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Balmer Series. In 1885, when Johann Balmer observed a spectral series in the visible spectrum of hydrogen, he made the following observations: The longest wavelength is 656.3 nm. The second longest wavelength is 486.1 nm. And the third is 434.1 nm. Also, as the wavelength decreases the lines appear closer together and weak in intensity.
The following are the series in the hydrogen spectrum: – a) Lyman Series: – when the electron jumps from any higher stationary orbit to first stationary orbit, the spectral lines falls in the Lyman series. For Lyman series, n i = 1. Now putting n f = 1 in the relation given by Bohr, w = R h (1 1 2 − 1 n 2 i), n i = 2, 3, 4....
The wave number of the Lyman series is given by, ¯ ¯ ¯ v = R (1 − 1 n 2 2) (ii) Balmer series : When the electron jumps from any of the outer orbits to the second orbit, we get a spectral series called the Balmer series. All the lines of this series in hydrogen have their wavelength in the visible region. Here n 1 = 2, n 2 = 3, 4, 5...
Learn the concepts of Class 12 Physics Atoms with Videos and Stories. Emission lines in the spectrum of hydrogen. Discuss Absorption spectrum. Analyse spectrum from a hot gas. Discuss hydrogen spectral series - Balmer series, Give Lyman series, Paschen series, Brackett series, Pfund series; 7. Balmer formula in terms of frequency of light.
Calculate the shortest and longest wavelengths in hydrogen spectrum of Lyman series. Solution. Verified by Toppr. For Lyman series, n1 =1. For shortest wavelength in Lyman series (i.e., series limit), the energy difference in two states showing transition should be maximum, i.e., n2 =∞. So, 1 λ = RH[1 12 − 1 ∞2] =RH.
Lyman series of hydrogen atom lies in ultraviolet region, Balmer series lies in visible region while Pfund and Paschen series lie in infrared region.
Draw a neat and labelled energy level diagram and explain Balmer series and Brackett series of spectral lines for hydrogen atom. The work function for a metal surface is 2.2 eV. If light of wavelength 5000 o A is incident on the surface of the metal, find the threshold frequency and incident frequency.
Every series of hydrogen spectrum has an upper and lower limit in wavelength. The spectral series which has an upper limit of wavelength equal to 18752 ∘ A is
Lyman series of hydrogen atom lies in the ultraviolet region, Balmer series lies in visible region, Paschen series lies in near infrared region whereas Bracket, Pfund as well as Humphrey series lie in far infrared region of electromagnetic spectrum. Was this answer helpful?
The ratio of shortest wavelength of two spectral series of hydrogen spectrum is found to be about 1: 9. The possible spectral series is/are: The possible spectral series is/are: View Solution
