What is the ratio of the shortest wavelength of the balmer to the shoretst of the lyman series ?

Correct Answer - A
For a balmer series `(1)/(lambda_B)=R[(1)/(2^2)-(1)/(n^2)]`........(i)
where n=3,4,....
by putting `n=oo` in equation (i), we obtain the series limit of the balmer series. This is the shortest wavelength of the balmer series.
or `lambda_B=(4)/(R)` ........(i)
For a lyman series
`(1)/(lambda_L)=R[(1)/(1^2)-(1)/(n^2)]`........(iii)
where n=2,3,4,......
By putting `n=oo` in equation (iii), we obtain the series limit of the lyman series. This is the shortest wavelength of the lyman series.
or `lambda_L(1)/(R)`........(iv)
Dividing (ii) by (iv),we get
`(lambda_B)/(lambda_L)=(4)/(1)`