Question

The angular speed of an electron revolving round the H-nucleus is proportional to

• A. $\frac{1}{r}$
• B. $\frac{1}{r^{\frac{3}{2}}}$
• C. $\frac{1}{r^2}$
• D. $r^{\frac{3}{2}}$

Answer 1

The correct option is B $\frac{1}{r^{\frac{3}{2}}}$

Angular speed $=\frac{V}{2\pi r}$
$v$ of an electron $=\frac{k}{n}$ --------(i)
$r$ of an electron $\Rightarrow r=c{n}^2$
$\Rightarrow \sqrt{\frac{r}{c}}=n$--------(ii)

From (i) and (ii),
$V=\frac{K\sqrt{c}}{\sqrt{r}}=\frac{a}{\sqrt{r}}$
Angular speed $={\left(\frac{2\pi r}{K}\sqrt{\frac{r}{c}}\right)}^{-1}$
$=(ar\sqrt{r}{)}^{-1}=\frac{r^{-3/2}}{a}$ where a is a constant

So, $\omega \propto r^{-3/2}$