A hydrogen like ion having wavelength difference between first Balmer and Lyman series equals to $593Å$, has $\mathrm{Z}$equal to :

first line's wave length in the Balmer series,

$\frac{1}{{\lambda }_{\mathrm{B}}}={\mathrm{Z}}^2{\mathrm{R}}_{\mathrm{H}}\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5}{36}{\mathrm{R}}_{\mathrm{H}}{\mathrm{Z}}^2$

Or ${\lambda }_{\mathrm{B}}=\frac{36}{5{\mathrm{R}}_{\mathrm{H}}{\mathrm{Z}}^2}$

The first line in the Lyman series' wavelength is,

$\begin{array}{r}\frac{1}{{\lambda }_{\mathrm{L}}}={\mathrm{Z}}^2{\mathrm{R}}_{\mathrm{B}}\left[\frac{1}{1^2}-\frac{1}{2^2}\right]\\ \text{ or }{\lambda }_{\mathrm{L}}=\frac{4}{3\times {\mathrm{R}}_{\mathrm{H}}{\mathrm{Z}}^2}\end{array}$

Difference ${\lambda }_{\mathrm{B}}-{\lambda }_{\mathrm{L}}=593\times {10}^{-7}=\frac{36}{5{\mathrm{R}}_{\mathrm{H}}{\mathrm{Z}}^2}-\frac{4}{3{\mathrm{R}}_{\mathrm{H}}{\mathrm{Z}}^2}$

$=\frac{1}{R_H{Z}^2}\left[\frac{36}{5}-\frac{4}{3}\right]$

$Z^2=\frac{88}{593\times {10}^{-7}\times 109678\times 15}=9.0$

$\text{ or }Z=3$

Hydrogen-like species is ${\mathrm{Li}}^{2+}$.