Q. The de-Broglie wavelength of an electron in the first Bohr orbit is
A) Equal to one fourth the circumference of the first orbit
B) Equal to half the circumference of the first orbit
C) Equal to twice the circumference of the first orbit
D) Equal to the circumference of the first orbit
Explanation
(d) $mvr_{n} = \frac{nh}{2\pi} \Rightarrow pr_{n} = \frac{nh}{2\pi} \Rightarrow \frac{h}{\lambda} \times r_{n} = \frac{nh}{2\pi}$
$\Rightarrow \lambda = \frac{2\pi r_{n}}{n}$, for first orbit $n = 1$ so $\lambda = 2\pi r_{1}$
$=$ circumference of first orbit
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.