Question

On the basis of Bohr's theory, derive an expression for the radius of the of the $n^{th}$ orbit of an electron of hydrogen atom.

Answer 1

Let e, m and v be respectively the charge, mass and velocity of the electron and r the radius of the orbit. The positive charge on the nucleus is Ze, where Z is the atomic number (in case of hydrogen atom Z = 1). As the centripetal force is provided by the electrostatic force of attraction. We have
$\frac{m{v}^2}{r}=\frac{1}{4\pi {\epsilon }_0}\frac{\left(Ze\right)\times e}{r^2}$
$m{v}^2=\frac{Z{e}^2}{4\pi {\epsilon }_{0}r}$ ....(i)
From the first postulate, the angular momentum of the electron is
$mvr=n\frac{h}{2\pi }$ ....(ii)
where n (= 1, 2, 3, ...) is quantum number. Squaring eq. (ii) and dividing by eq. (i), we
get
$r=n^2\frac{h^2{\epsilon }_0}{\pi mZ{e}^2}$
$Z=1$
Since
$r=n^2\frac{h^2{\epsilon }_0}{\pi m{e}^2}$