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Let energy, magnitude of linear momentum and angular frequency of an electron in hydrogen atom be $E, p$ and $\omega$ respectively. If $n$ be the corresponding quantum number then
Option: 1
$\frac{E}{\omega} \alpha\ n$

Option: 2
$\frac{E p}{\omega} \, \alpha \ \ n^2$

Option: 3
$p\, \omega \, \, \alpha \, \, n^{1/2}$

Option: 4
$Ep\, \omega \, \, \alpha \ n^{\circ}$

Answers (1)

best_answer

Energy is $E\, \alpha \ \frac{1}{n^2} \text {. }$

Angular momentum $L=m v r\, \, \alpha \ n$

radius $r_n \, \, \alpha \ n^2$ $\Rightarrow p \, \, \alpha \ 1 / n .$

Time period $T \, \alpha \ \, n^{3}$

angular freq. $\omega \, \, \alpha \ \frac{1}{n^{3}}$

speed of electron $v \propto \frac{1}{n} .$

$\Rightarrow \frac{E}{\omega} \, \alpha \ n$

$\frac{E p}{\omega} \, \, \alpha \ n^0$ $; p \cdot \omega=\alpha \ \frac{1}{n^4} \text {. }$

$E \, p \, \omega \, \, \alpha \ \frac{1}{n^6} \text {. }$

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