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## Main Question or Discussion Point

I'm trying to have a guest on the neutral Helium energy level diagram using the Bohr theory approach (just with n, does not include 2nd quantum number or spin). I think it should give me a close result. If not, I have to understand why. I went to NIST website:

http://physics.nist.gov/PhysRefData/ASD/levels_form.html

And see the energy level for He I (neutral He) and H I (neutral Hydrogen)

My approach is to use the Bohr radius.

For the ground state one (n=1), it would be a radius with one electron seeing 2 positive charge on the inside, so it would be the same as with radius of a ground-state-electron-of-a-singly-ionized-Helium (He II).

For the excited one in n=2, it would see the inside as 1 positive charge (because the ground state electron cancel one charge) and have the same radius as with the n=2 electron in Hydrogen of Bohr.

I calculated the difference between the 2 radius, use the Coulomb equation to get the repulsion energy, and the repulsion come out as about 7.5 eV.

So I say, this mean that to get to the ground state, the outside electron will have to overcome this much extra repulsion energy.

Is my approach wrong? and I can't seem to explain why the experimental energy of NIST is higher than mine, and my approach seems to me too simple. Maybe it's just a coincidence that my result get close.

Could someone please help me...? Thank you very much!!!

http://physics.nist.gov/PhysRefData/ASD/levels_form.html

And see the energy level for He I (neutral He) and H I (neutral Hydrogen)

__in the unit of eV__. I discover something: The energy level of Helium is just energy level of Hydrogen plus 10 eV. My hypothesis is that this extra energy comes from the repulsion energy between the 2 electrons of Helium when one in excited state and one in ground state.My approach is to use the Bohr radius.

For the ground state one (n=1), it would be a radius with one electron seeing 2 positive charge on the inside, so it would be the same as with radius of a ground-state-electron-of-a-singly-ionized-Helium (He II).

For the excited one in n=2, it would see the inside as 1 positive charge (because the ground state electron cancel one charge) and have the same radius as with the n=2 electron in Hydrogen of Bohr.

I calculated the difference between the 2 radius, use the Coulomb equation to get the repulsion energy, and the repulsion come out as about 7.5 eV.

So I say, this mean that to get to the ground state, the outside electron will have to overcome this much extra repulsion energy.

Is my approach wrong? and I can't seem to explain why the experimental energy of NIST is higher than mine, and my approach seems to me too simple. Maybe it's just a coincidence that my result get close.

Could someone please help me...? Thank you very much!!!