Q. De-Broglie wavelength of an electron orbiting in the $n = 2$ state of hydrogen atom is close to (Given Bohr radius = 0.052 nm)
A) 0.067 nm
B) 0.67 nm
C) 1.67 nm
D) 2.67 nm
Explanation
[Solution by Claude - not from source] $2\pi r_n=n\lambda$ with $r_2=4a_0$: $\lambda=\frac{2\pi\times4a_0}{2}=4\pi a_0=4\pi\times0.052\approx0.65$ nm, closest to 0.67 nm. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.