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4 June 2012

- Balmer's formula

- Bohr's original atom

- Bohr's atom as a binary system

- Balmer series hydrogen spectral lines

- Ionization

The

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*ve ^{2}/re = ce^{2}/(4*pi*e0*re^{2}),** The electron centrifugal force equals the electron-proton Coulomb force but

**Solve for the electron velocity ve:**

**ve ^{2} = c^{2}*rc/re,** from

**Solve for the electron radius of orbit re:**

**ve ^{2} = c^{2}*rc/re**, from

**Substitute values of ve and re in electron momentum:**

**me*ve*re = n*hp/(2*pi)**

**me *c*alpha/n *n ^{2}*rc/alpha^{2} = n*hp/(2*pi),** substituted

**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 *ve ^{2}**, substitute for hp, frequency and ve

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?

- 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.

**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.

**me*ve = mp*vp**, the momentum of the electron and proton are equal.**me*re = mp*rp**, the mass times distance products are equal and balanced.**ve/re = vp/rp**, this is**(1)/(2)**, the angular velocity and orbital periods are equal.**me*ve**this is^{2}/re = mp*vp^{2}/rp,**(1)*(3)**, the centrifugal forces are equal to each other.**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**,

**cd = rc/alpha**, from^{2}*n^{2}= 5.29178E-11_m *n^{2}**(D)**, but**cd**must vary in an elliptical orbit.

**re = cd *mp/(mp+me) *n**, the distance between the electron and the center of mass.^{2}= rc/alpha^{2}*mp/(mp+me)) *n^{2}= 5.28889E-11_m *n^{2}**rp = cd *me/(mp+me) *n**, the distance for orbit^{2}= rc/alpha^{2}*me/(mp+me) *n^{2}= 2.88042E-14_m *n^{2}**n**between the proton and the center of mass.**ve = c*alpha/n = c/(137.036*n) = 2187691.56_m/s *1/n,**from**(C)**, the velocity for orbit**n**of the electron around the center of mass.**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.- ve/(2*pi*re) = frequency = 6.58327E15_1/s *1/n
^{3} **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:**

**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*alpha ^{2})**. When value of

where

and

**We can isolate k as follows:**

where

where

and

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 *ve ^{2},** substitute for hp, frequency and

Wavelength table in nano-meters | ||||
---|---|---|---|---|

n | base | measured | Bohr | Bohr*(1+(me/mp)) |

3 | 2 | 656.2852_nm | 656.1128_nm | 656.4702_nm |

3 | 2 | 656.2720_nm | 656.1128_nm | 656.4702_nm |

4 | 2 | 486.1330_nm | 486.0095_nm | 486.2742_nm |

5 | 2 | 434.0470_nm | 433.9372_nm | 434.1734_nm |

6 | 2 | 410.1740_nm | 410.0705_nm | 410.2938_nm |

7 | 2 | 397.0072_nm | 396.9078_nm | 397.1239_nm |

8 | 2 | 388.9049_nm | 388.8076_nm | 389.0193_nm |

9 | 2 | 383.5384_nm | 383.4426_nm | 383.6514_nm |

**First three columns**;**n, base, measured wavelength**from hyperphysics.

**Fourth column**,**calculated wavelength**from Bohr, Balmer or Rydberg.

**wavelength = 91.126705_nm *base**, these wavelengths are a little too short so the transitions show too much energy.^{2}*n^{2}/(n^{2}- base^{2})

**Fifth column**,**calculated wavelength**from Bohr *(1+me/mp) considering the hydrogen atom as a binary system.

**wavelength = 91.176410_nm *base**, 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.^{2}*n^{2}/(n^{2}- base^{2})

**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*m ^{2}/s^{2} = 2.55041819174_eV**, which is the correct value for the 4 to 2 transition line in the Balmer series in the hydrogen spectrum.

Here

**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*velocity ^{2} = 3/2 *k_{B}*T = 0.037892_eV**, the kinetic energy caused by thermal velocity at

**proton kinetic energy = keP = .5*mp*vp ^{2}**,

where

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:**

**ce ^{2}/(4*pi*e0*cd),** this is the energy in the electrostatic field for the electron-proton pair,

substitute

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/c ^{2} = 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

The mass of an electron is

The mass of a proton plus an electron is

The mass of ionization for the electron and proton is

The mass of a hydrogen atom is

The mass of ionization for the electron only is **_____13.5910_eV/c ^{2}**

The mass of ionization for the proton only is

The mass of ionization for the electron and proton is

**The virial theorem:** says the kinetic energy equals half the gravitational potential energy.

**m*vt ^{2}/r = G*m*M/r^{2},** the centrifugal force equals the gravitational force.

**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.

This is a wonderful book,

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