When the electron orbiting in hydrogen atom goes from one orbit to another orbit (principal quantum number \( = n \)), the de-Broglie wavelength (\( \lambda \)) associated with it is related to \( n \) as}
Show Hint
The circumference of the $n^{th}$ orbit is exactly $n$ de-Broglie wavelengths.
Step 1: Bohr's Postulate
$mvr = n\frac{h}{2\pi} \implies 2\pi r = n \left( \frac{h}{mv} \right)$.
Step 2: de-Broglie Relation
$\lambda = \frac{h}{mv}$. Substituting this into the postulate: $2\pi r = n\lambda$.
Step 3: Analysis
For Hydrogen atom, $r_n \propto n^2$ and $v_n \propto \frac{1}{n}$.
From $\lambda = \frac{h}{mv}$, we get $\lambda \propto \frac{1}{1/n} \implies \lambda \propto n$.
Final Answer: (D)