Question

Calculate the shortest and longest wavelength of the Balmer series of the ${H_2}$ spectrum. Given $R = 1.097 \times {10^7}\,{m^{ - 1}}$ .

Answer 1

Electrons revolve around the nucleus with certain quantized energies. When there is a transition of an electron from one shell to another, there is either an absorption or a release of energy. This energy is in the form of a wave and can be found using the concept of Balmer series.

Formula Used: $\dfrac{1}{\lambda } = R(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}})$

Complete step by step answer:
When a transition of an electron takes place, a corresponding energy line can be measured using Balmer’s method. The collection of such lines for the various energy transformations is known as Balmer’s series. The wavelength of these energy transformations can be calculated by the following formula:
$\dfrac{1}{\lambda } = R(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}})$
Where $\lambda$ is the wavelength of the energy, R is the Rydberg constant, ${n_1}$ is the orbit number of the electron from where it is moving, ${n_2}$ is the orbit number of the electron to where it is moving. This formula is also known as Rydberg’s Formula.
Now the shortest wavelength is because of the electron transition from 2nd orbit to infinity. Therefore, the wavelength will be,

$$\dfrac{1}{\lambda } = R(\dfrac{1}{{{2^2}}} - \dfrac{1}{{{\infty ^2}}}) \\\ \dfrac{1}{\lambda } = 1.097 \times {10^7} \times (\dfrac{1}{4} - 0) \\\ \dfrac{1}{\lambda } = \dfrac{{1.097 \times {{10}^7}}}{4} \\\ \dfrac{1}{\lambda } = 0.274 \times {10^7} \\\ \lambda = 3.64 \times {10^{ - 7}}m \\\ \lambda = 364.6\,nm \\\$$

And the longest wavelength is because of the electron transition from 2nd orbit to 3rd orbit. Therefore , the wavelength will be,

$$\dfrac{1}{\lambda } = R(\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}) \\\ \dfrac{1}{\lambda } = 1.097 \times {10^7} \times (\dfrac{5}{{36}}) \\\ \dfrac{1}{\lambda } = \dfrac{{1.097 \times {{10}^7} \times 5}}{{36}} \\\ \dfrac{1}{\lambda } = 0.1523 \times {10^7} \\\ \lambda = 65.65 \times {10^{ - 7}}m \\\ \lambda = 656.5\,nm \\\$$

The shortest and longest wavelength of the Balmer series of ${H_2}$ spectrum are, 364.6nm and 656.6 nm respectively.

Note: The first emission line represents the transition from orbit number 3 to orbit number 2. Balmer series is specific to only one element, i.e. Hydrogen. Balmer series provides us with the values of the electronic spectral energy of only Hydrogen