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Line Spectra Of Hydrogen Atom - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Line spectra of hydrogen atom is considered one the most difficult concept.
63 Questions around this concept.

Solve by difficulty

EASY

Arrange the following electromagnetic radiations per quantum in the order of increasing energy :

A : Blue light B : Yellow light
C : X-ray D : Radiowave.

D, B, A, C

A, B, D, C

C, A, B, D

B, A, D, C

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The hydrogen atom is excited from the ground state to another state with a principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be :

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Concepts Covered - 1

Line spectra of hydrogen atom

Line spectra of a Hydrogen atom -

According to Bohr, when an atom makes a transition from a higher energy level to a lower energy level, it emits a photon with energy equal to the energy difference between the initial and final levels. If E_i, the initial energy of the atom before such a transition, E_f is its final energy after the transition, then conservation of energy gives the energy of emitted photon-

$\\ \mathrm{h} v=\frac{\mathrm{hc}}{\lambda}=E_{i}-E_{\mathrm{f}} \\ \frac{hc}{\lambda }=\frac{-13.6}{n^{2}_{i}}eV-\frac{-13.6}{n^{2}_{f}}eV=13.6eV\left ( \frac{1}{n^{2}_{f}} -\frac{1}{n^{2}_{i}}\right ) \\ Rch=13.6 eV= 1 \text{Rydberg energy} \\ \\ \Rightarrow \frac{1}{\lambda }=R\left ( \frac{1}{n^{2}_{f}} -\frac{1}{n^{2}_{i}}\right ) \\ \text{where} \ R=Rydberg's \ constant= 1.097\times 10^{7} m^{-1}$

For Hydrogen-like atoms, the wavelength of an emitted photon during transition from n_f orbit to n_i orbit is $\\ \frac{1}{\lambda}=R z^{2}\left (\frac{1}{n^{2}_{f}} -\frac{1}{n^{2}_{i}} \right )$

Because of this photon, spectra of hydrogen atom will emit and which is studied by various scientists. One such scientist named Balmer found a formula that gives the wavelengths of these lines for all the transitions taking place to the 2^nd orbit.

The Balmer series is a series of spectral emission lines of the hydrogen atom that result from electron transitions from higher levels down to the energy level with principal quantum number 2

Balmer observed the spectra and found the formula for the visible range spectra which is obtained by Balmer's formula is-

$\frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right) \dots (1)$

Here, n=3,4,5,. . . . etc. ,

$R= Rydberg \ constant =1.097 \times 10^{7} \mathrm{m}^{-1}$
and $\lambda$ is the wavelength of the light photon emitted during the transition.

Since Balmer had found the formula for n = 2, but we can obtain different spectra for different values of n. For n = $\infty$ , we get the smallest wavelength of this series, which is equal to = 3646 $\dot{A}$ . We can also obtain the value of wavelength for Balmer's series by putting different values of 'n' in the equation (1). Similarly, we can obtain the wavelength of the different spectra like the Lyman, Paschen series. Since the Balmer series is in the visible range but the Lyman series is in the Ultraviolet range and the Paschen, Brackett, and Pfund are in the Infrared range.

$\begin{array}{l}{\text { Lyman series: } \frac{1}{\lambda}=R\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right), n=2,3,4, \ldots} \\ \\ {\text { Balmer series: } \frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right), n=3,4,5, \ldots}\\ \\ {\text { Paschen series: } \frac{1}{\lambda}=R\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right), n=4,5,6, \ldots}\\ \\ {\text { Brackett series: } \frac{1}{\lambda}=R\left(\frac{1}{4^{2}}-\frac{1}{n^{2}}\right), n=5,6,7, \ldots}\\ \\ {\text { Pfund series: } \frac{1}{\lambda}=R\left(\frac{1}{5^{2}}-\frac{1}{n^{2}}\right), n=6,7,8}\end{array}$