SOLUTION —
By one of the postulates of Bohr theory angular momentum $(m v r)=\frac{n h}{2 \pi}$
and we know that $\quad r=\frac{n^{2} h^{2}}{4 \pi^{2} m e^{2}}, \quad \therefore \quad m v \times \frac{n^{2} h^{2}}{4 \pi^{2} m e^{2}}=\frac{n h}{2 \pi}$
$\therefore \quad v=\frac{n h}{2 \pi} \times \frac{4 \pi^{2} m e^{2}}{m \times n^{2} h^{2}}=\frac{2 \pi e^{2}}{n / h}$
In gencral, $\quad v_{n}=\frac{2 \pi e^{2}}{h} \times \frac{1}{n} ; \quad$
For first orbit $\quad(n=1), \quad v=\frac{2 \pi e^{2}}{h}$
$\therefore \quad$ velocity $(v)=\frac{2 \times 3 \cdot 1416 \times\left(4 \cdot 80 \times 10^{-10}\right)^{2}}{6 \cdot 62 \times 10^{-27}}=2 \cdot 18 \times 10^{8}$ cm per second.
