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

According to Bohr’s second postulate, electron angular momentum in the \( n \)-th orbit is quantized: \[m v r = \frac{nh}{2 \pi},\] where \( m \) denotes the electron's mass, \( v \) its velocity, \( r \) the orbital radius, and \( n \) the principal quantum number. Furthermore, applying Coulomb's law and the centripetal force equation yields: \[\frac{m v^2}{r} = \frac{e^2}{4 \pi \epsilon_0 r^2},\] where \( e \) is the electron's charge and \( \epsilon_0 \) is the permittivity of free space. Solving these equations provides 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 radius is known as the Bohr radius \( a_0 \): \[a_0 = \frac{6.63 \times 10^{-34} \times 8.854 \times 10^{-12}}{3.14 \times 9.1 \times 10^{-31} \times (1.6 \times 10^{-19})^2} = 5.29 \times 10^{-11} \, \text{m} = 0.53 \, \text{Å}.\]