The velocity of an electron in excited state of H-atom is $1.093\times {10}^6\mathrm{m}/\mathrm{s}$. What is the circumference of this orbit?

$$\begin{gathered} V_n=2.186\times {10}^6\times \frac{z}{n} (z=1) \\ n=\frac{2.186\times {10}^6}{1.093\times {10}^6}=2 \end{gathered}$$

$$\begin{gathered} r=0.529\times \frac{n^2}{z}\stackrel{\circ }{A} \\ r=0.529\times \frac{2^2}{1}\stackrel{\circ }{A}=2.1 \stackrel{\circ }{A} \end{gathered}$$

$$\begin{gathered} \mathrm{circumference} = 2\mathrm{\pi r} \\ \mathrm{circumference} = 2\times 3.14\times 2.1 \stackrel{\circ }{\mathrm{A}} \\ \mathrm{circumference} = 13.30 \stackrel{\circ }{\mathrm{A}} \\ \mathrm{circumference} = 13.30\times {10}^{-10} \mathrm{m} \end{gathered}$$