me*ve2, where ve = c*alpha*mp/(me+mp)
keE = .5*me*c2*alpha2*mp2/(me+mp)2 /n2 = 2.17749984211E-18_kg*m2/s2 *1/n2 = 13.5908791202_eV/n2.

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. It is sometimes said that the Bohr atom give 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.

Many decimal places are required to cope with the smallness of 13.5984_eV = 2.42E-35_kg 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 so,
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, multiply 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.

Reference:
This is a wonderful book, Hydrogen The Essential Element by John H. Rigden.

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