Instead of angular momentum quantisation a student gives predicts… - Vidyadip

MCQ

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

A
The radius of the electron orbit is $r \propto \sqrt{n}$.
B
The speed of the electron is $v \propto \sqrt{n}$.
C
The angular speed of the electron is $\omega \propto \frac{1}{n}$.
✓
The angular momentum of the electron is $L \propto \sqrt{n}$.

✓

Answer

Correct option: D.

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

d
$(d)$ As in a circular orbit, the electrostatic force provides the centripetal force, so

$\frac{m v^{2}}{r}=\frac{K z e^{2}}{r^{2}}$

or $\quad \frac{1}{2} m v^{2}=\frac{K z e^{2}}{2 r}= KE$

and $\quad PE =-\frac{K z e^{2}}{r}$

$\therefore$ Total energy $= KE + PE$

$=\frac{K z e^{2}}{2 r}-\frac{K z e^{2}}{r}=-\frac{1}{2} \frac{K z e^{2}}{r}$

Given, $\quad E=\frac{-E_{0}}{n}$

Comparing given equation, we get

$r \propto n$

$\text { As, } \quad KE \propto \frac{1}{r} \propto \frac{1}{n} (\because r \propto n)$

$\Rightarrow v^{2} \propto \frac{1}{n} \text { or } \quad v \propto \frac{1}{\sqrt{n}}$

$\text { Since, angular momentum, } L=m v r$

$\text { or } L \propto v r$

$\text { or } L \propto \frac{1}{\sqrt{n}} n$ $\quad(\because r \propto n)$

$\text { or } L \propto \sqrt{n}$

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