The ratio of wavelengths of the last line of the Balmer series and the last line of the Lyman series is

Taking Rydberg's constant $R_H = 1.097 \times 10^7 \ m^{-1}$,the first and second wavelengths of the Balmer series in the hydrogen spectrum are:

Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is :-

In the hydrogen spectrum,the ratio of the wavelengths of the last line of the Lyman series and that of the last line of the Balmer series is:

In the hydrogen emission spectrum,for any series,the principal quantum number of the higher energy level is $n+1$ and the lower energy level is $n$. The corresponding maximum wavelength $\lambda$ is ($R=$ Rydberg's constant).

$A$ doubly ionised $Li$ atom is excited from its ground state $(n = 1)$ to $n = 3$ state. The wavelengths of the spectral lines are given by $\lambda_{32}, \lambda_{31}$ and $\lambda_{21}$. The ratios $\lambda_{32}/\lambda_{31}$ and $\lambda_{21}/\lambda_{31}$ are,respectively: