Hydrogen-like ion has the wavelength difference between the first lines of Balmer and lyman series equal to 33.4nm. The atomic number of that ion equal to $\left[{\mathrm{R}}_{\mathrm{H}}=1.1\times {10}^7{\mathrm{m}}^{-1}\right]$

Given,

${\lambda }_{\text{B}}-{\lambda }_{\text{L}}=334\text{\hspace{0.17em}nm}$

Atomic number, Z = ?

Use the formula,$\frac{1}{\lambda }=R_{\text{H}}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\times Z^2$

Where,  

l wave length of radiation

R_H = Rydberg's constant

n₂ = Higher energy level = 3

n₁ = lower energy level = 2

$\Rightarrow \frac{1}{{\lambda }_1}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)Z^2$

$\Rightarrow {\lambda }_1=\frac{36}{5R{Z}^2}$

$\frac{1}{{\lambda }_2}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$

$\therefore {\lambda }_2=\frac{4}{3R{Z}^2}$

$\therefore \Delta \lambda =\left({\lambda }_1-{\lambda }_2\right)=\frac{36}{5R_H{Z}^2}-\frac{4}{3R_H{Z}^2}$

$\Rightarrow 334=\left(\frac{36}{5}-\frac{4}{3}\right)\times \frac{1}{R_H}\times \frac{1}{Z^2}$

$\Rightarrow Z^2=\frac{5.86\times 912}{334}$

=16

$\therefore Z=4$