Question

# The first emission line in the H-atom spectrum in the Balmer series will have a wave number of:A.$\dfrac{{5R}}{{36}}c{m^{ - 1}}$B.$\dfrac{{3R}}{4}c{m^{ - 1}}$C.$\dfrac{{7R}}{{144}}c{m^{ - 1}}$D.$\dfrac{{9R}}{{400}}c{m^{ - 1}}$

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

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:
$\dfrac{1}{\lambda } = R(\dfrac{5}{{36}})$
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.
The first emission line represents the transition from orbit number 3 to orbit number 2. Hence,
${n_1}$= 4, ${n_2}$=9
Substituting these values in Rydberg’s formula, we get
$\dfrac{1}{\lambda } = R(\dfrac{1}{{2_{}^2}} - \dfrac{1}{{3_{}^2}})$
$\dfrac{1}{\lambda } = R(\dfrac{1}{4} - \dfrac{1}{9})$
$\dfrac{1}{\lambda } = R(\dfrac{{9 - 4}}{{36}})$
$\dfrac{1}{\lambda } = R(\dfrac{5}{{36}})$
Wave number is basically the inverse of wavelength
Hence,
Wave number = $R(\dfrac{5}{{36}})$

Hence, Option A is the correct option.

Note:
Balmer series is specific to only one element, i.e. Hydrogen. The Balmer series provides us with the values of the electronic spectral energy of only Hydrogen.