The kinetic energy of the electron in an orbi | vedclass

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(C) In a hydrogen atom, the electrostatic force of attraction between the nucleus (charge $+e$) and the electron (charge $-e$) provides the necessary

Vedclass · Accepted Answer

$e^2/2r$

Vedclass · Answer

(C) In a hydrogen atom, the electrostatic force of attraction between the nucleus (charge $+e$) and the electron (charge $-e$) provides the necessary centripetal force for the electron to move in a circular orbit of radius $r$.According to Coulomb's law, the electrostatic force is $F = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$.The centripetal force required for circular motion is $F = \frac{mv^2}{r}$.Equating these, we get $\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$.Multiplying both sides by $r/2$, we get $\frac{1}{2}mv^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{2r}$.The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2}mv^2 = \frac{e^2}{8\pi\epsilon_0 r}$.Since $\epsilon_0$ is a constant, the kinetic energy is proportional to $e^2/r$.

$A$. Speed of revolution of electron	$i$. $\frac{1}{4 \pi \varepsilon_0} \frac{2 \pi Z e^2}{n h}$
$B$. Kinetic energy	$ii$. $-\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$C$. Total energy	$iii$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$D$. Frequency	$iv$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}$

The kinetic energy of the electron in an orbit of radius $r$ in a hydrogen atom is proportional to ($e$ = electronic charge).

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