Question

State Bohr's postulate to explain stable orbits in a hydrogen atom. Prove that the speed with which the electron revolves in n^th orbit is proportional to ${\displaystyle \left(\frac{1}{\text{n}}\right)}$.

Answer 1

The stationary orbits are those orbits for which the angular momentum of the electron is an integral multiple of ${\displaystyle \frac{\text{h}}{2\pi }}$, where h is Planck's constant.

Electron in an atom revolves because of the balance of Coulomb force of attraction between the protons and electrons and the centripetal force.

In the n^th orbit,

${\displaystyle \frac{{\text{mv}}_{\text{n}}^2}{{\text{r}}_{\text{n}}}=\frac{1}{4\pi {\epsilon }_0}\frac{{\text{e}}^2}{{\text{r}}_{\text{n}}^2}}$

∴ v_n = ${\displaystyle \frac{\text{e}}{\sqrt{4\pi {\epsilon }_0{\text{mr}}_{\text{n}}}}}$  ......(i)

Again, r_n = ${\displaystyle \frac{{\epsilon }_0{\text{h}}^2{\text{n}}^2}{{\text{e}}^2\pi \text{m}}}$  .....(ii)

Putting in equation (i)

v_n = ${\displaystyle \frac{\text{e}}{\sqrt{4\pi {\epsilon }_0\text{m}\frac{{\epsilon }_0{\text{h}}^2{\text{n}}^2}{{\text{e}}^2\pi \text{m}}}}}$

= ${\displaystyle \frac{{\text{e}}^2}{2{\epsilon }_0\text{hn}}}$

∴ ${\displaystyle {\text{v}}_{\text{n}}\propto \frac{1}{\text{n}}}$