Q. According to de-Broglie, the de-Broglie wavelength for electron in an orbit of (radius $5.3 \times 10^{- 11}m$) hydrogen atom is $10^{- 10}m.$ The principle quantum number for this electron is
A) 1
B) 2
C) 3
D) 4
Explanation
(c) $2\pi r = n\lambda \Rightarrow n = \frac{2\pi r}{\lambda} = \frac{2 \times 3.14 \times 5.3 \times 10^{- 11}}{10^{- 10}} = 3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.