# 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 $IE_2=-E_n=54.4eV$. Now the energy for removing both electron $IE_{net}=IE_1+IE_2$. Using $IE_{net}=79eV$and $IE_2=54.4eV$ we get, $$IE_1=24.6eV$$ which corresponds to option (A), which I assume is typed wrong by you in the question.