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Line Spectra Of Hydrogen Atom - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Line spectra of hydrogen atom is considered one the most difficult concept.

  • 49 Questions around this concept.

Solve by difficulty

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

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

 

Given the value of Rydberg constant is 107 m-1, the wave number of the last line of the Balmer series in hydrogen spectrum will be:

In the spectrum oh hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer Series is

Concepts Covered - 1

Line spectra of hydrogen atom

Line spectra of Hydrogen atom - 

Accordiag 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 Ei, the initial energy of the atom before such a transition, Ef is its final energy after the transition, then conservation of energy gives -

                                                                                    \mathrm{h} v-\frac{\mathrm{hc}}{\lambda}-E_{i}-E_{\mathrm{f}}

                                                                        This is the energy of the emitted electron.

Because of this photon, spectra of hydrogen atom will emits and which is studied by various scienticst.

One scientist named as Balmer, found a formula that gives the wavelienglts of these lines. This is now called the Balmer series. Balmer obsereved the spectra and found the formula for the visible range spectra which is obtained by the 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 light photon emitted during transition.

Since the 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 equals 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 wavelength of the different spectra like 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 is 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 { 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}

 

                                                         

                              

                                                                         This is for the hydrogen spectrum

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Line spectra of hydrogen atom

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