The Lyman series of the hydrogen spectrum can be represensted by the equation `v = 3.2881 xx 10^(15)s^(-1)[(1)/((1)^(2)) - (1)/((n)^(2))] [where n = 2,3,…….)`
Calculate the maximum and minimum wavelength of the lines in this series

Explanation

`bar v = (1)/(lambda) = (v)/(c ) = (3.2881 xx 10^(15))/(3 xx 10^(8)) m^(-1)[(1)/((1)^(2)) - (1)/(n^(2))]`
Wavelength is maximum `(bar v_(min))` when n is minimum so that `1//n^(2) `is maximum
: `bar v_(min) = (1)/(lambda_(max)) = (3.2881 xx 10^(15))/(3 xx 10^(8)) ((1)/(1)^(2) - (1)/((2)^(2)))`
`:. lambda_(max) = (3 xx 10^(8))/(3.2881 xx10^(15)) xx (4)/(3)`
`= 12165 xx 10^(-7) m = 121.67 nm `
Wavelength is minimum `(bar v_(max))` when n is `oo` i.e.series converges
`:. v_(max) = (1)/(lambda_(max)) = (3.2881 xx 10^(15))/(3 xx 10^(8))`
`:. lambda_(max) = 0.9124 xx 10^(-2) m = 91.24 nm`