Q. The number of revolutions per second made by an electron in the first Bohr orbit of hydrogen atom is of the order of
A) $10^{20}$
B) $10^{19}$
C) $10^{17}$
D) $10^{15}$
Explanation
(d) $mvr = \frac{h}{2\pi}$ (for first orbit)
$\Rightarrow m\omega r^{2} = \frac{h}{2\pi} \Rightarrow m \times 2\pi v \times r^{2} = \frac{h}{2\pi} \Rightarrow v = \frac{h}{4\pi^{2}mr^{2}}$
$= \frac{6.6 \times 10^{- 34}}{4(3.14)^{2} \times 9.1 \times 10^{- 31} \times \left( 0.53 \times 10^{- 10} \right)^{2}} = 6.5 \times 10^{15}\frac{rev}{s}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.