Bohr radius for the hydrogen atom $$\left( {n = 1} \right)$$  is approximately $$0.530\mathop {\text{A}}\limits^{\text{o}} .$$   The radius for the first excited state $$\left( {n = 2} \right)$$  is $$\left( {{\text{in}}\,\mathop {\text{A}}\limits^{\text{o}} } \right)$$

Solution :
$$\eqalign{ & r \propto {n^2}/Z \cr & {\text{where, }}n = {\text{number of orbit}} \cr & Z = {\text{atomic number}} \cr & \because \,\,{{\text{r}}_1} \propto n_1^2 \cr & {r_2} \propto n_2^2\,\,\left( {Z = 1{\text{ for }}H{\text{ - }}atom} \right) \cr & {\text{So,}}\,\,\,\frac{{{r_1}}}{{{r_2}}} = \frac{{n_1^2}}{{n_2^2}}\,\, \cr & \frac{{0.530}}{{{r_2}}} = \frac{{{1^2}}}{{{2^2}}} \cr & \therefore \,\,\,\,{r_2} = 0.530 \times 4 = 2.120\,\mathop {\text{A}}\limits^{\text{o}} \cr} $$