Q141 / 150 · MHT CET 2024 (03 May Shift 2) · Physics
When the electron orbiting in hydrogen atom in its ground state moves to the third excited state, the de-Broglie wavelength associated with it
- Abecomes zero.
- Bremains unchanged.
- Cwill decrease.
- Correct answerDwill increase.
Solution
As $\mathrm{v} \propto \frac{1}{\mathrm{n}}$, we can also write
$\therefore \quad m v \propto \frac{1}{n} \quad \Rightarrow p \propto \frac{1}{n}$
Also, $\lambda=\frac{\mathrm{h}}{\mathrm{p}} \quad \Rightarrow \lambda \propto \mathrm{n}$ $\therefore \quad$ Wavelength will increase.
Also, $\lambda=\frac{\mathrm{h}}{\mathrm{p}} \quad \Rightarrow \lambda \propto \mathrm{n}$ $\therefore \quad$ Wavelength will increase.
Hence, the correct answer is option: 'D'.