Obtain the ratio of the shortest wavelength of spectral line in the Lyman series to the longest wavelength of spectral line in the Balmer series.

Explanation

`barv=(1)/(lamda)=R((1)/(n_(f)^(2)-n_(1)^(2)))`
The Lyman series of spectral lines arises due to the transitions to `n_(f)=1`. The shortest wavelength `(lamda_(L oo))` line in this series arises due to the transition from `n_(1)=oo`. The Balmer series of spectral lines arises due to the transitions to `n_(f)=2.` The longest wavelength `(lamda_(Hz))` line in this series arises due to the transition from `n_(1)=3.`
`therefore(1)/(lamdaLoo)=R((1)/(1)-0)=R`
and `(1)/(lamda_(Hz))=R((1)/(4)-(1)/(9))=(5R)/(36)`
`therefore` The required ratio is
`(lamda_(L oo))/(lamda_(Hz))=(5R//36)/(R)=(5)/(36)`