To which of the following is the angular velo | vedclass

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(D) According to the Bohr's atomic model, the angular momentum $L$ is given by:$L = mvr = \frac{nh}{2\pi} \dots (i)$Since angular velocity $\omega = \

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$\frac{1}{n^{3}}$

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(D) According to the Bohr's atomic model, the angular momentum $L$ is given by:$L = mvr = \frac{nh}{2\pi} \dots (i)$Since angular velocity $\omega = \frac{v}{r}$, we have $v = r\omega$.Substituting $v = r\omega$ into equation $(i)$:$m(r\omega)r = \frac{nh}{2\pi} \Rightarrow m\omega r^2 = \frac{nh}{2\pi} \Rightarrow \omega = \frac{nh}{2\pi mr^2} \dots (ii)$The radius of the electron in the $n$-th orbit is given by:$r = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2} \dots (iii)$Substituting the expression for $r$ from equation $(iii)$ into equation $(ii)$:$\omega = \frac{nh}{2\pi m} \left( \frac{\pi m Z e^2}{n^2 h^2 \epsilon_0} \right)^2$$\omega = \frac{nh}{2\pi m} \cdot \frac{\pi^2 m^2 Z^2 e^4}{n^4 h^4 \epsilon_0^2}$$\omega = \frac{\pi m Z^2 e^4}{2 h^3 \epsilon_0^2} \cdot \frac{1}{n^3}$Therefore, $\omega \propto \frac{1}{n^3}$.

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

To which of the following is the angular velocity of the electron in the $n$-th Bohr orbit proportional?

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