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

Explanation

`(1)/(lambda)=R[(1)/(n_(f)^(2))-(1)/(n_(i)^(2))]`
To calculate longest wavelength `n_(f)=2 and n_(i)=3`
`(1)/(lambda)=1.0977xx10^(7)[(1)/(2^(2))-(1)/((3)^(2))]`
`lambda=6.561xx10^(-7) m`
`(1)/(lambda)=R[(1)/(n_(f)^(2))-(1)/(n_(i)^(2))]`
To calculate shortest wavelength `n_(f)=2` and `n_(i)= oo`
`(1)/(lambda)=1.097xx10^(7)[(1)/(2^(2))-(1)/((3)^(2))]`
`lambda=3.647xx10^(-7)m`