What would be the wavelength and name of series respectively for the emission transition for $$H$$ - atom if it starts from the orbit having radius 1.3225$$\,nm$$  and ends at 211.6$$\,pm?$$

Solution :
$${\text{Radius of}}\,\,{n^{th}}\,\,{\text{orbit}}$$     $$ = \frac{{0.529 \times {n^2}}}{Z}\mathop {\text{A}}\limits^{\text{o}} = \frac{{52.9 \times {n^2}}}{Z}pm$$
$${\text{Radius}}\left( {{r_2}} \right) = 1.3225\,nm$$     $$ = 1322.5\,pm = \frac{{52.9\,n_2^2}}{Z}$$
$${\text{Radius}}\left( {{r_1}} \right) = 211.6\,pm = \frac{{52.9\,n_1^2}}{Z}$$
$$\eqalign{ & \therefore \,\,\,\frac{{{r_2}}}{{{r_1}}} = \frac{{1322.5}}{{211.6}} = \frac{{n_2^2}}{{n_1^2}} \cr & \Rightarrow 6.25 = \frac{{n_2^2}}{{n_1^2}} = {\left( {\frac{{{n_2}}}{{{n_1}}}} \right)^2} \cr & \Rightarrow \frac{{{n_2}}}{{{n_1}}} = \sqrt {6.25} = 2.5 \cr} $$
$$\therefore \,\,\,{n_1} = 2,{n_2} = 5$$     thus, the transition is from 5^th orbit to 2^nd orbit. It belongs to Balmer series.
$$\eqalign{ & \bar \upsilon = R\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right) \cr & \Rightarrow \bar \upsilon = 1.097 \times {10^7}\left( {\frac{1}{{{2^2}}} - \frac{1}{{{5^2}}}} \right) \cr & = 1.097 \times {10^7}\left( {\frac{1}{4} - \frac{1}{{25}}} \right) \cr & = 1.097 \times {10^7} \times \frac{{21}}{{100}} \cr} $$
$$\lambda = \frac{1}{{\bar \upsilon }} = \frac{{100}}{{1.097 \times 21 \times {{10}^7}}}m$$       $$ = 4.34 \times {10^{ - 7}}\,m = 434 \times {10^{ - 9}}\,m$$
$$\eqalign{ & \lambda = 434\,nm \cr & {\text{Thus, it lies in the visible region}}{\text{.}} \cr} $$