Question

Prove that, in the Bohr model of the hydrogen atom, the time period of revolution of an electron in the \( n \)-th orbit is proportional to \( n^3 \).

Hint:

In the Bohr model, the time period of revolution of an electron is derived from the balance between centripetal force and electrostatic force, and from the quantization of angular momentum.

Answer 1

In the Bohr model, the centripetal force required for an electron to revolve in a circular orbit is provided by the electrostatic force of attraction between the electron and the nucleus. The centripetal force is given by: \[ F_{\text{centripetal}} = \frac{m v^2}{r} \] where \( m \) is the mass of the electron, \( v \) is its speed, and \( r \) is the radius of the orbit. The electrostatic force is given by Coulomb's law: \[ F_{\text{electrostatic}} = \frac{k e^2}{r^2} \] where \( e \) is the charge of the electron, and \( k \) is Coulomb's constant. Equating these two forces: \[ \frac{m v^2}{r} = \frac{k e^2}{r^2} \] From this, we can solve for \( v \) and \( r \) in terms of \( n \), the principal quantum number. Using Bohr's quantization condition: \[ m v r = n h \quad \Rightarrow \quad v = \frac{n h}{2 \pi m r} \] Substitute \( v \) into the force equation: \[ \frac{m \left( \frac{n h}{2 \pi m r} \right)^2}{r} = \frac{k e^2}{r^2} \] Solving for \( r \) gives: \[ r \propto n^2 \] Now, the time period \( T \) is the time taken for one complete revolution, and it is related to the velocity \( v \) and the radius \( r \) by: \[ T = \frac{2 \pi r}{v} \] Substituting \( v \) and \( r \) into this equation: \[ T \propto n^3 \] Thus, the time period of revolution of the electron in the \( n \)-th orbit is proportional to \( n^3 \).