Calculate the wave number for the longest wave length transition in the. Balmer series of atomic hydrogen.

Calculate the wave rumber of yellow light of wavelength 585 nm.

Calculate the wavelength of the first and the last line in the Balmer series of the hydrogen spectrum.

For one-electron species, the wave number of radiation emitted during the transition of electron from a higher energy state (n_(2)) to a lower energy state (n_(1)) is given by: bar v =(1)/(lamda)=R_(H) xx Z^() ((1)/(n_(1)^(2)) (1)/(n_(2)^(2))) where R_(H)=(2 pi m_(s) k^(2) c^(4))/(h^(3) c) is Rydberg constant for hydrogen atom. Now, considering nuclear motion, the accurate measurement would be obtained by replacing mass of electron (m_(e)) by the reduced mass (mu) in the above expression, defined as mu =(m_(n)xx m_(e))/(m_(n) +m_(e)) where m_(n) = mass of nucleus. For Lyman series, n_(t) =1 (fixed for all the lines) while n_(2) = 2, 3, 4 .... For Balmer series: n_(1) = 2 (fixed for all the lines) while n_(2) = 3,4,5 .... The ratio of the wave numbers for the highest energy transition of electron in Lyman and Balmer series of hydrogen atom is

The Balmer series of lines in the hydrogen spectrum appear in the visible region of the electron magnetic spectrum . Calculate the wave number of the second line in the Balmer series. ( Rydberg constant for hydrogen is 109677 cm^(-1)

Longest wavelengths in Balmer series of He^(+1) is

Unit of wave number is

Calculate the wavelength of the limiting line in Balmer series.

Calculate the wavelength of the second line in Balmer series.