Question:medium

Write the mathematical forms of three postulates of Bohr’s theory of the hydrogen atom. Using them, prove that for an electron revolving in the \( n^{\text{th}} \) orbit:
(a) the radius of the orbit is proportional to \( n^2 \), and
(b) the total energy of the atom is proportional to \( \left( \frac{1}{n^2} \right) \).

Show Hint

To prove \( r \propto n^2 \) and \( E \propto -1/n^2 \), combine Coulomb's force law with Bohr's angular momentum quantization: Use \( mvr = \frac{nh}{2\pi} \) and \( \frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{r^2} \)
Updated On: Jan 13, 2026
Show Solution

Solution and Explanation

Bohr's Postulates:
  1. Electrons orbit the nucleus in specific circular paths without losing energy.
  2. The angular momentum of an electron is quantized according to the formula: \[ mvr = \frac{nh}{2\pi} \]
  3. Energy is emitted or absorbed when an electron transitions between orbits, described by: \[ E = hu = E_i - E_f \]
(a) Orbit Radius \( r \propto n^2 \) The centripetal force required for the electron's motion is supplied by the electrostatic Coulomb force: \[ \frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{r^2} \Rightarrow mv^2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{r} \quad \cdots(1) \] According to Bohr's quantization condition: \[ mvr = \frac{nh}{2\pi} \Rightarrow v = \frac{nh}{2\pi mr} \quad \cdots(2) \] Substituting equation (2) into equation (1): \[ m \left( \frac{nh}{2\pi mr} \right)^2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{r} \Rightarrow \frac{n^2 h^2}{4\pi^2 m r^2} = \frac{e^2}{4\pi\varepsilon_0 r} \] Solving for \( r \): \[ r \propto \frac{n^2 h^2 \varepsilon_0}{\pi m e^2} \Rightarrow r_n \propto n^2 \] (b) Total Energy \( E \propto -\frac{1}{n^2} \) Kinetic Energy: \[ K = \frac{1}{2} mv^2 = \frac{e^2}{8\pi\varepsilon_0 r} \quad \text{(derived from centripetal force equation)} \] Potential Energy: \[ U = -\frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{r} \] Total Energy: \[ E = K + U = \frac{e^2}{8\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r} = -\frac{e^2}{8\pi\varepsilon_0 r} \] Since \( r \propto n^2 \), it follows that \( E \propto -\frac{1}{n^2} \).
Was this answer helpful?
0