Using ${\mathbf{E}}_{\mathbf{gs}}\mathbf{=}\mathbf{-}\mathbf{79}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathbf{eV}$ for the ground state energy of helium, calculate the ionization energy (the energy required to remove just one electron). Hint: First calculate the ground state energy of the helium ion, ${\mathbf{He}}^{\mathbf{+}}$, with a single electron orbiting the nucleus; then subtract the two energies.

A harmonic oscillator is a system that experiences a restoring forceFproportionate to the displacement xwhen it is moved from its equilibrium position, where k is a positive constant.

Subtract Helium's ground state energy from its first excited state yields the first ionization energy (energy necessary to remove one electron).

The energy of the Helium ion's${\mathrm{He}}^{+}$ground state is equal to the energy of the first excited state.

So, calculate the ground state energy of${\mathrm{He}}^{+}$as:

${\mathrm{He}}^{+}$ is a Hydrogen like atom. It contains just one electron but two protons in its nucleus, which indicates it has only one electron.

Hydrogen atom energy,

${\mathrm{E}}_{\mathrm{H}}=-13.6\hspace{0.33em}\mathrm{eV}\frac{{\mathrm{Z}}^2}{{\mathrm{n}}^2}$

ForHydrogen ground state n = 1 , and Z = 1.

For helium, Z = 2, then the expression will be :

$$\begin{gathered} {\mathrm{E}}_{\mathrm{He}}=-13.6\hspace{0.33em}\mathrm{eV}\left(\frac{2^2}{1^2}\right) \\ {\mathrm{E}}_{\mathrm{He}}=-54.4\hspace{0.33em}\mathrm{eV} \end{gathered}$$

Ground state energy for helium is:

$E_{gs}=-79.0\hspace{0.33em}eV$

Subtracting$E_{He}$ and$E_{gs}$ ground state energy as:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{join}}=\left(-54\mathrm{eV}\right)-\left(-79\mathrm{eV}\right) \\ =24.6\mathrm{eV} \end{gathered}$$

Thus, the first ionization energy is $24.6\hspace{0.33em}\mathrm{eV}$.