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: Understanding the Concept:
The third excited state is \(n=4\) (ground \(n=1\), first excited \(n=2\), second excited \(n=3\)). In Bohr's model the orbit radius grows as \(n^2\) and the speed falls as \(\dfrac1n\).

Step 2: de Broglie wavelength:
\[ \lambda=\frac h{mv} \]
Since \(v\propto\dfrac1n\), \(\lambda\propto n\). Equivalently \(2\pi r=n\lambda\) gives \(\lambda=\dfrac{2\pi r}{n}\propto\dfrac{n^2}{n}=n\).

Step 3: Compare:
From \(n=1\) to \(n=4\), \(\lambda\) becomes 4 times. So it will increase, option (B).

Final Answer:
The wavelength becomes 4 times larger. \[ \boxed{B:\ \text{will increase}} \]