# Energy required for ionizing Helium

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 (i.e., to remove one electron)?

(A) 24.6 eV
(B) 39.5 eV
(C) 51.8 eV
(D) 54.4 eV
(E) 65.4 eV


I can easily say that the answer will be (A) because single electron removing requires less than half the energy of removing both electrons. But if it is not multiple choice, I have to determine the answer by using an equation. What equation would I use to get this answer?

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$$He \longrightarrow He^+ +e^- \ \ \ \ \ .....\Delta H_1$$ $$He^+\longrightarrow He^{+2} +e^-\ \ \ \ .......\Delta H_2$$
We know $\Delta H_2$ by the Bohr's model of H-like atom .
Ionization energy of a H-like atom with atomic number $Z$ is $\Delta H_2=13.6Z^2/n^2$ ($e^-$ initially in $n^{th}$ orbit)
Also, $\Delta H_1 +\Delta H_2=79\text{ eV}$
So, we can get $\Delta H_1$.
The line of reasoning here is instructive. We don't have closed form solution for the two-electron atom which seems like a show-stopper, but we have been given the figure for fully ionizing Helium (which is presumably obtained either experimentally or numerically) and the $\mathrm{He}^{+1}$ ion is hydrogen-like which means we do have a theory for it, and that splits the problem down the middle. –  dmckee May 8 '13 at 18:13