Calculate the shortest and longest wavelength of Balmer series of hydrogen atom. Given `R=1.097 xx 10^(7)m^(-1)`.

Explanation

Shortest wavelength(highest frequency) is for the last line in the series. For last line in the Balmer.s series,`n_(1)=2,n_(2)=00`
We have `(1)/(lambda)=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
`(1)/(lambda)=1.097xx10^(7)[(1)/(2)^(2)-(1)/(00)]`
`=1.097xx10^(7)((1)/(4))`
`lambda=(4)/(1.097xx10)^(7)=3.646xx10^(-7)`
Longest wavelength (smallest frequency) is for the first line in the series. For first line in the Balmer series, `n_(1) = 2, n_(2) =3`.
we have `(1)/(lambda)=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))=1.097xx10^(7)((1)/(2^(2))-(1)/(3^(2)))`
`=1.097xx10^(7)((1)/(4)-(1)/(9))`
`=1.097xx10^(7)((9-4)/(36))` `(1)/(lambda)=0.1523xx10^(7)`
`lambda=(1)/(0.1523xx10^(7))`
`lambda=6.566xx10^(-7)`
`lambda=6566A^(@)`