Question

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

• A. $3$
• B. $2$
• C. $1$
• D. $4$

Answer 1

Answer: (B)

Velocity of electrons $=\frac{1}{275}\times velocityoflight$
$=\frac{1}{275}\times 3\times 10^{10}=1.09\times 10^{8}cm{s}^{-1}$
Since $v_n=\frac{2\pi e^2}{nh}$
$\therefore 1.09\times 10^8=\frac{2\times 3.14\times {\left(4.803\times 10^{-10}\right)}^2}{6.626\times 10^{-27}\times n}$
$\therefore n=2$