The ratio of longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen is

Options

A. `4/3`
B. 25
C. `525/376`
D. `900/11`

Answer

D

Explanation

For lyman series , `n_i = oo , n_f = 2`
`1/lamda = R [1/1^1 = 1/n^2 ] = R [1/1^1 = 1/oo] =R`
`therefore lamda_("shortest") = 1/R`
For Pfund series , `n_f = 5 " and " n_i=6`
`therefore 1/lamda = R [ 1/5^2 = 1/6^2]`
`=R((36-25))/(900) = (11R)/(900)`
`therefore lamda_S = (900)/(11R)`
`therefore (lamda_L)/(lamda_S) = (900)/(11R) xx R/1 = 900/11`