What is Balmer Series ???

2 Answers
Oct 26, 2015

Answer:

The Balmer series or Balmer lines in atomic physics, is the designation of one of a set of six named series describing the spectral line emissions of the hydrogen atom

Explanation:

The Balmer series is the name given to 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#.

There are four transitions that are visible in the optical waveband that are empirically given by the Balmer formula.

The generalisation of this is the Rydberg formula, which also gives the other lines of hydrogen outside the optical region of the electromagnetic spectrum:

#1/(lamda_0) = R_H(1/n_1^2 - 1/n_2^2)#

Oct 26, 2015

The Balmer series is seen in atomic spectra, commonly discussed for the hydrogen atom.

You have the Lyman (#Ly-alpha#), Balmer (#Ba-alpha#), Paschen (#Pa-alpha#), Brackett (#Br-alpha#), etc. for electronic transitions downwards from #n = N# to #n = 1#, #n = 2#, #n = 3#, or #n = 4#, respectively, where #N# is some high energy level.

For the Balmer series, the transitions always land on #\mathbf(n = 2)# and may start at #\mathbf(n = 3, 4, 5, ... , N)#.

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Each energy level converges as #n# increases, and the wavelength of each transition can be calculated from the Rydberg formula:

#color(blue)(1/lambda = R_H (1/(n_1^2) - 1/(n_2^2)))#

where #lambda# is the wavelength of the transition, the Rydberg constant is #R_H = "10,973,731.6 m"^(-1)#, #n_1# is the lower energy level, and #n_2# is the higher energy level. These could differ by #1, 2, 3, ... , n#.

So if we wanted to know what transition corresponds to the wavelength of #"656 nm"# (red), we would use the equation like so:

#1/("656 nm")#

#= "10973731.6" cancel("m"^(-1)) xx cancel"1 m"/("10"^9 "nm") xx (1/(n_1^2) - 1/(n_2^2))#

#= "0.0109737316 nm"^(-1) xx (1/(2^2) - 1/(n_2^2))#

#0.1389 = 1/4 - 1/n_2^2#

#n_2^2 = 9.0019#

#color(blue)(n = 3.0003 ~~ 3)#

So the transition must be from #n = 3# to #n = 2#.