The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number $$Z$$ of hydrogen like ion is

Solution :
For Lyman series for $$H$$-atom
$$\frac{{hc}}{\lambda } = Rhc\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right)$$
and for $$H$$-like ion,
$$\eqalign{ & \frac{{hc}}{\lambda } = {Z^2}Rhc\left( {\frac{1}{{{2^2}}} - \frac{1}{{{4^2}}}} \right) \cr & \therefore \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) = {Z^2}\left( {\frac{1}{4} - \frac{1}{{16}}} \right) \cr & \left( {1 - \frac{1}{4}} \right) = {Z^2}\left( {\frac{1}{4} - \frac{1}{{16}}} \right) \cr & Z = 2 \cr} $$