The angular momentum of an electron in a hydr | vedclass

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Question

(A) The centripetal force required for the electron to orbit the nucleus is provided by the electrostatic force of attraction: $F_{C} = F_{e}$.Using t

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$\sqrt{r}$

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(A) The centripetal force required for the electron to orbit the nucleus is provided by the electrostatic force of attraction: $F_{C} = F_{e}$.Using the formula for centripetal force $F_{C} = \frac{mv^2}{r}$ and Coulomb's law $F_{e} = \frac{kZe^2}{r^2}$,we have: $\frac{mv^2}{r} = \frac{kZe^2}{r^2}$.Multiplying both sides by $mr^2$,we get: $m^2v^2r^2 = mkZe^2r$.Since angular momentum $L = mvr$,we can write: $L^2 = mkZe^2r$.Taking the square root of both sides,we find: $L = \sqrt{mkZe^2r}$.Since $m$,$k$,$Z$,and $e$ are constants,the angular momentum is proportional to the square root of the radius: $L \propto \sqrt{r}$.

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

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