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Q.
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
Structure of Atom
Solution:
Velocity of electrons $=\frac{1}{275}\times velocity\, of \,light$
$=\frac{1}{275}\times3\times10^{10}=1.09\times10^{8}\,cm\,s^{-1}$
Since $v_{n}=\frac{2\pi e^{2}}{nh}$
$\therefore 1.09\times10^{8}=\frac{2 \times 3.14 \times \left(4.803 \times 10^{-10}\right)^{2}}{6.626 \times 10^{-27} \times n}$
$\therefore n=2$