Question

The ground state energy of a hydrogen atom is -13.6 ev.
$(a)$ What is the potential energy of an electron in the 3rd excited state?
$(b)$ If the electron jumps to the ground state from the 3rd excited state, calculate the wavelength of the photon emitted.

Answer 1

Hint – In this question use the direct formula for potential energy of an electron in ${n^{th}}$ state that is $P.E = \dfrac{{2{Z^2}.{E_g}}}{{{n^2}}}$, where ${E_g}$ is the ground state energy. For the second part use the concept that Energy in photon = change in potential energy from ground state. Use the formula for energy that is $\dfrac{{hc}}{\lambda }$. This will help getting the answer.
Formula used - As we know now that the potential energy of an electron in ${n^{th}}$ state is given as $P.E = \dfrac{{2{Z^2}.{E_g}}}{{{n^2}}}$ Where Z is the atomic number of a given electron and ${E_g}$ is the ground state energy.

Complete step-by-step solution -
Now n is starting from 1, 2.....
Where 0 is for ground state, so for 3rd excited state n = 3.
And for hydrogen Z = 1 and ${E_g}$ = - 13.6 eV (given).
$\left( A \right)$
Therefore P.E in his 3rd excited state is
\[ \Rightarrow P.E = \dfrac{{2{{\left( 1 \right)}^2}.\left( { - 13.6} \right)}}{{{3^2}}} = - 3.022\] eV.
$\left( B \right)$ Wavelength of photon emitted
Therefore, Energy in photon = change in potential energy from ground state.
As we know energy in photon = $\dfrac{{hc}}{\lambda }$, where h = Planck's constant = $6.626 \times {10^{ - 34}}$, c = speed of light = $3 \times {10^8}$ m/s, and $\lambda $ = wavelength of the photon.
$ \Rightarrow \dfrac{{hc}}{\lambda } = - 2{E_g}{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
Where ${n_1} = $ ground state and ${n_2} = $ 3rd excited state.
Now substitute the values in above equation we have,
$ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}\left( {3 \times {{10}^8}} \right)}}{\lambda } = - 2\left( { - 13.6 \times 1.6 \times {{10}^{ - 19}}} \right){\left( 1 \right)^2}\left( {\dfrac{1}{1} - \dfrac{1}{9}} \right)$, $\left[ {\because 1ev = 1.6 \times {{10}^{ - 19}}} \right]$
Now simplify this we have,
$ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}\left( {3 \times {{10}^8}} \right)\left( 9 \right)}}{{27.2\left( 8 \right)\left( {1.6 \times {{10}^{ - 19}}} \right)}} = \lambda $
$ \Rightarrow \lambda = \dfrac{{178.902 \times {{10}^{ - 26}}}}{{348.16 \times {{10}^{ - 19}}}} = 0.513 \times {10^{ - 7}}$
$ \Rightarrow \lambda = 0.0517\mu m$, $\left[ {\because 1\mu m = 1 \times {{10}^{ - 6}}m} \right]$
So this is the required answer.

Note – The trick point here was that when an electron jumps from one state to another either energy is required to make this transaction or energy is released in this process. This energy is always equivalent to the change in potential energy from the ground state.