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

Concept:
Bohr's Second Postulate of Quantization states that the angular momentum ($L$) of an electron in a stable orbit is an integral multiple of $\frac{h}{2\pi}$. $$L = mvr = \frac{nh}{2\pi}$$ De Broglie's Hypothesis states that any moving particle has an associated wave nature, with a wavelength ($\lambda$) given by: $$\lambda = \frac{h}{p} = \frac{h}{mv}$$
Step 1: Express momentum in terms of the de Broglie wavelength.
From the de Broglie relation, we can isolate the momentum ($mv$): $$mv = \frac{h}{\lambda}$$
Step 2: Substitute momentum into Bohr's quantization condition.
Take Bohr's angular momentum condition: $$(mv)r = \frac{nh}{2\pi}$$ Substitute $mv = \frac{h}{\lambda}$: $$\left(\frac{h}{\lambda}\right) r = \frac{nh}{2\pi}$$
Step 3: Solve for the wavelength $\lambda$.
Cancel Planck's constant ($h$) from both sides: $$\frac{r}{\lambda} = \frac{n}{2\pi}$$ Rearrange the equation to isolate $\lambda$: $$2\pi r = n\lambda$$ $$\lambda = \frac{2\pi r}{n}$$