Question

How do you calculate the frequency of the light emitted by a hydrogen atom during a transition of its electron from the $n = 6$ to the $n = 3$ principal energy level? Recall that for hydrogen${E_n} = - 2.18x{10^{ - 18}}J\left( {\dfrac{1}{{{n^2}}}} \right)$ ?

Answer 1

The electrons undergo transition from $n = 6$ to $n = 3$. The electron moves from a higher energy orbit to a lower energy orbit around the nucleus. The frequency of the emitted light is calculated from the wavelength or the wave number using the Rydberg-Balmer formula.

Complete step by step answer:
The Balmer formula for the transition of an electron from one energy level to another energy level is given by
$\overline \nu = {R_H}\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right]$
Where $\overline \nu$ is the wave number, ${R_H}$ is the Rydberg constant, ${n_1}$ is the initial energy level and ${n_2}$ is the final energy level.
The wave number is related to the wavelength of the emitted radiation. Thus the equation is
$\overline \nu = \dfrac{1}{\lambda }$
The wavelength is related to the frequency of the light emitted by the following equation:
$\nu = \dfrac{c}{\lambda }$ .
The energy of an electron is related to the frequency of the emitted light by the Planck relation
$E = h\nu$, where $h$ is the Planck’s constant.
The energy of the electron for transition of an electron in hydrogen atom is
${E_n} = - 2.18x{10^{ - 18}}J\left( {\dfrac{1}{{{n^2}}}} \right)$, where ${E_n}$, is the energy of the electron and $n$ is the principal quantum number.
So we need to determine at first the energy difference between the energy levels $n = 6$and $n = 3$. The energy thus obtained is used to calculate the frequency of the electromagnetic radiation which this corresponds to.
Thus the change in energy $\Delta E = - 2.18 \times {10^{ - 18}}\left[ {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{6^2}}}} \right]$
$\Delta E = 2.18 \times {10^{ - 18}}\left[ {\dfrac{1}{9} - \dfrac{1}{{36}}} \right]$
$\Delta E = 0.1816 \times {10^{ - 18}}J$
Thus the frequency of the emission light is
$E = h\nu$
$\nu = \dfrac{E}{h}$
$\nu = \dfrac{{0.1816 \times {{10}^{ - 18}}}}{{6.626 \times {{10}^{ - 34}}}} = 2.74 \times {10^{14}}{s^{ - 1}}$.

Additional Information: The corresponding wavelength is calculated as
$\lambda = \dfrac{c}{\nu }$
$\lambda = \dfrac{{3 \times {{10}^8}m{s^{ - 1}}}}{{2.74 \times {{10}^{14}}{s^{ - 1}}}}$
$\lambda = 1.095 \times {10^{ - 6}}m$
$\lambda = 1095nm$.
The obtained wavelength $1095nm$ indicates that the transition occurs in the infrared region of the electromagnetic spectrum. This transition is part of the Paschen Series.

Note:
The magnitude of the energy shows that the electron undergoes transition from higher to lower energy level. So the electrons attains more stability and the energy lost during the emission has negative value.