Question

What is the Balmer series?

Answer 1

In a hydrogen gas, the emission spectrum appears as distinct lines of different wavelengths when an electron returns to a lower energy level from a high energy level. Some of these lines are in the visible range of the electromagnetic spectrum and comprise a series that is known as the Balmer series.

Complete answer:
Niels Bohr, in 1913, discovered a model for the hydrogen and hydrogen like atoms that states that each orbit has discrete energy that is associated with the electrons present in the orbit. When the state of an electron is changed, that is when an electron moves from orbit to another, it either absorbs or releases energy in doing so.
The excitation of electrons from a lower state to a higher state requires the absorption of energy. And when the electron spontaneously returns to a lower energy state, it emits energy in the form of radiation of light. The wavelengths of emitted light ultimately depend on the energy difference between the two levels. These emitted lights are obtained as a line in the emission spectrum in which some lines are in the visible range of the electromagnetic spectrum, while some are in the ultraviolet or infrared range.
The excited state or the higher energy level of an electron is considered as the initial level $\left( {{\text{n}}_{\text{i}}} \right)$ while the lower level is considered as the final level $\left( {{\text{n}}_{\text{f}}} \right)$.
The series of visible lines in the hydrogen atom spectrum are named the Balmer series. This occurs when the electron transitions from any high-energy level $\left( \text{n}=3,4,5,6..... \right)$ to the lower energy level $\text{n}=2$.
The general formula for calculation of wave number for the hydrogen emission spectrum is given by Johannes Rydberg:
$\bar{\nu }=109677\left( \dfrac{1}{\text{n}_{\text{f}}^{2}}-\dfrac{1}{\text{n}_{\text{i}}^{2}} \right)$
Where $109677\text{ c}{{\text{m}}^{-1}}$ is Rydberg’s constant.
In the case of the Balmer series, $\left( {{\text{n}}_{\text{f}}} \right)$ is equal to 2. Thus, the equation becomes:
$\bar{\nu }=109677\left( \dfrac{1}{{{\text{2}}^{2}}}-\dfrac{1}{\text{n}_{\text{i}}^{2}} \right)$

Note:
The wave number is just an inverse of wavelength. Using this relation, and the relation between energy and wavelength, we can also calculate the energy required to move an electron from one energy level to another using the below equation:

& \bar{\nu }=\dfrac{1}{\text{ }\\!\\!\lambda\\!\\!\text{ }}=109677\left( \dfrac{1}{\text{n}_{\text{f}}^{2}}-\dfrac{1}{\text{n}_{\text{i}}^{2}} \right) \\\ & \because \Delta \text{E}=\dfrac{\text{hc}}{\text{ }\\!\\!\lambda\\!\\!\text{ }} \\\ & \therefore \Delta \text{E}=\text{hc}\left\\{ 109677\left( \dfrac{1}{\text{n}_{\text{f}}^{2}}-\dfrac{1}{\text{n}_{\text{i}}^{2}} \right) \right\\} \\\ & \text{where h}=6.626\times {{10}^{-34}}\text{ and c}=3\times {{10}^{8}}\text{ m }{{\text{s}}^{-1}} \\\ \end{aligned}$$