Question

The ratio of radius of two different orbits in a $H-$atom is 4:9, then the ratio of the frequency of revolution of electron in these orbits is:

• A. 9:4
• B. 27:8
• C. 3:2
• D. 8:27

Answer 1

Answer: (A)

9:4

Solution:-

The relation between frequency and radius for an electron in an orbit is-

$\vartheta =\frac{v}{2\pi r}$, i.e.,

$\vartheta \propto \frac{1}{r}$

$\Rightarrow {\vartheta }_1{\vartheta }_2=\frac{r_2}{r_1}$

$\Rightarrow \frac{{\vartheta }_1}{{\vartheta }_2}=\frac{1}{\left(\frac{r_1}{r_2}\right)}.....\left(1\right)$

Given that:-

$\frac{r_1}{r_2}=\frac{4}{9}.....\left(2\right)$

From equation $\left(1\right)\&\left(2\right)$, we have

$\frac{{\vartheta }_1}{{\vartheta }_2}=\frac{1}{\left(\frac{4}{9}\right)}=\frac{9}{4}$

Hence the ratio of the frequency of revolution of electron is $9:4$.