Question
When the electron orbiting in hydrogen atom in its ground state moves to the third excited state, show how the de Broglie wavelength associated with it would be affected.

Answer

de Broglie wavelength $\lambda = \frac{\text{h}}{\text{mv}}$

$\therefore \lambda \propto\frac{1}{v} ; \text{v} \propto\frac{1}{\text{n}}$

$\therefore \lambda\propto\text{n}$

$\therefore$ 𝑑𝑒 Broglie wavelength will increase

Alternate Answer

As $2\pi\text{r}_{n} = \text{n}\lambda ; \lambda = \frac{2\pi\text{r}_{n}}{\text{n}}(\lambda\propto\frac{\text{r}_{n}}{\text{n}})$

$\text{r}_{n}\propto\text{n}^{2}$

$\therefore \lambda\propto\frac{\text{n}^{2}}{\text{n}}\Rightarrow\lambda\propto\text{n}$

$\therefore$ 𝑑𝑒 Broglie wavelength will increase.

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