What hydrogen like ion has wavelength difference between the first lines of Balmer and Lyman series equal to 59.3 nm?(`R_H=109678 cm^-1`)

What is the ratio of wavelength of ii line of balmer series and i line of Lyman series ?

Calculate the wave length of two lines H_alpha and H_beta of Balmcr series. R= 109670 cm^-1

The shortest lambda for the Lyman series of hydrogen atom is _______(given R_H=109678 cm^-1 )

Calculate the wavelength of the spectral line with n_(2)=3 in Lyman series of hydrogen atom.

Determine the wavelengths of H_(alpha) and H_(beta) lines in the Balmer series. [R=109677cm^(-1)]

What is the maximum wavelength of line of Balmer series of hydrogen spectrum (R=1.09xx10^(7)m^(-1)) -

The atomic spectrum of hydrogen contains a series of four lines havinng wavelengths 656.5, 486.3, 434.2 and 410.3 nm. Determine the wavelength of the next line in the same series [R_(H)=109678cm^(-1)] .

The Rydberg constant for hydrogen is 109737 cm^(-1) . Find out the lowest wavelength among the spectral lines of Balmer series.

For balmer series in the spectrum of atomic hydrogen the wave number of each line is given by barv=R_H(1/(n_1^2)-1/(n_2^2)) where R_H is a constant and n_1 and n_2 are integers. Which of the following statement is correct? As wavelength decreases the lines in the series converge. The integer n_1 is equal to 2 (iii) The ionisation energy of hydrogen can be calculated from the wave number of these lines. (iv) The line of longest wavelength corresponds to n_2=3

The wavelength of second Balmer line in Hydrogen spectrum is 600 nm. The wavelength for its third line in Lymann series is