Find the ratio of longest wavelangth in Paschen series to shortest wavelength in Balmer series.

Explanation

`1/lambda = R(1/n_(f)^(2) - 1/n_(1)^(2))`
For the Paschen series, `n_(f)=3`. For the longest wavelength `(lambda_(palpha)), n_(i) =4`.
`therefore 1/lambda_(px) = R(1/9-1/16) = (7R)/144`
`therefore lambda_(p alpha) = 144/(7R)`
For the Balmer series, `n_(f)=2`. For the short wavelength limit `(lambda_(Binfty)), n_(i) = infty`.
`therefore 1/(lambdaBinfty)= R(1/4-1/infty)=R/4`
`therefore lambda_(Binfty) = 4/R`
`therefore lambda_(P alpha)/(lambda_(B infty)) = 144/(7R) xx R/4 = 36/7 = 5.143`