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 electrostatic attraction between the electron and the nucleus provides the centripetal force necessary for the electron's circular orbit. The centripetal force is defined as: \[ F_{\text{centripetal}} = \frac{m v^2}{r} \] where \( m \) represents the electron's mass, \( v \) its speed, and \( r \) the orbit's radius. Coulomb's law describes the electrostatic force: \[ F_{\text{electrostatic}} = \frac{k e^2}{r^2} \] where \( e \) is the electron's charge, and \( k \) is Coulomb's constant. Equating these forces yields: \[ \frac{m v^2}{r} = \frac{k e^2}{r^2} \] From this relationship, \( v \) and \( r \) can be expressed in terms of \( n \), the principal quantum number. Employing Bohr's quantization condition: \[ m v r = n h \quad \Rightarrow \quad v = \frac{n h}{2 \pi m r} \] Substituting \( v \) 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 \( r \) reveals: \[ r \propto n^2 \] The time period \( T \) is the duration of one complete revolution, related to \( v \) and \( r \) by: \[ T = \frac{2 \pi r}{v} \] Substituting the expressions for \( v \) and \( r \) results in: \[ T \propto n^3 \] Consequently, the time period for an electron's revolution in the \( n \)-th orbit is directly proportional to \( n^3 \).