Question

What hydrogen-like ion has the wavelength difference between the first lines of Balmer and Lyman series equal to $59.3nm$ $(R_H=109678{cm}^{-1})$?

Answer 1

Wavelength of $1$st line in Balmer series,
$\frac{1}{{\lambda }_B}=Z^2{R}_H[\frac{1}{2^2}-\frac{1}{3^2}]=\frac{5}{36}{R}_H{Z}^2$
or ${\lambda }_B=\frac{36}{5R_H{Z}^2}$
Wavelength of $1$st line in Lyman series is,
$\frac{1}{{\lambda }_L}=Z^2{R}_B[\frac{1}{1^2}-\frac{1}{2^2}]$
or ${\lambda }_L=\frac{4}{3\times R_H{Z}^2}$
Difference ${\lambda }_B-{\lambda }_L=593\times {10}^{-7}=\frac{36}{5R_H{Z}^2}-\frac{4}{3R_H{Z}^2}$
$=\frac{1}{R_H{Z}^2}[\frac{36}{5}-\frac{4}{3}]$
$Z^2=\frac{88}{593\times {10}^{-7}\times 109678\times 15}=9.0$
or $Z=3$
Hydrogen-like species is ${Li}^{2+}$.