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.
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Answer
de Broglie wavelength $\lambda = \frac{\text{h}}{\text{mv}}$ $\therefore \lambda \propto\frac{1}{v} ; \text{v} \propto\frac{1}{\text{n}}$ $\therefore \lambda\propto\text{n}$ $\therefore$ ?? Broglie wavelength will increaseAlternate Answer
As $2\pi\text{r}_{n} = \text{n}\lambda ; \lambda = \frac{2\pi\text{r}_{n}}{\text{n}}(\lambda\propto\frac{\text{r}_{n}}{\text{n}})$ $\text{r}_{n}\propto\text{n}^{2}$ $\therefore \lambda\propto\frac{\text{n}^{2}}{\text{n}}\Rightarrow\lambda\propto\text{n}$ $\therefore$ ?? Broglie wavelength will increase.
As $2\pi\text{r}_{n} = \text{n}\lambda ; \lambda = \frac{2\pi\text{r}_{n}}{\text{n}}(\lambda\propto\frac{\text{r}_{n}}{\text{n}})$ $\text{r}_{n}\propto\text{n}^{2}$ $\therefore \lambda\propto\frac{\text{n}^{2}}{\text{n}}\Rightarrow\lambda\propto\text{n}$ $\therefore$ ?? Broglie wavelength will increase.
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