Question

The de Broglie wavelength $\lambda_n$ of the electron in the $n^{\text{th}}$ orbit of hydrogen atom is

• A. inversely proportional to $n$
• B. proportional to $n^2$
• C. proportional to $n$
• D. inversely proportional to $n^2$
• E. inversely proportional to radius of the orbit in the $n^{th}$ state

Hint:

This relationship explains why orbits are stable; only orbits where the electron forms a stationary standing wave (an integer number of wavelengths) are allowed.

Answer 1

The Correct Option is C

Concept:
According to Bohr's quantization condition, the circumference of the $n^{th}$ orbit must be an integral multiple of the de Broglie wavelength: \[ 2\pi r_n = n\lambda_n \implies \lambda_n = \frac{2\pi r_n}{n} \]

Step 1: Relate radius to $n$.
In a hydrogen atom, the radius of the $n^{th}$ orbit is $r_n \propto n^2$.

Step 2: Determine the proportionality for $\lambda_n$.
\[ \lambda_n \propto \frac{n^2}{n} \implies \lambda_n \propto n \] Thus, the de Broglie wavelength is directly proportional to the principal quantum number $n$.