Question

In Balmer series of hydrogen spectrum, which electronic transition causes the third line
A.Fifth Bohr orbit to second one
B.Fifth Bohr orbit to first one
C.Fourth Bohr orbit to second one
D.Fourth Bohr orbit to first one

Answer 1

The proof of atoms electronic structure is the hydrogen spectrum. On passing electric discharge through gaseous hydrogen molecules results in dissociation of the molecule which results in emission of electromagnetic radiation by excited hydrogen atoms.

Complete step by step answer:
Let's discuss the hydrogen spectrum in detail. On passing electric discharge through hydrogen gas which is enclosed in a discharge tube at low pressure results in emission of light and a spectroscope does the analysis of emitted light. The obtained spectrum is composed of a large number of lines and these are grouped in different series. The series appear in different regions of light named after their discoveries. The complete spectrum is termed as hydrogen spectrum.

Rydberg’s formula is used to calculate the wavelengths of the spectral lines in different series of hydrogen spectrum.
$\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$

Where, $\lambda$ is wavelength, ${R_H}$ is Rydberg constant, ${n_1}$ and ${n_2}$ are quantum numbers of initial state and final state respectively.

Now, come to the question. We have to identify the third line of the Balmer series. The Balmer series is the only series of lines which lies in the visible region of the electromagnetic spectrum. This series is the part of the hydrogen spectrum responsible for excitation of electrons from the 2nd shell to any other shell.

For Balmer series, ${n_1}$=2 , ${n_2}$=3,4,5,6…… in Rydberg’s formula.

So,
First line of Balmer series is 3rd orbit to 2nd orbit
Second line of Balmer series is 4th orbit to second orbit
Third line of Balmer series is 5th orbit to second orbit

So, the correct answer is Option A.

Note: In Rydberg formula, the value of ${n_1}$ is constant while that of ${n_2}$ changes. For example,
For the Lyman series, ${n_1}$=1 , ${n_2}$=2,3,4,5……
For the Balmer series, ${n_1}$=2 , ${n_2}$=3,4,5,6……
For the Paschen series,${n_1}$=3 ,${n_2}$=4,5, 6,7……
For the Brackett series, ${n_1}$=4 , ${n_2}$=5,6,7,8…..
For the Lyman series, ${n_1}$=5 , ${n_2}$=6,7,8,9……

Thus, by substituting the values of ${n_1}$ and ${n_2}$ in the Rydberg formula, wavelengths of different spectral lines can be calculated. When ${n_1}$=2, the Rydberg expression is termed as Balmer’s formula.