Question

A hydrogen atom initially in the ground level absorbs a photon, which excites it to the $n = {4^{th}}$ level. Determine the wavelength and frequency of photons.

Answer 1

The energy required for an electron for the excitation of an electron from the ground state to higher energy levels will be equal to the energy difference between those levels. This energy difference will be useful in calculating the wavelength and frequency of a photon.
${E_n} = \dfrac{{{E_0}}}{{{n^2}}}$
${E_n}$ is energy of ${n^{th}}$ level
${E_0}$ is energy in zero energy level is $- 13.6eV$
n is the number of energy levels.

Complete answer:
Hydrogen is an element with atomic number $1$ and has only one electron in its shell. When light energy is given to an atom, it absorbs a photon and excites from ground state to higher energy levels. Here in the hydrogen atom, the ground state will be $n = 1$ , and the excited energy level is $n = 4$ .
${E_1} = \dfrac{{ - 13.6}}{{{1^2}}} = - 13.6eV$
The energy of fourth energy level will be
${E_4} = \dfrac{{ - 13.6}}{{{4^2}}} = \dfrac{{ - 13.6}}{{16}}eV$
Thus, energy of a photon will be $E = {E_1} - {E_4}$
But the energy will be equal to $E = \dfrac{{hc}}{\lambda }$
The frequency will be calculated from Planck’s constant $\left( {6.6 \times {{10}^{ - 34}}} \right)$ and velocity of light $\left( {3 \times {{10}^8}} \right)$
$\lambda = \dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{13.6 - \dfrac{{13.6}}{{16}}}} = 9.8 \times {10^{ - 8}}m = 98nm$
The wavelength of a photon is $98nm$
The frequency will be calculated as $\vartheta = \dfrac{c}{\lambda } = \dfrac{{3 \times {{10}^8}}}{{9.8 \times {{10}^{ - 8}}}} = 3.1 \times {10^{15}}Hz$
Thus, the frequency of a photon is $3.1 \times {10^{15}}Hz$ .

Note:
The wavelength can be expressed in different units like nanometres and meters. But, one nanometre will be equal to ${10^{ - 9}}$ metres. Frequency can be expressed in second inverse or hertz, and these both are equal. The velocity of light and Planck’s constant are constant values.