```HYDROGEN ATOM
If an electric discharge is passed through hydrogen gas taken in a discharge tube under low pressure, and the emitted radiation is analyzed with the help of spectrograph, it is found to consist of a series of sharp lines in the UV, visible and IR regions. This series of lines is known as line or atomic spectrum of hydrogen. The lines in the visible region can be directly seen on the photographic film.
Each line of the spectrum corresponds to a light of definite wavelength. The entire spectrum consists of six series of lines, each series, known after their discoverer as the Balmer, Paschen, Lyman, Brackett, Pfund and Humphrey series. The wavelength of all these series can be expressed by a single formula.
1/ λ  v- = R  (1/ n12  - 1/n22 )
Where, v- = wave number
λ = wave length
R = Rydberg constant (109678 cm–1)
n1 and n2 have integral values as follows

Series

n1

n2

Main spectral lines

Lyman
Balmer
Paschen
Brackett
Pfund

1
2
3
4
5

2, 3, 4, etc
3, 4, 5 etc
4, 5, 6 etc
5, 6, 7 etc
6, 7, etc

Ultra – violet
Visible
Infra – red
Infra – red
Infra – red

Note:
All lines in the visible region are of Balmer series but reverse is not true. i.e., all Balmer lines will not fall in visible region.The pattern of lines in atomic spectrum is characteristic of hydrogen.
Illustration: A series of lines in the spectrum of atomic hydrogen lies at wavelengths 656.46, 482.7, 434.17, 410.29 nm. What is the wavelength of next line in this series?
Solution:
The given series of lines are in the visible region and thus appears to be Balmer series
Therefore n1 = 2 and n2=? For next line
If l = 410.29 ´10–7 cm and n1 = 2
n2 may be calculated for the last line
1/ λ  = R  [ 1/ n12  - 1/n22 ]
1 / 410.29 x 10-7 =  109673  [ 1/ 22 - 1/n22 ]
n2 = 6
Thus next line will be obtained during the jump of electron from 7th to 2nd shell i.e.
1 /  λ  = R  [ 1 / 22 - 1/72 ]  = 109673  [ 1/4 - 1/49 ]
λ = 397.2 x 10–7 cm = 397.2 nm```
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