Question

Using Bohr’s postulates, derive the expression for the radius of the \( n \)-th orbit of an electron in a hydrogen atom. Also find the numerical value of Bohr’s radius \( a_0 \).

Hint:

Bohr’s radius gives the size of the smallest orbit for an electron in a hydrogen atom.

Answer 1

From Bohr’s second postulate, the angular momentum of the electron in the \( n \)-th orbit is quantized: \[ m v r = \frac{nh}{2 \pi}, \] where \( m \) is the mass of the electron, \( v \) is its velocity, \( r \) is the radius of the orbit, and \( n \) is the principal quantum number. Also, from Coulomb's law and centripetal force, we have: \[ \frac{m v^2}{r} = \frac{e^2}{4 \pi \epsilon_0 r^2}, \] where \( e \) is the charge of the electron and \( \epsilon_0 \) is the permittivity of free space. By solving these equations, we get the expression for the radius of the \( n \)-th orbit: \[ r = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}. \] For \( n = 1 \), this is the Bohr radius \( a_0 \): \[ a_0 = \frac{6.63 \times 10^{-34} \times 8.854 \times 10^{-12}}{3.14 \times 9.1 \times 10^{-31} \times (1.6 \times 10^{-19})^2} = 5.29 \times 10^{-11} \, \text{m} = 0.53 \, \text{Å}. \]