The third line of the Balmer series spectrum of a hydrogen like ion of atomic number $$Z$$ equals to $$108.5\,nm.$$  Then $$Z$$ is

Solution :
For the third line of Balmer series,
$$\eqalign{ & {n_1} = 2,{n_2} = 5 \cr & \therefore \frac{1}{\lambda } = R{Z^2}\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right) \cr & = R{Z^2}\left( {\frac{1}{{{2^2}}} - \frac{1}{{{5^2}}}} \right) = \frac{{21R{Z^2}}}{{100}} \cr & E = - 13.6\,eV \cr & {Z^2} \times \frac{{21}}{{100}} = \frac{{hc}}{\lambda } = \frac{{1242\,eVnm}}{{108.5\,nm}} \cr & {Z^2} = \frac{{1242 \times 100}}{{108.5 \times 21 \times 13.6}} = 4 \cr & \Rightarrow Z = 2 \cr} $$