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Calculate the wave number for the lo...

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

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Radius of orbit like hydrogen species `=r-(n)`
`r_(n) =(n^(2)a_(0))/( Z) = n^(2) (52.9 pm)`
`r_(1) =1.3225 nm = 1325 pm=(52.9)/(Z) n_(1)^(2)`
`(n_(1)^(2))/(n_(2)^(2)) = (1322.5)/( 211.6) =6.25`
`(n_(1))/(n_(2)) = squr(6.25 ) =2.5`
If `n_(1) = 2 ` so `n_(2) = 2.5 (2) =5`
so transition occurs in `n_(2) = 5 ` to `n_(1) = 2` and Balmer series and spectrum line in visible region.
Calculation of wavelength in `n_(5) to n_(2)` transition
for Balmer series.
`vec( v) = (1)/(lambda) = 1.09677 xx 10^(7) ((1)/(n_(l )) - (1)/(n_(f)))`
`lambda= (100)/( 1.09677xx 10^(7) xx 21)`
`4.342 xx 10^(7) xx 10^(9)nm`
`=434.2 nm`
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