Calculating second ionization energy of He

  • Thread starter BobRoss
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  • #1

Homework Statement

The equation En=-Z2RH/n2 gives the energy (according to the Bohr model) of the energy levels available to an electron in any hydrogen-like atom. Calculate the second ionization energy of He(in J). The second ionization energy of He is the ionization energy of He+.

Homework Equations

En=-Z2RH/n2 where RH=2.178x10-18J


The Attempt at a Solution

I don't understand which value I am supposed to use for n. When I look up the second ionization energy of He I find that the answer is 54.4 eV or 8.712x10-18J, which is the answer I get when I input n=1 into the above equation. But it's asking for the second ionization energy of He, so why would I use n=1?

Answers and Replies

  • #2
Gold Member
n=1 is the orbit number of He. He has configuration of 1s2, meaning that both 2 electrons are in 1st orbit. 2nd Ionization energy refers to removal of 2nd electron, or removal of only electron of He+ ion, in this case.

Z refers to the atomic number of the species. For He, Z = 2.
  • #3
Thanks for the question.
  • #4
I think there is some problem in understanding here.
The formula En=-Z2RH/n2 can be used only for Hydrogen like atoms i.e. we can use this formula for He+ and not for ground state of Helium.
Ground state electron configuration of Helium is 1s2 while the configuration for He+ is 1s1 same as of Hydrogen.
Therefore, the energy given by En=-Z2RH/n2 will be the ionization energy for He+

The first ionization energy of He will be E1=24.5 eV and it is difficult to calculate that because of the presence of strong interactions.
Then, with this energy E1, we get He+ which is Hydrogen like atom and now we will use that formula and we get 54.4 eV.

Therefore, the process of removal of 2nd electron from He takes place in two steps ad hence the total energy required to knock off two electrons will be 24.5 eV +54.4 eV = 78.9 eV
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