Question

When an electron in hydrogen atom jumps from third excited state to the ground state, the de-Broglie wavelength associated with the electron becomes

• A. $(1/2)^{th}$
• B. $(1/4)^{th}$
• C. $(1/8)^{th}$
• D. $(1/6)^{th}$

Hint:

Remember: "n-th excited state" means $n+1$ in the Bohr formula. Always add 1 to the excitation level to get the correct quantum number $n$.

Answer 1

The Correct Option is B

Step 1: Link wavelength to the orbit number.
The de Broglie wavelength is $\lambda = \tfrac{h}{p}$. In Bohr orbits the speed falls as $\tfrac1n$, so the momentum also falls as $\tfrac1n$, which makes $\lambda \propto n$.

Step 2: Identify the two levels.
The ground state is $n=1$. The third excited state is $n=4$ (ground, then first, second, third excited give $1,2,3,4$).

Step 3: Take the ratio.
Since $\lambda \propto n$, going from $n=4$ to $n=1$ scales the wavelength by $\tfrac{1}{4}$.

Step 4: State the change.
\[ \boxed{\lambda \to \tfrac14\ \text{of its value}} \]