Question
When an electron in hydrogen atom jumps from the third excited state to the ground state, how would the de Broglie wavelength associated with the electron change? Justify your answer.
✓
Answer
$\lambda = \frac{\text{h}}{\text{P}} = \frac{\text{h}}{\sqrt{2\text{mK}}}$
$\frac{\lambda_{1}}{\lambda_{4}} = \sqrt{\frac{\text{K}_{4}}{\text{K}_{1}}}$
But $\text{K}_{n}( = - \text{E}_{n})\propto\frac{1}{\text{n}^{2}}$
Hence , $\frac{\lambda_{1}}{\lambda_{4}} = \sqrt{\frac{1}{16}}$
$\therefore \frac{\lambda_{1}}{\lambda_{4}} = \frac{1}{4}$
$\lambda_{4} = 4 \lambda_{1}$
i.e. $\lambda_{4} > \lambda_{1}$
Alternate Answer
$\lambda_{n} = \frac{\text{h}}{\text{P}_{n}} = \frac{\lambda}{\text{mv}_{n}}$
Velocity of electron in $n^{th}$ state $\upsilon_{n}\propto\frac{1}{n}$
$\lambda_{n}\propto\frac{1}{\upsilon_{n}}\therefore\lambda\propto\text{n}$
$\therefore \frac{\lambda_{4}}{\lambda_{1}} = \frac{\text{n}_{4}}{\text{n}_{1}} = \frac{4}{1}$
$\frac{\lambda_{1}}{\lambda_{4}} = \sqrt{\frac{\text{K}_{4}}{\text{K}_{1}}}$
But $\text{K}_{n}( = - \text{E}_{n})\propto\frac{1}{\text{n}^{2}}$
Hence , $\frac{\lambda_{1}}{\lambda_{4}} = \sqrt{\frac{1}{16}}$
$\therefore \frac{\lambda_{1}}{\lambda_{4}} = \frac{1}{4}$
$\lambda_{4} = 4 \lambda_{1}$
i.e. $\lambda_{4} > \lambda_{1}$
Alternate Answer
$\lambda_{n} = \frac{\text{h}}{\text{P}_{n}} = \frac{\lambda}{\text{mv}_{n}}$
Velocity of electron in $n^{th}$ state $\upsilon_{n}\propto\frac{1}{n}$
$\lambda_{n}\propto\frac{1}{\upsilon_{n}}\therefore\lambda\propto\text{n}$
$\therefore \frac{\lambda_{4}}{\lambda_{1}} = \frac{\text{n}_{4}}{\text{n}_{1}} = \frac{4}{1}$
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