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To which of the following the angular velocity of the electron in the $n$ -th Bohr orbit is proportional?
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Verified Answer
The correct answer is:
$\frac{1}{n^{3}}$
According to the Bohr's atomic model.
$\text {Angular momentum, } L=m v=\frac{n h}{2 \pi}....(i)$
since, angular velocity $\omega=\frac{v}{r}$
$\Rightarrow v=r \omega$
From Eq. (i), we get
$\begin{aligned}
m(r \omega) r &=\frac{n h}{2 \pi} \\
m \omega r^{2} &=\frac{n h}{2 \pi} \\
\omega &=\frac{n h}{2 \pi m r^{2}}....(ii)
\end{aligned}$
since, the radius of the electron in $n^{t h}$ orbit of
Bohr's atomic model is given as, $r=\frac{n^{2} h^{2} \varepsilon_{0}}{\pi m z e^{2}} \ldots$ (iii)
Squaring the Eq. (iii) and substituting its value in Eq. (ii), we get
$\omega=\frac{n h(\pi n z e)^{2}}{2 \pi m\left(n^{4} h^{4} \varepsilon_{0}^{2}\right)} \Rightarrow \omega \propto \frac{1}{n^{3}}$
$\text {Angular momentum, } L=m v=\frac{n h}{2 \pi}....(i)$
since, angular velocity $\omega=\frac{v}{r}$
$\Rightarrow v=r \omega$
From Eq. (i), we get
$\begin{aligned}
m(r \omega) r &=\frac{n h}{2 \pi} \\
m \omega r^{2} &=\frac{n h}{2 \pi} \\
\omega &=\frac{n h}{2 \pi m r^{2}}....(ii)
\end{aligned}$
since, the radius of the electron in $n^{t h}$ orbit of
Bohr's atomic model is given as, $r=\frac{n^{2} h^{2} \varepsilon_{0}}{\pi m z e^{2}} \ldots$ (iii)
Squaring the Eq. (iii) and substituting its value in Eq. (ii), we get
$\omega=\frac{n h(\pi n z e)^{2}}{2 \pi m\left(n^{4} h^{4} \varepsilon_{0}^{2}\right)} \Rightarrow \omega \propto \frac{1}{n^{3}}$
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