Question

The frequency of revolution of the electron in Bohr’s orbit varies with \( n \), the principal quantum number as:

• A. \( \frac{1}{n^3} \)
• B. \( \frac{1}{n^4} \)
• C. \( \frac{1}{n} \)
• D. \( \frac{1}{n^2} \)

Hint:

For Bohr’s model, remember that the frequency of revolution decreases with the cube of the principal quantum number \( n \).

Answer 1

The Correct Option is A

The frequency of revolution is inversely proportional to \( n^3 \), as the energy of the electron in Bohr’s model depends on the quantum number \( n \).

Answer 2

Step 1: Understanding the concept.
In Bohr’s model of the hydrogen atom, the electron revolves around the nucleus in discrete circular orbits characterized by the principal quantum number \( n \). The frequency of revolution of the electron depends on \( n \).

Step 2: Expression for velocity of the electron.
According to Bohr’s theory, the velocity of the electron in the \( n^{th} \) orbit is given by:
\[ v_n = \frac{e^2}{2\varepsilon_0 h} \cdot \frac{1}{n} \] Thus, the velocity of the electron is inversely proportional to \( n \):
\[ v_n \propto \frac{1}{n} \]

Step 3: Expression for radius of the orbit.
The radius of the \( n^{th} \) Bohr orbit is:
\[ r_n = n^2 a_0 \] where \( a_0 \) is the Bohr radius. Hence,
\[ r_n \propto n^2 \]

Step 4: Frequency of revolution.
The frequency of revolution \( f \) is the number of revolutions per second made by the electron, 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.
The frequency of revolution of the electron in Bohr’s orbit varies with \( n \) as:
\[ \boxed{f \propto \frac{1}{n^3}} \]