Question

How can you calculate the excited state energy level$?$

Answer 1

Hint: Excited state energy level is basically known for Hydrogen like atoms $($i.e. having only one electron$)$like $H{e^ + },L{i^{2 + }}$. For these type of atoms the energy of the ${n^{th}}$ level can be given by the expression, ${E_n}\,\, = \,\, - {Z^2} \times \dfrac{{13.61\,eV}}{{{n^2}}}$ where $Z$is the atomic number of the atom. Using this, calculate the excited state energy level.

Complete step-by-step answer:Bohr found out that the energy associated with the ${n^{th}}$ energy level of Hydrogen like atoms $($i.e. having only one electron$)$like $H{e^ + },L{i^{2 + }}$ can be given by:
${E_n}\, = \, - \dfrac{{m{e^4}{Z^2}}}{{8{\varepsilon _ \circ }^2{n^2}{h^2}}}$
where, $m$is the mass of an electron and is equal to $9.1 \times {10^{ - 31}}\,kg$,
$e$is the charge of an electron and is equal to $1.602 \times {10^{ - 19}}\,C$,
$Z$ is the atomic number of the atom,
${\varepsilon _ \circ }$ is the permittivity of free space and is equal to $8.85 \times {10^{ - 12}}\,{m^{ - 3}}k{g^{ - 1}}{s^4}{A^2}$,
$n$is the quantum level , and
$h$is the Planck’s constant and is equal to $6.626 \times {10^{ - 34}}\,{m^2}\,kg\,{s^{ - 1}}$.
Putting all the values we get,
${E_n}\,\, = \,\, - {Z^2} \times \dfrac{{13.61\,eV}}{{{n^2}}}$.
For any atom containing one electron, electronic configuration is given by $1{s^1}$ and is given by,
${E_1}\,\, = \,\, - {Z^2} \times \dfrac{{13.61\,eV}}{{{1^2}}}$
So the first excited state energy level would correspond to $1{s^0}2{p^1}$ configuration.
Therefore the energy associated with it is given by,
${E_2}\,\, = \,\, - {Z^2} \times \dfrac{{13.61\,eV}}{{{2^2}}}$.
For example if we consider $H{e^ + }$ion, the atomic number $Z\, = \,2$.
Therefore, the energy associated with excited state is,
${E_2}\,\, = \,\, - {2^2} \times \dfrac{{13.61\,eV}}{{{2^2}}}\, = \, - 13.61\,eV$.
The energy associated with ground state of $H{e^ + }$ion is,
${E_1}\,\, = \,\, - {2^2} \times \dfrac{{13.61\,eV}}{{{1^2}}}\, = \, - 54.44\,eV$.
The energy difference between the two energy levels $\Delta E\, = \,{E_2} - {E_1}\,\, = \,\,\left\{ { - 13.61 - \left( { - 54.44} \right)} \right\}eV$
$ = \,\left( { - 13.61 + 54.44} \right)eV\,\, = \,\,40.83\,eV$.
Therefore the first excited state lies $40.83\,eV$above its ground state.

Note: You should always remember this expression of the energy associated with the ${n^{th}}$ energy level is only true for hydrogen-like atoms and not for any other type of atom. Also you must take proper care of the units, do not mix up between the SI and CGS units.