Verify that the ground state energy \({{\rm{E}}_{\rm{0}}}\) is \({\rm{13}}{\rm{.6eV}}\) by using

\({{\rm{E}}_{\rm{0}}}{\rm{ = }}\frac{{{\rm{2}}{{\rm{\pi }}^{\rm{2}}}{\rm{q}}_{\rm{e}}^{\rm{4}}{{\rm{m}}_{\rm{e}}}{{\rm{k}}^{\rm{2}}}}}{{{{\rm{h}}^{\rm{2}}}}}\)

Energy: The ability to perform work in physics. It can take many different forms, including potential, kinetic, thermal, electrical, chemical, nuclear, and others.

Considering the given information:

The ground state energy\({{\rm{E}}_{\rm{o}}}{\rm{ = }}\frac{{{\rm{2}}{{\rm{\pi }}^{\rm{2}}}{{\rm{q}}_{\rm{e}}}^{\rm{4}}{{\rm{m}}_{\rm{e}}}{{\rm{k}}^{\rm{2}}}}}{{{{\rm{h}}^{\rm{2}}}}}\).

Apply the formula:

The ground state energy is denoted as:

\({{\rm{E}}_{\rm{o}}}{\rm{ = }}\frac{{{\rm{2}}{{\rm{\pi }}^{\rm{2}}}{{\rm{q}}_{\rm{e}}}^{\rm{4}}{{\rm{m}}_{\rm{e}}}{{\rm{k}}^{\rm{2}}}}}{{{{\rm{h}}^{\rm{2}}}}}\)

Where\({{\rm{E}}_{\rm{o}}}{\rm{ = }}\)Ground state energy

h= Planck's Constant

\({{\rm{m}}_{\rm{e}}}{\rm{ = }}\)Mass of electron

k= Electrostatic Constant

\({{\rm{q}}_{\rm{e}}}{\rm{ = }}\)Charge of electron

We have

\(\begin{aligned}{}{\rm{h = 6}}{\rm{.626 \times 1}}{{\rm{0}}^{{\rm{ - 34}}}}\;{\rm{\;J}}{\rm{.s}}\\{{\rm{m}}_{\rm{e}}}{\rm{ = 9}}{\rm{.11 \times 1}}{{\rm{0}}^{{\rm{ - 31}}}}{\rm{\;kg}}\\{{\rm{q}}_{\rm{e}}}{\rm{ = 1}}{\rm{.60 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}{\rm{C}}\\{\rm{k = 9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{\;N}}{\rm{.}}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}\end{aligned}\)

As a result, the energy of Ground Stats is calculated as,

\(\begin{aligned}{}{E_o} &= \frac{{2{\pi ^2}{{\left( {1.60 \times {{10}^{ - 19}}{\rm{C}}} \right)}^4}\left( {9.11 \times {{10}^{ - 31}}\;{\rm{kg}}} \right){{\left( {9 \times {{10}^9}N.{m^2}/{{\rm{C}}^2}} \right)}^2}}}{{{{\left( {6.626 \times {{10}^{ - 34}}\;{\rm{J}}.{\rm{s}}} \right)}^2}}}\\{E_o} &= 2.172 \times {10^{ - 18}}\;{\rm{J}} \times \frac{{{\rm{leV}}}}{{1.60 \times {{10}^{ - 19}}\;{\rm{J}}}}\\{E_o} \approx 13.6{\rm{eV}}\end{aligned}\)

Hence, it is proved that the ground state energy \({{\rm{E}}_{\rm{o}}}\) is \({\rm{13}}{\rm{.6eV}}\).