The wavelength of the first line of Lyman series for $\mathrm{H}$ - atom is equal to that of the second line of Balmer series for a $\mathrm{H}$-like ion. The atomic number $\mathrm{Z}$ of $\mathrm{H}$-like ion is

The wavelength $\lambda$ for a transition in a hydrogen-like ion is given by the Rydberg formula: $\dfrac{1}{\lambda} = R Z^2 \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$.

For the first line of the Lyman series in H-atom ($Z=1$): $n_1 = 1$, $n_2 = 2$.
$\dfrac{1}{\lambda_1} = R(1)^2 \left( \dfrac{1}{1^2} - \dfrac{1}{2^2} \right) = R \left( 1 - \dfrac{1}{4} \right) = \dfrac{3R}{4}$.

For the second line of the Balmer series in a H-like ion with atomic number $Z$: $n_1 = 2$, $n_2 = 4$.
$\dfrac{1}{\lambda_2} = R Z^2 \left( \dfrac{1}{2^2} - \dfrac{1}{4^2} \right) = R Z^2 \left( \dfrac{1}{4} - \dfrac{1}{16} \right) = R Z^2 \left( \dfrac{4-1}{16} \right) = \dfrac{3R Z^2}{16}$.

Equating the wavelengths $\lambda_1 = \lambda_2$, we have $\dfrac{1}{\lambda_1} = \dfrac{1}{\lambda_2}$.
$\dfrac{3R}{4} = \dfrac{3R Z^2}{16}$.

Simplifying the equation:
$\dfrac{1}{4} = \dfrac{Z^2}{16} \Rightarrow Z^2 = 4 \Rightarrow Z = 2$.

Answer: 2