The radius of electron orbit and the speed of electron in the ground state of hydrogen atom is $5.30 \times {10}^{-11} m$and$2.2 \times {10}^6 m{s}^{-1}$ respectively, then the orbital period of this electron in first excited state will be:

Here, $r_1=5.30\times {10}^{-11} m; {\nu }_1= 2.2\times {10}^6 m{s}^{-1}$

In the first excited state, $r_n=n^2{r}_1, {\nu }_n=\frac{v_1}{n}$

$\therefore r_2=4r_1=4\times 5.30\times {10}^{-11} \mathrm{m} =2.12\times {10}^{-10} \mathrm{m}$

and $v_2=\frac{v_1}{2}=\frac{2.2\times {10}^6}{2} m s^{-1}=1.1\times {10}^6 m s^{-1}$

Since, orbit period 

$$\begin{gathered} T=\frac{2{\mathrm{\pi r}}_2}{v_2}=\frac{2\times 3.14\times 2.12\times {10}^{-10}}{1.1\times {10}^6}=\frac{13.31\times {10}^{-16}}{1.1} \\ =1.21\times {10}^{-15} \mathrm{s}. \end{gathered}$$