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(D) In a circular orbit,the electrostatic force provides the centripetal force:$\frac{m v^{2}}{r} = \frac{K Z e^{2}}{r^{2}}$This implies the kinetic e

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The angular momentum of the electron is $L \propto \sqrt{n}$.

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(D) In a circular orbit,the electrostatic force provides the centripetal force:$\frac{m v^{2}}{r} = \frac{K Z e^{2}}{r^{2}}$This implies the kinetic energy $KE = \frac{1}{2} m v^{2} = \frac{K Z e^{2}}{2 r}$.The potential energy is $PE = -\frac{K Z e^{2}}{r}$.Thus,the total energy $E = KE + PE = \frac{K Z e^{2}}{2 r} - \frac{K Z e^{2}}{r} = -\frac{K Z e^{2}}{2 r}$.Given $E = \frac{-E_{0}}{n}$,comparing this with $E = -\frac{K Z e^{2}}{2 r}$,we get $r \propto n$.Since $KE = \frac{1}{2} m v^{2} \propto \frac{1}{r} \propto \frac{1}{n}$,we have $v^{2} \propto \frac{1}{n}$,which means $v \propto \frac{1}{\sqrt{n}}$.The angular momentum is $L = m v r$.Substituting the proportionalities $v \propto n^{-1/2}$ and $r \propto n$,we get $L \propto n^{-1/2} \cdot n = n^{1/2} = \sqrt{n}$.Therefore,$L \propto \sqrt{n}$.

$A$. 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	$iv$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}$

Instead of angular momentum quantisation,a student predicts that energy is quantised as $E = \frac{-E_{0}}{n}$,$(E_{0} > 0)$ and $n$ is a positive integer. Which of the following options is correct?

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