The velocity of an electron in a certain Bohr orbit of $$H$$ - atom bears the ratio 1 : 275 to the velocity of light. The quantum number $$(n)$$  of the orbit is

Solution :
$${\text{Velocity}}\,\,{\text{of electrons}}$$     $$ = \frac{1}{{275}} \times {\text{velocity}}\,\,{\text{of light}}$$
$$\eqalign{ & = \frac{1}{{275}} \times 3 \times {10^{10}} = 1.09 \times {10^8}\,cm\,{s^{ - 1}} \cr & {\text{Since}}\,\,{\nu _n} = \frac{{2\pi {e^2}}}{{nh}} \cr} $$
$$\therefore \,\,1.09 \times {10^8}$$    $$ = \frac{{2 \times 3.14 \times {{\left( {4.803 \times {{10}^{ - 10}}} \right)}^2}}}{{6.626 \times {{10}^{ - 27}} \times n}};$$
$$\therefore \,\,n = 2$$