Question

Show that angular speed of electron in ${n^{th}}$ Bohr’s orbit is equal to $\dfrac{{\pi m{e^4}}}{{2{\varepsilon _0}^2{h^3}{n^3}}}$ or frequency revolution, $f = \dfrac{{\pi m{e^4}}}{{2{\varepsilon _0}^2{h^3}{n^3}}}$ .

Answer 1

Electrons are the negative subatomic particles of an atom. They are held in a circular orbit by electrostatic attraction. The speed of an electron due to its rotation is known as angular speed. It is donated by ${\omega _n}$. Electrons when rotate do not radiate energy and do not fall into the nucleus of an atom. The energy in the ${n^{th}}$ orbit acts as the binding energies of a highly excited atom with one electron in a large circular orbit around the rest of the atom.

Complete step by step answer:
In ${n^{th}}$orbit the angular speed gives relation with linear speed as ${V_n} = {r_n}{\omega _n}$,
Where ${\omega _n}$= angular velocity
${r_n}$ = the radius of atom
${V_n}$ = the linear velocity of an electron in ${n^{th}}$ orbit.
Therefore, the angular speed of electron is given by,
${\omega _n} = \dfrac{{{V_n}}}{{{r_n}}}$
For ${n^{th}}$ Bohr atom ${V_n}$ and ${r_n}$is,
${V_n} = \dfrac{{2\pi kz{e^2}}}{{nh}}$ and ${r_n} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}kz{e^2}m}}$
By substituting the values, we have,
${\omega _n} = \dfrac{{\dfrac{{2\pi kz{e^2}}}{{nh}}}}{{\dfrac{{{n^2}{h^2}}}{{4{\pi ^2}kz{e^2}m}}}}$
$\Rightarrow{\omega _n} = \dfrac{{2\pi kz{e^2}}}{{nh}} \times \dfrac{{4{\pi ^2}kz{e^2}m}}{{{n^2}{h^2}}}$
$\Rightarrow{\omega _n} = \dfrac{{8{\pi ^3}{k^2}{z^2}{e^4}m}}{{{n^3}{h^3}}}$
Where, $k = \dfrac{1}{{4\pi {\varepsilon _0}}}$ and $z = 1$
Now, substituting the values, we have,
${\omega _n} = \dfrac{{8{\pi ^3}m{e^4}{{\left( 1 \right)}^2}}}{{{n^3}{h^3}}} \times \dfrac{1}{{{{\left( 4 \right)}^2}{\pi ^2}{\varepsilon _0}^2}}$
$\Rightarrow{\omega _n} = \dfrac{{8{\pi ^3}m{e^4}}}{{{n^3}{h^3}}} \times \dfrac{1}{{16{\pi ^2}{\varepsilon _0}^2}}$
$\therefore{\omega _n} = \dfrac{{\pi m{e^4}}}{{2{\varepsilon _0}^2{h^3}{n^3}}}$

Hence, proved that angular speed of electron in ${n^{th}}$ Bohr’s orbit is equal to $\dfrac{{\pi m{e^4}}}{{2{\varepsilon _0}^2{h^3}{n^3}}}$.

Note: Electrons in atoms orbit the nucleus. The electrons can be only stably without radiating, in certain orbits at a certain discrete set of distances from the nucleus. These orbits are associated with definite energies and also called energy shells. In these orbits, an electron’s acceleration does not result in radiation and energy loss as required by classical electromagnetic theory. Electrons can only gain or lose energy by jumping from one allowed orbit to another but in the ${n^{th}}$ orbit it acts as the binding energies of a highly excited atom with one electron in a large circular orbit around the rest of the atom.