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, $E_n=-Z^2\times {\displaystyle \frac{13.61eV}{n^2}}$ 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:
$E_n=-{\displaystyle \frac{m{e}^4{Z}^2}{8{{\epsilon }_{\circ }}^2{n}^2{h}^2}}$
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,
${\epsilon }_{\circ }$ 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,
$E_n=-Z^2\times {\displaystyle \frac{13.61eV}{n^2}}$.
For any atom containing one electron, electronic configuration is given by $1s^1$ and is given by,
$E_1=-Z^2\times {\displaystyle \frac{13.61eV}{1^2}}$
So the first excited state energy level would correspond to $1s^{0}2p^1$ configuration.
Therefore the energy associated with it is given by,
$E_2=-Z^2\times {\displaystyle \frac{13.61eV}{2^2}}$.
For example if we consider $H{e}^{+}$ion, the atomic number $Z=2$.
Therefore, the energy associated with excited state is,
$E_2=-2^2\times {\displaystyle \frac{13.61eV}{2^2}}=-13.61eV$.
The energy associated with ground state of $H{e}^{+}$ion is,
$E_1=-2^2\times {\displaystyle \frac{13.61eV}{1^2}}=-54.44eV$.
The energy difference between the two energy levels $\mathrm{\Delta }E=E_2-E_1=\left\{-13.61-\left(-54.44\right)\right\}eV$
$=\left(-13.61+54.44\right)eV=40.83eV$.
Therefore the first excited state lies $40.83eV$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.