Q. For the Bohr’s second orbit of circumference $2\pi r$, the de-Broglie wavelength of revolving electron will be
A) $2\pi r$
B) $\pi r$
C) $\frac{1}{2\pi r}$
D) $\frac{1}{4\pi r}$
Explanation
(b) $mvr = \frac{nh}{2\pi}$ , according to Bohr’s theory
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ 2\pi r = n\left( \frac{h}{mv} \right) = n\lambda$ for n=2, $\lambda = \pi r$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.