When the electron orbiting in hydrogen atom in its ground state moves to the third excited state, the…

When the electron orbiting in hydrogen atom in its ground state moves to the third excited state, the de-Broglie wavelength associated with it
  1. becomes zero.
  2. remains unchanged.
  3. will decrease.
  4. will 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.

Asked in: MHT CET 2024 (03 May Shift 2)

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