Question

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 \( n \)-th orbit,
(a) the radius of the orbit is proportional to \( n^2 \), and
(b) the total energy of the atom is proportional to \( \frac{1}{n^2} \).

Hint:

For Bohr’s model problems: - Use the quantization condition \( m v r = \frac{n h}{2\pi} \) to relate \( v \) and \( r \). - Total energy in the Bohr model is always negative, and its magnitude is proportional to \( \frac{1}{n^2} \).

Answer 1

Step 1: State the three postulates of Bohr’s theory.
1. The electron revolves in circular orbits, with quantized angular momentum: \[ m v r = \frac{n h}{2\pi}, \] where \( m \) is the electron’s mass, \( v \) is its velocity, \( r \) is the radius, \( n \) is the quantum number, and \( h \) is Planck’s constant.
2. The electron does not radiate energy while in these allowed orbits.
3. The electron transitions between orbits by emitting or absorbing a photon: \[ \Delta E = h \nu, \] where \( \nu \) is the frequency of the photon. (a): Prove the radius \( r \propto n^2 \).
From the first postulate: \[ m v r = \frac{n h}{2\pi} \quad \Rightarrow \quad v = \frac{n h}{2\pi m r}. \] The centripetal force equals the Coulomb force: \[ \frac{m v^2}{r} = \frac{k e^2}{r^2}. \] Substitute \( v \): \[ \frac{m}{r} \left( \frac{n h}{2\pi m r} \right)^2 = \frac{k e^2}{r^2} \quad \Rightarrow \quad \frac{n^2 h^2}{4\pi^2 m r^3} = \frac{k e^2}{r^2} \quad \Rightarrow \quad r = \frac{n^2 h^2}{4\pi^2 m k e^2}. \] Thus, \( r \propto n^2 \). (b): Prove the total energy \( E \propto \frac{1}{n^2} \).
Total energy: \[ E = \frac{1}{2} m v^2 - \frac{k e^2}{r}. \] From the force balance, \( \frac{m v^2}{r} = \frac{k e^2}{r^2} \), so: \[ \frac{1}{2} m v^2 = \frac{1}{2} \frac{k e^2}{r}. \] Thus: \[ E = \frac{1}{2} \frac{k e^2}{r} - \frac{k e^2}{r} = -\frac{1}{2} \frac{k e^2}{r}. \] Substitute \( r = \frac{n^2 h^2}{4\pi^2 m k e^2} \): \[ E = -\frac{1}{2} \frac{k e^2}{\frac{n^2 h^2}{4\pi^2 m k e^2}}} = -\frac{2\pi^2 m k^2 e^4}{n^2 h^2}.\] Thus, \( E \propto -\frac{1}{n^2} \), or the magnitude \( |E| \propto \frac{1}{n^2} \).