Question

Derive an expression for the radius of the \( n \)-th Bohr orbit of the electron in hydrogen atom.

Hint:

Balance centripetal and electrostatic forces, then apply Bohr’s quantization to find orbit radius.

Answer 1

For a hydrogen atom, the electron revolves in a circular orbit. The centripetal force is provided by the electrostatic force between the proton and electron.
Step 1: Electrostatic force: \[ F = \frac{1}{4\pi \epsilon_0} \frac{e^2}{r^2}, \] where \( e \) is the electron charge, \( r \) is the radius, and \( \epsilon_0 \) is the permittivity of free space.
Centripetal force: \[ \frac{m v^2}{r} = \frac{1}{4\pi \epsilon_0} \frac{e^2}{r^2} \quad \Rightarrow \quad m v^2 = \frac{e^2}{4\pi \epsilon_0 r}, \quad \cdots (1) \] where \( m \) is the electron mass, \( v \) is velocity.
Step 2: Bohr’s quantization (third postulate): \[ m v r = \frac{n h}{2\pi}, \quad \Rightarrow \quad v = \frac{n h}{2\pi m r}, \quad \cdots (2) \] where \( n \) is the principal quantum number, \( h \) is Planck’s constant.
Step 3: Substitute \( v \) into (1): \[ m \left( \frac{n h}{2\pi m r} \right)^2 = \frac{e^2}{4\pi \epsilon_0 r}. \] \[ \frac{n^2 h^2}{4\pi^2 m r^2} = \frac{e^2}{4\pi \epsilon_0 r} \quad \Rightarrow \quad n^2 h^2 = 4\pi^2 m r \cdot \frac{e^2}{4\pi \epsilon_0} \quad \Rightarrow \quad r = \frac{n^2 h^2 4\pi \epsilon_0}{4\pi^2 m e^2}. \] \[ r = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}. \] Answer: \( r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2} \).