Question

When the electron orbiting in hydrogen atom in its ground state moves to the third excited state, the de-Broglie wavelength associated with it

• A. will decrease
• B. will increase
• C. will remain same
• D. will be zero

Hint:

Speed falls with n, so lambda = h/mv rises.

Answer 1

The Correct Option is B

Step 1: Standing wave idea:
The circumference holds $n$ de Broglie waves: $2\pi r_n=n\lambda_n$.

Step 2: Ground state:
$\lambda_1=2\pi r_1$.

Step 3: Fourth state:
$r_4=16r_1$, so $\lambda_4=\dfrac{2\pi\cdot16r_1}{4}=4\lambda_1$. It increases. Option (B).

Final Answer:
lambda is proportional to n. \[ \boxed{B} \]