Question:medium

The frequency of revolution of an electron in the $n^{\text{th}$ orbit of hydrogen atom is

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Radius $n^2$, Velocity $1/n$, Time Period $n^3$, Frequency $1/n^3$.
Updated On: May 14, 2026
  • directly proportional to $n^2$
  • inversely proportional to $n^2$
  • directly proportional to $n^3$
  • inversely proportional to $n^3$
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
According to Bohr's model, the velocity and orbital radius of an electron depend on the principal quantum number $n$.
The frequency of revolution is the inverse of the time period, which depends on radius and velocity.
Step 2: Key Formula or Approach:
Radius of $n^{\text{th}}$ orbit: $r_n \propto n^2$.
Velocity of electron in $n^{\text{th}}$ orbit: $v_n \propto \frac{1}{n}$.
Time period $T_n = \frac{2\pi r_n}{v_n}$.
Frequency $f_n = \frac{1}{T_n} = \frac{v_n}{2\pi r_n}$.
Step 3: Detailed Explanation:
Using the proportionalities: \[ f_n \propto \frac{v_n}{r_n} \] Substitute the dependencies on $n$: \[ f_n \propto \frac{1/n}{n^2} \] \[ f_n \propto \frac{1}{n \cdot n^2} \] \[ f_n \propto \frac{1}{n^3} \] This means the frequency is inversely proportional to the cube of the principal quantum number $n^3$.
Step 4: Final Answer:
The frequency is inversely proportional to $n^3$.
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