Name: Bohr Model of the Hydrogen Atom
Uploaded: 2019-07-19
Duration: 4 min 50 s
Description: In this video, we will discuss the Bohr model of the hydrogen atom and electron transition in the hydrogen atom. Get in this to explore more.

Question

Calculate the wavelength of the first lines of Balmer, Lyman and Paschen series.

Accepted Answer

For the first line of the Balmer series $n_{21} = 2$ and $n_{22} = 3$

The wave number $ν$ of the associated wave will be:
$ν = R_{H} [\frac{1}{{n^{2}}_{21}} - \frac{1}{{n^{2}}_{22}}] {cm}^{- 1}$

Where, Rydberg Constant, $R_{H} = 109677 {cm}^{- 1}$

Substitute the values in the equation,
$= 109677 [\frac{1}{2^{2}} - \frac{1}{3^{2}}] {cm}^{- 1}$
$= 109677 \times \frac{5}{36} {cm}^{- 1}$
$= 15232.9 {cm}^{- 1}$
$∵ λ = \frac{1}{ν}$

$= 6.565 \times 10^{- 5} cm$

$= 6565 \overset{o}{A}$

For the first line of the Lyman series $n_{11} = 1$ and $n_{12} = 2$

The wave number $ν$ of the associated wave will be:
$ν = R_{H} [\frac{1}{{n^{2}}_{11}} - \frac{1}{{n^{2}}_{12}}] {cm}^{- 1}$

Substitute the values in the equation,
$= 109677 [\frac{1}{1^{2}} - \frac{1}{2^{2}}] {cm}^{- 1}$
$= 109677 \times \frac{3}{4} {cm}^{- 1}$
$= 82257.75 {cm}^{- 1}$
$∵ λ = \frac{1}{ν}$

$= 1.216 \times 10^{- 5} cm$

$= 1216 \overset{o}{A}$

For the first line of the Paschen series $n_{31} = 3$ and $n_{32} = 4$

The wave number $ν$ of the associated wave will be:
$ν = R_{H} [\frac{1}{{n^{2}}_{31}} - \frac{1}{{n^{2}}_{32}}] {cm}^{- 1}$

Substitute the values in the equation,
$= 109677 [\frac{1}{3^{2}} - \frac{1}{4^{2}}] {cm}^{- 1}$
$= 109677 \times \frac{7}{144} {cm}^{- 1}$
$= 5331.52 {cm}^{- 1}$
$∵ λ = \frac{1}{ν}$

$= 1.8756 \times 10^{- 5} cm$

$= 18756 \overset{o}{A}$

Calculate the wavelength of the first lines of Balmer, Lyman and Paschen series.

Important Questions on Atom, Origin of Spectra : Bohr's Theory of Hydrogen Atom

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