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The wavelength of the first spectral lin...

The wavelength of the first spectral line in the Balmer series of hydrogen atom is `6561 A^(@)` . The wavelength of the second spectral line in the Balmer series of singly - ionized helium atom is

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According to Rydberg equation,
`1/(lambda) = R (1/(n_1^2) - 1/(n_2^2))`
For first line in Balmer series , `n_1 = 2, n_2 = 3`
`:. 1/(6561) = R (1/(2^2) - 1/(3^2)) = R (5/36) …..(i)`
For second line Balmer series `n_1 = 2, n_2 = 4`
`:. 1/lambda = R (1/(2^2) - 1/(4^2)) = R (3/16)...(ii)`
Dividing eq. (i) by (ii)
`lambda/(6561) = (5)/36 x 16/3`
`:. = (6561 xx 5 xx 16)/(36 xx 3) = 4860 Å`
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