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The H (1s) electron BE=13.6eV. What about the nucleus etc? |
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| Jan22-05, 09:17 PM | #1 |
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The H (1s) electron BE=13.6eV. What about the nucleus etc?
For a ground state Hydrogen atomic system (1 electron and 1 proton - not molecular di-hydrogen), what energies are being exerted by the proton to hold that electron? Does TE = KE + PE apply?
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| Jan24-05, 01:56 AM | #2 |
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OK. If we use Pearson's value of 7.2 eV for the electronegativity of Hydrogen, and assume it represents unused Binding Energy (BE) then the total energy of the nucleus must be 20.8 eV (the sum of 13.6 and 7.2 eV). Naturally the energy of the nucleus is opposite in sign.
So the nucleus has at least 20.8 eV of energy to use to hold electrons in orbit. Is this the total amount of energy that the nucleus exhibits in its effort to hold onto the 1s electron and attract other electrons? Centripetal force is said to be the main force that keeps the electron from being pulled into physical contact with the nucleus. If this is true, then because Force x Distance = Work and Work is the equivalent of Energy, we should be able to determine the Energy of the Centripetal force. This represents another bit of energy that the nucleus must deal with in its effort to hold onto the 1s electron and attact other electrons. Another question is: What is the distance factor that needs to be used to calculate this Energy due to the centripetal force? Is it the atomic radius? PS I am deliberately ignoring the magnetic field effect, corrected center of mass values and self-energy factors for the moment. |
| Jan24-05, 02:07 AM | #3 |
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| Jan24-05, 02:31 AM | #4 |
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The H (1s) electron BE=13.6eV. What about the nucleus etc?And, why is it that we can use Coulomb's Force equation, that was derived from a large macro-scale torsion balance, to describe the electrostatics (or EM fields) that exist between the electron and the proton of hydrogen. Many thanks for your help! |
| Jan24-05, 02:53 AM | #5 |
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