In hydrogen atom the kinetic energy of electron is 3.4 eV . The distance of that electron from the nucleus

The formula of energy is

${\mathrm{E}}_{\mathrm{n}}=\frac{-13.6\mathrm{eV}/\mathrm{atom}\times {\mathrm{Z}}^2}{{\mathrm{n}}^2}$

For hydrogen atom Z =1.

Kinetic energy of electron K.E = 3.4eV

Then total energy of electron ${\mathrm{E}}_{\mathrm{total}}=-\mathrm{K}.\mathrm{E}$

So, ${\mathrm{E}}_{\mathrm{total}}=-\left(3.4\mathrm{eV}\right)$

Then $-3.4\mathrm{eV}=\frac{-13.6\mathrm{eV}\times {\left(1\right)}^2}{{\mathrm{n}}^2}; \mathrm{Then}, \mathrm{n}=2$

If n = 2 , then the radius of orbit is

${\mathrm{r}}_{\mathrm{n}}=0.529\times {\mathrm{n}}^2{\mathrm{A}}^{°}$

${\mathrm{r}}_2=0.529\times 4{\mathrm{A}}^{°}$

$\therefore \mathrm{r} \mathrm{of} 2^{\mathrm{nd}} \mathrm{orbit}=2.116{\mathrm{A}}^{°}$