JEE Physics 2025 Question with Solution

Given: The frequency of revolution of the electron in Bohr’s orbit is to be related with the principal quantum number $n$.

Find: How the frequency varies with $n$.

In Bohr’s model, the velocity of the electron in the $n^{\text{th}}$ orbit varies as

$$v_n \propto \frac{1}{n}$$

and the radius of the $n^{\text{th}}$ orbit varies as

$$r_n \propto n^2$$

The frequency of revolution is

$$f = \frac{v_n}{2\pi r_n}$$

Substituting the proportionalities,

$$f \propto \frac{1/n}{n^2} = \frac{1}{n^3}$$

Therefore, the frequency of revolution varies as $\frac{1}{n^3}$. The correct option is A.

Given: In Bohr’s model of the hydrogen atom, the electron revolves in discrete circular orbits characterized by the principal quantum number $n$.

Find: The dependence of frequency of revolution on $n$.

Step 1: Understanding the concept. In Bohr’s model, the frequency of revolution depends on the electron speed and the orbit radius.

Step 2: Expression for velocity of the electron. According to Bohr’s theory, the velocity of the electron in the $n^{\text{th}}$ orbit is given by

$$v_n = \frac{e^2}{2\varepsilon_0 h} \cdot \frac{1}{n}$$

Thus,

$$v_n \propto \frac{1}{n}$$

Step 3: Expression for radius of the orbit. The radius of the $n^{\text{th}}$ Bohr orbit is

$$r_n = n^2 a_0$$

Hence,

$$r_n \propto n^2$$

Step 4: Frequency of revolution. The frequency of revolution $f$ is given by

$$f = \frac{v_n}{2\pi r_n}$$

Substituting the proportionalities of $v_n$ and $r_n$,

$$f \propto \frac{1/n}{n^2} = \frac{1}{n^3}$$

Step 5: Final Answer. Therefore, the frequency of revolution of the electron in Bohr’s orbit varies as $\frac{1}{n^3}$. The correct option is A.