Calculate the longest wavelengths belonging to Lyman and Balmer series. Which of these wavelengths will lie in the visible region ?

Explanation

From Rydberg.s formula , we have
`1/lambda_(if)=R[1/n_f^2-1/n_i^2]`
For longest wavelength in Lyman series,
`n_i`=2 and `n_f` =1
`therefore 1/lambda_21=R[1/1^2-1/2^2]=3/4R`
`rArr lambda_21=4/(3R)=4/(3xx1.1xx10^7)`
`=1.21xx10^7` m =121 nm
For longest wavelength in Balmer series, `n_i`=3 and `n_f`=2
`therefore 1/lambda_32=R[1/2^2-1/3^2]=(5R)/36`
`rArr lambda_32=36/(5R)=36/(5xx1.1xx10^7)=6545xx10^(-10)` m
=654.5 nm
Between the two wavelengths , only 655 nm lies in the visible region.