Question

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

Answer 1

Let,

λ_S = shortest wavelength

λ_L = longest wavelength

`1/lambda=R(1/p^2-1/n^2)`

Longest wavelength in Paschen series is obtained when p = 3, n = 4

For longest wavelength,

`therefore 1/lambda_l=R[1/3^2-1/4^2]`

`therefore 1/lambda_l=R[1/9-1/16]`

`therefore 1/lambda_l=R[(16-9)/144]`

`therefore 1/lambda_l=(7R)/144`

`therefore lambda_l=144/(7R)`

Shortest wavelength in Balmer series is obtained when p = 2, n =∞ For shortest  wavelength,

`therefore 1/lambda_s=R[1/2^2-1/oo^2]`

`therefore 1/lambda_5=4/R`

`therefore lambda_l/lambda_s=144/(7R)xxR/4`

`therefore lambda_l/lambda_s=5.131`

The ratio of longest wavelength in Paschen serires to shortest wavelength in Balmer series is 5.131.