Chemistry - Energy required to remove an electron from He

Solution 1:

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.

Solution 2:

Taking half the total value of both ionization energies isn't a bad initial approach. You were just one step shy of finding the answer (without calculations). Each additional ionization energy will be greater than the last because the potential will be greater; the fewer the electrons, the less electron-electron repulsive force. Thus, the first ionization energy is less than half the total. Only answer A satisfies that requirement, so that must be the answer.