Bohr radius for the hydrogen atom (n = 1) is approximately $0.\text{530}\stackrel{\text{°}}{\text{A}}$. The radius for the first

            excited state $(n=2)is\left(in\stackrel{°}{\text{A}}\right)$

$r{\text{\hspace{0.17em}α\hspace{0.17em}n}}^{\text{2}}/\text{\hspace{0.17em}Z\hspace{0.17em}}$

where, n = number of orbit Z = atomic number

$\because {\text{\hspace{0.17em}r}}_{\text{1}}{\text{\hspace{0.17em}α\hspace{0.17em}n}}_{\text{1}}^{\text{2}}$

${\text{r}}_{\text{2\hspace{0.17em}}}{\text{α\hspace{0.17em}n}}_{\text{2}}^{\text{2}}(\text{Z\hspace{0.17em}=\hspace{0.17em}1\hspace{0.17em}for\hspace{0.17em}H\hspace{0.17em}}-\text{\hspace{0.17em}atom})$

$\text{So,\hspace{0.17em}}\frac{{\text{r}}_{\text{1}}}{{\text{r}}_{\text{2}}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{n}}_{\text{1}}^{\text{2}}}{{\text{n}}_{\text{2}}^{\text{2}}}\Rightarrow \frac{\text{0.530}}{{\text{t}}_{\text{2}}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{1}}^{\text{2}}}{{\text{2}}^{\text{2}}}$

$\therefore {\text{\hspace{0.17em}t}}_{\text{2}}\text{\hspace{0.17em}=\hspace{0.17em}0.530\hspace{0.17em}×\hspace{0.17em}4\hspace{0.17em}=\hspace{0.17em}2.120\hspace{0.17em}}\stackrel{\text{°}}{\text{A}}$