Question

For an electron moving in the $n^{th}$ Bohr orbit the de-Broglie wavelength of an electron is

• A. $n\pi r$
• B. $\frac{\pi r}{n}$
• C. $\frac{nr}{2\pi}$
• D. $\frac{2\pi r}{n}$

Hint:

Logic Tip: A conceptual way to arrive at this answer instantly is to visualize an electron as a standing wave around the nucleus. For the orbit to be stable (constructive interference), the total circumference of the orbit ($2\pi r$) must be equal to exactly $n$ complete wavelengths ($n\lambda$). Thus, $2\pi r = n\lambda \implies \lambda = \frac{2\pi r}{n}$.

Answer 1

The Correct Option is D