Step 1: Understanding the Concept:
Bohr's model of the atom describes electrons as moving in circular orbits around a central nucleus.
The frequency of revolution (\(f\)) represents how many times the electron orbits the nucleus in one second.
Frequency is the inverse of the time period (\(T\)), where \(T\) is the time taken for one full revolution.
To determine how \(f\) depends on the principal quantum number \(n\), we need to look at the dependencies of orbital velocity (\(v\)) and orbital radius (\(r\)) on \(n\).
Step 2: Key Formula or Approach:
The frequency is given by the relation:
\[ f = \frac{1}{T} = \frac{v}{2\pi r} \]
From Bohr's postulates for a hydrogen atom (\(Z=1\)):
1. Orbital velocity (\(v_n\)) is proportional to \(1/n\).
2. Orbital radius (\(r_n\)) is proportional to \(n^2\).
Step 3: Detailed Explanation:
Let's derive the exact dependencies. Bohr's quantization of angular momentum states:
\[ mvr = \frac{nh}{2\pi} \implies v \propto \frac{n}{r} \]
The centripetal force is provided by the electrostatic force:
\[ \frac{mv^2}{r} = \frac{ke^2}{r^2} \implies v^2 \propto \frac{1}{r} \]
Combining these, we find that \(r \propto n^2\) and \(v \propto 1/n\).
Now, let's substitute these proportionalities into the frequency formula:
\[ f \propto \frac{v}{r} \]
\[ f \propto \frac{1/n}{n^2} \]
\[ f \propto \frac{1}{n^3} \]
This means that as the orbit number \(n\) increases, the frequency of revolution decreases significantly.
For example, an electron in the \(n=2\) orbit revolves \(2^3 = 8\) times slower than an electron in the \(n=1\) orbit.
The full expression for frequency is:
\[ f = \frac{m Z^2 e^4}{4 \epsilon_0^2 h^3 n^3} \]
From this complete formula, it is clear that \(f \propto 1/n^3\).
Step 4: Final Answer:
The frequency of revolution of an electron in Bohr's orbit is inversely proportional to the cube of the principal quantum number, i.e., \(f \propto 1/n^3\).