The angular momentum of an electron in a hydrogen atom is proportional to : (Where $r$ is the radius of orbit of electron)

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{F}_{\mathrm{C}}=\frac{\mathrm{mv}^2}{\mathrm{r}} \ & \frac{\mathrm{Kq}_1 \mathrm{q}_2}{\mathrm{r}^2}=\frac{\mathrm{mv}^2}{\mathrm{r}} \ & \mathrm{mv}^2 \mathrm{r}^2=K \mathrm{~K}_1 \mathrm{q}_2 \mathrm{r} \ & \frac{\mathrm{L}^2}{\mathrm{~m}}=K \mathrm{Kq}_1 \mathrm{q}_2 \mathrm{r} \ & \mathrm{L} \propto \sqrt{\mathrm{r}}\end{aligned}$

The angular momentum of an electron in a hydrogen atom is proportional to : (Where $r$ is the radius of…

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