Question

Prove that the period of revolution (T) of electrons in stable orbits of the atom is directly proportional to the cube of the principal quantum number ($n$), on the basis of Bohr’s atom model.

Hint:

In Bohr’s atom: $r \propto n^2$, $v \propto 1/n$, hence $T \propto n^3$.

Answer 1

Step 1: Bohr’s quantization condition.
According to Bohr’s model, angular momentum of electron in $n^{th}$ orbit is: \[ m v r = n \hbar, \] where $m$ = electron mass, $v$ = velocity, $r$ = radius.
Step 2: Radius of $n^{th$ orbit.}
From Bohr’s theory: \[ r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2 Z}. \] So, \[ r_n \propto n^2. \]
Step 3: Velocity of electron.
Velocity is given by: \[ v_n = \frac{Z e^2}{2 \epsilon_0 h} \cdot \frac{1}{n}. \] So, \[ v_n \propto \frac{1}{n}. \]
Step 4: Period of revolution.
Period is time taken for one complete revolution: \[ T = \frac{2 \pi r_n}{v_n}. \] Substituting dependencies: \[ T \propto \frac{n^2}{1/n} = n^3. \]
Step 5: Conclusion.
Thus, the period of revolution of electrons in Bohr’s model is proportional to $n^3$.