The neutral carbon has six electrons, in which two 1S electrons are very close to the nucleus (see above C3+ section), and four outer-shell electrons are thought to be arranged tetrahedrally. In the neutral carbon (C), how are four outer-shell electrons moving ? This means that it is almost correct to say that the three electrons of the carbon ion (C3+) are moving like Fig.1. (This small error is probably caused by using the "approximate" orbit of the 2S electron.) So the calculation error is only 1.87 eV ( 0.20 % ). This results shows that when the ground state energy of C3+ ion is -944.70 eV, the orbital length becomes just 1.00000 × de Broglie's wavelength. Results of r1 and WN in which last VY is zero (C3+). 1 shows the results in which the last VY ( y component of electron 1's velocity after moving one quarter of its orbit ) is the closest to zero in C3+. Then we compute the number of de Broglie's waves contained in one quarter of the 1S orbit like C4+ In the same way, the y component of the acceleration (m/sec 2) of the electron 1 is, The x component of the acceleration (m/sec 2) of the electron 1 is, Where ra is the distance between the electron 1 and the nucleus, rb is between the two 1S electrons, and rc is between the electron 1 and 2S electron, respectively. So the distances among the particles are, Like C4+ (see this page ), when the electron 1 is at (x, y, 0), the electron 2 is at ( -x, 0, y ). shows the relative positions, But actually, this 2S electron is rotating around the nucleus slower than the 1S electrons.) The two electrons of 1S state are avoiding 2S electron, so we fix the 2S electron at ( 0, - Rb / √2, - Rb / √2 ) as shown in Fig. So we suppose the carbon ion (C3+) model as follows, In the 2S electron of C3+, we substitute n=2 and z=4 into this equation. The energy levels of the hydrogen-like atom is known to be, (Here we use the new units, 1 MM = 10 -14 meter, 1 SS = 10 -22 second, 1 MM/SS = 10 8 m/s ) In 2S electron of C3+, the positive charge of the nucleus (z) is supposed to be 4+ (see above), and n is 2. (when z =1 and n = 1, this Rb becomes "Bohr radius.") (This means that we can use the equations of the hydrogen-like atom in 2S electron approximately.)Īccording to the Bohr hydrogen-like model, the radius (Rb) of the n × de Broglie's wavelength orbit is, So approximately, we can suppose the 2S electron is moving around the 4e+ nucleus ( +6e - 2e = +4e ) on the circular orbit of the two de Broglie's wavelength. The two electrons of 1S state are attracted to the 6e+ nucleus strongly, so they are much closer to the nucleus than the 2S electron. Here we try the three-electron Carbon ion (C3+).Ĭarbon ion (C3+) has two electrons in 1S orbital, and one electron in 2S orbital. Visualization of four valence electrons in Carbon atom (C).Surprisingly, this new atomic structure of the Bohr's helium can be applied to other two-electron atoms (ions ), Lithium ion (Li+), Beryllium (Be2+), Boron (B3+) and Carbon (C4+) ions, too.įurthermore, the ionization (ground state) energy of the three-electron atom lithium (Li) can be calculated correctly using the approximate "2S" Bohr orbit. This model can explain the phenomena of Pauli exclusion principle correctly, because there is no space for the third electron to enter this new two-electron Bohr atom. In this new successful Bohr model, the two electrons of the helium atom (He) are moving on the orbits of just one de Broglie's wavelength which are perpendicular to each other. Our new Bohr model has succeeded in calculating the Helium ionization energy more correctly than the quantum mechanical variational methods as shown in the Top page. Top page (correct Bohr model including the two-electron atoms). (OilerLagrangian, CC BY-SA 4.Bohr's Carbon (ion) New Bohr model Carbon (C) \): Two balls moving from a state of higher potential energy to a lower state of potential energy.
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