To what series does the speciral lines of atomic hydrogen belong if its wavelength is equal to the difference between the wavenumber of the folowing two lines of the Balmer series `486.1` and `419.2 nm`? What is the waveeath of thqat line ?

Explanation

Given
`lambda_(1) = 486.1 xx 10^(-9)m = 486.1 xx 10^(-7) cm`
`lambda_(2) = 410.2 xx 10^(-9)m = 410.2 xx 10^(-7) cm`
` bar v = bar v_(2) - barv _(1)`
`= (1)/(lambda_(2)) - (1)/(lambda_(1))`
`= R_(H) [(1)/(2^(2)) - (1)/(n_(2)^(2))] - R_(H) [(1)/(2^(2)) - (1)/(n_(1)^(2))]`
`bar v = R_(H)[1/n_(1)^(2)-1/n_(1)^(2)]`........(i)
For 1 case of balance series
`(1)/(lambda_(1)) = R_(H) [(1)/(2^(2)) - (1)/(n_(1)^(2))] = 109678 [(1)/(2^(2)) - (1)/(n_(1)^(2))]`
or `(1)/(4186.1 xx 10^(-5)) = 109678[(1)/(2^(2)) - (1)/(n_(1)^(2))]`
`:. n_(1) = 4`
For II case of Balmer series
`(1)/(lambda) = (1)/(410.2 xx 10^(-7)) = 10978 [(1)/(2^(2)) - (1)/(n_(2)^(2))]`
`:. n_(2) = 6`
THus giuven transition occurs from sixth level to four level .Also equation (i) `bar v = (1)/(lambda) = 109678 [(1)/(4^(2)) - (1)/(6^(2))]`
`:. lambda = 2.63 xx 10^(-4) cm `