The H (1s) electron BE=13.6eV. What about the nucleus etc?

In summary, for a ground state Hydrogen atomic system (1 electron and 1 proton - not molecular di-hydrogen), the total energy of the nucleus is 20.8 eV and the energy of the centripetal force is also a factor in holding onto the electron. However, classical mechanics cannot be used to describe the hydrogen atom and quantum mechanics must be used instead. The only interaction between the electron and proton is electrostatic, and the discrete bound states are a result of the mathematics. Coulomb's Force equation can be used to describe the electrostatics between the electron and proton because it has been experimentally shown to work on any scale where quantization can be neglected.
  • #1
what_are_electrons
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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  • #2
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.
 
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  • #3
what_are_electrons said:
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?
TE=KE+PE applies, but you cannot use classical mechanics to describe the hydrogen atom. Rather, you must use quantum mechanics; nonrelativistic (Schroedinger) will do nicely for this problem and gives that the binding energy is 13.6eV. The only interaction between electron and proton is assumed to be electrostatic (Coulomb) in this picture; one can add corrections to this for improved accuracy.
 
  • #4
zefram_c said:
TE=KE+PE applies, but you cannot use classical mechanics to describe the hydrogen atom. Rather, you must use quantum mechanics; nonrelativistic (Schroedinger) will do nicely for this problem and gives that the binding energy is 13.6eV. The only interaction between electron and proton is assumed to be electrostatic (Coulomb) in this picture; one can add corrections to this for improved accuracy.

Please refresh my memory about what force, QM based or CM based, prevents the electron cloud from touching the nucleus.

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!
 
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  • #5
what_are_electrons said:
Please refresh my memory about what force, QM based or CM based, prevents the electron cloud from touching the nucleus.
Actually, no "force" prevents it. When one solves the Schroedinger equation, the requirement that physically acceptable solutions be normalizable results in discrete bound states (ie only some binding energies are permitted, the electron can be bound by 13.6eV or 3.4eV but nothing in between) for all attractive potentials that I know of. So a physical explanation would be that if you try to localize the electron in a small volume close to the nucleus, the resulting uncertainty in its momentum (and thence energy) eventually overcomes the Coulomb attraction and prevents further localization. The discrete bound states are a result of the mathematics.
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.
I suppose the best answer to that is "because it works" - experimentally, we find that the Maxwell equations work on any scale at which quantization of the EM field can be neglected.
 

Related to The H (1s) electron BE=13.6eV. What about the nucleus etc?

1. What does the "H (1s) electron BE=13.6eV" mean?

The notation "H (1s) electron" refers to a hydrogen atom with one electron in its first energy level, or "1s" orbital. The "BE=13.6eV" indicates the binding energy of this electron, which is the amount of energy required to remove the electron from the atom.

2. How is the binding energy of the H (1s) electron determined?

The binding energy of an electron in an atom is determined by the energy levels of the atom's orbitals and the attractive force between the nucleus and the electron. In the case of the H (1s) electron, its binding energy is equal to the energy of the first energy level, which is 13.6 electron volts (eV) for a hydrogen atom.

3. What is the significance of the H (1s) electron's binding energy being 13.6eV?

The value of 13.6eV for the binding energy of the H (1s) electron is a fundamental constant in quantum mechanics known as the Rydberg constant. It is used to calculate the binding energies of electrons in other atoms and molecules, and is an important factor in understanding atomic structure and chemical bonding.

4. How does the nucleus affect the H (1s) electron's binding energy?

The nucleus, which contains the protons and neutrons of an atom, exerts a strong attractive force on the electrons orbiting it. This attraction is what keeps the electron in its energy level and determines its binding energy. In the case of the H (1s) electron, the attractive force from the single proton in the nucleus results in the binding energy of 13.6eV.

5. Can the binding energy of the H (1s) electron change?

Yes, the binding energy of an electron can change if the atom undergoes a chemical reaction or is exposed to external energy. For example, if a hydrogen atom gains or loses an electron, the binding energy of the remaining electron in the atom will change. Additionally, if the atom is exposed to high energy radiation, such as x-rays, the electron's binding energy can increase or decrease depending on the intensity of the radiation.

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