Energy required to ionize helium

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Hi,
It requires 79.0 eV to remove both electrons from He in the ground state. The energy to remove a single electron is 24.6 eV. What is the relationship between these two numbers?
Thanks.
 
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Maybe the question should be is there a relationship between these two numbers? Or better yet, should there be a relationship between those two numbers?
What do you think yxgao?
 
If you are given the energy to remove both electrons, is there a way to calculate the energy to just remove one electron? I am thinking that there is.
 
yxgao said:
If you are given the energy to remove both electrons, is there a way to calculate the energy to just remove one electron? I am thinking that there is.

Lithium requires almost the same for both electrons 81 but only 5.4 for a single. There doesn't seem to be a trend.
 
Is there an analytical model from which one can calculate 24.6 eV and 79.0 eV in some way?
 
yxgao said:
Is there an analytical model from which one can calculate 24.6 eV and 79.0 eV in some way?

Of course, it's just quantum mechanics applied to a three body (nucleus, electron, elctron) system. Obviously, there is no analytical solution,but one can work out the energy levels in perturbation theory. It's quite involved, as you can imagine. The potential energy from the interactions between the electrons is a major contribution.

So yes it can be calculated, but it's not given by a simple formula.

However, notice that after removing the first electron, the second electron is in a hydrogenlike atom with Z=2. Then the energy required to remove the second electron is simply Z^2 times 13.6 eV = 54.4 eV. That's the only number easy to get.

As for the 24.6 eV for the first electron, it comes out of a messy calculation. However, it's possible to get some intuitive feel for it. If the first electron was completely shielding one of the protons in the nucleus, the first electron would require 13.6 eV to be extracted. The shielding is not perfect so it takes a bit more to pull it out. Beyond that, one needs to work out some perturbation theory integrals to get a precise value.

Pat
 
This is question 18 on the Physics GRE (GR0177):

The energy required to remove both electrons from the helium atom in its ground state is 79.0 eV. How much energy is required to ionize helium (ie. to remove one electron) ?
a) 24.6 eV
b) 39.5 eV
c) 51.8 eV
d) 54.4 eV
e) 65.4 eV
 
Have you read the posts here?
 
nrqed said:
However, notice that after removing the first electron, the second electron is in a hydrogenlike atom with Z=2. Then the energy required to remove the second electron is simply Z^2 times 13.6 eV = 54.4 eV. That's the only number easy to get.
Pat

Well isn't the total energy of 79.0 eV minus that energy of 54.4 eV the answer? That's obviously the way it's supposed to be done.

But I'm not seeing how you get to 54.4 in the first place. Can someone re-explain that please?
 
  • #10
Have you heard about Rydberg formula?
 
  • #11
i think it Rhz=[1/n12-1/n22]
am i right...
 
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