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: Understanding the Question:
We need to determine the change in the de-Broglie wavelength of an electron in a hydrogen atom when it transitions from the third excited state to the ground state.

Step 2: Key Formula or Approach:
1. de-Broglie wavelength: $\lambda = \frac{h}{p}$, where $p$ is momentum.
2. According to Bohr's quantization condition: $mvr = \frac{nh}{2\pi}$. Since $p=mv$, $p \propto \frac{1}{r}$. Also $r \propto n^2$.
3. More directly, the velocity $v \propto \frac{1}{n}$, so $p \propto \frac{1}{n}$. Thus, $\lambda = \frac{h}{p} \propto n$.

Step 3: Detailed Explanation:
4. The ground state is $n_1 = 1$.
5. The third excited state is $n_2 = 4$ (Ground: 1, 1st excited: 2, 2nd excited: 3, 3rd excited: 4).
6. Using the proportionality $\lambda \propto n$, the ratio of the wavelengths is $\frac{\lambda_{ground}}{\lambda_{excited}} = \frac{n_1}{n_2}$.
7. $\frac{\lambda_{final}}{\lambda_{initial}} = \frac{1}{4}$.

Step 4: Final Answer:
The wavelength becomes $(1/4)^{th}$ of its original value, which is option (B).