# Energy required to remove an electron from He

The energy required to remove both electrons from the helium atom in its ground state is $$79.0\;\mathrm{eV}$$. The energy required to ionize helium (i.e. to remove one electron) is -

(A) $$24.6\;\mathrm{eV}$$

(B) $$39.5\;\mathrm{eV}$$

(C) $$51.8\;\mathrm{eV}$$

(D) $$54.4\;\mathrm{eV}$$

My thoughts: What I actually did was to take half of the given energy value since energy required to remove electrons depends only on the principal quantum number(and also on atomic number). But, I found my answer to be wrong.

On the other hand just calculating the energy required to remove one electron from $$n=1$$ state gives the correct answer. Why is it so?

• because Helium is more stable in its ground state so energy required to remove first electron will be higher than second. – ashu Oct 6 '13 at 20:36

We can use Bohr's atomic model to calculate the ionization energy of $$\ce{He+}$$. The ionization energy will be the amount of energy we must give to the electron of singly ionized helium to remove it apart from the nucleus to infinity. That gives $$IE_2=E_{\infty}-E_1$$ where $$E_1$$ is the energy in the first bohr-orbit (principal quantum number=$$1$$). Conventionally taking $$E_{\infty}$$ as $$0$$ we use the Bohr's equation to find the energy $$E_1$$:- $$E_n=-\frac{Z^2me^4}{8n^2h^2\epsilon_0^2}$$ where $$Z$$= atomic number, $$e$$= charge on electron, $$n$$= principal quantum number, $$h$$= Planck's constant, $$m$$= mass of electron, $$\epsilon_0$$= permittivity of free space.
Using $$n=1$$, $$Z=2$$ and the appropriate values for other constants we get $$IE_2=-E_n=\pu{54.4eV}$$. Now the energy for removing both electron $$IE_{net}=IE_1+IE_2$$. Using $$IE_{net}=\pu{79eV}$$and $$IE_2=\pu{54.4eV}$$ we get, $$IE_1=\pu{24.6eV}$$ which corresponds to option (A), which I assume is typed wrong by you in the question.