Question

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.

Short answer

The first ionization energy is $24.6\hspace{0.33em}\mathrm{eV}$.

Step-by-step solution

Definition of the hyperfine harmonic oscillator

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.

The ionization energy

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 $24.6\hspace{0.33em}\mathrm{eV}$.