Question

If $\omega $ be the angular velocity of electron and $v$ be its speed in ${n^{th}}$ orbit of hydrogen atom. Then,
$\eqalign{
  & A.\omega \propto \dfrac{1}{{{n^2}}}{\text{ and }}v \propto \dfrac{1}{n} \cr
  & B.\omega \propto {n^{\dfrac{1}{2}}}{\text{ and }}v \propto \dfrac{1}{{{n^2}}} \cr
  & C.\omega \propto \dfrac{1}{{{n^3}}}{\text{ and }}v \propto \dfrac{1}{n} \cr
  & D.\omega \propto {n^{ - \dfrac{1}{2}}}{\text{ and }}v \propto \dfrac{1}{n} \cr} $

Answer 1

Hint: Use Bohr’s theory of atom to calculate the radius of ${n^{th}}$ orbit. From that calculate the velocity of the electron in ${n^{th}}$ orbit. From velocity of ${n^{th}}$ orbit calculate the angular speed of the ${n^{th}}$ orbit. And see how the velocity and angular velocity depends upon the value of n.

Formula used:
${\omega _n} = \dfrac{{{v_n}}}{{{r_n}}}$
${F_c} = \dfrac{{{m_e}v_n^2}}{{{r_n}}}$

Complete step-by-step answer:
Bohr’s atomic model:
Postulate I: According to Bohr’s atomic model the electron revolves around the nucleus in a circular orbit and the necessary centripetal force is provided by the electrostatic attraction between the electron and nucleus.
Postulate II: Electron can revolve only in those circular orbits for which the magnitude of angular momentum is integral multiple of $\dfrac{h}{{2\pi }}{\text{ or }}\hbar $. These orbits are known as stationary orbits.
i.e. $\left| {\overrightarrow L } \right| = {m_e}{v_n}{r_n} = \dfrac{{nh}}{{2\pi }}{\text{ where n = 1,2,3,}}...$ and
$\eqalign{
  & {m_e} = {\text{mass of electron}} \cr
  & {v_n} = {\text{velocity of electron in }}{{\text{n}}^{{\text{th}}}}{\text{ orbit}} \cr
  & {r_n} = {\text{radius of }}{{\text{n}}^{{\text{th}}}}{\text{ orbit}} \cr} $
Explanation:
So If the charge on the nucleus is $ + Ze$ and the charge on the electron is $ - e$ and mass ${m_e}$ . According to Bohr’s theory, the electron revolves around the nucleus in a circular orbit and the required centripetal force is provided by the coulombic attraction between nucleus and electron. Thus,
Centripetal force ${F_c} = \dfrac{{{m_e}v_n^2}}{{{r_n}}}$ and
Electrostatic force ,
So

Multiplying ${m_e}r_n^2$ on both sides of above equation then

But ${m_e}{v_n}{r_n} = \dfrac{{nh}}{{2\pi }}$ is the magnitude of angular momentum so

This gives the expression for radius of ${n^{th}}$ orbit.
So ${r_n} \propto {n^2}$
Also
$\eqalign{
  & {m_e}{v_n}{r_n} = \dfrac{{nh}}{{2\pi }} \cr
  & \Rightarrow {v_n} = \dfrac{{nh}}{{2\pi {m_e}{r_n}}} = \dfrac{h}{{2\pi {m_e}}} \times \dfrac{n}{{{r_n}}} \cr
  & \Rightarrow {v_n} \propto \dfrac{n}{{{r_n}}} \cr} $
But ${r_n} \propto {n^2}$
So
 $\eqalign{
  & {v_n} \propto \dfrac{n}{{{r_n}}} \cr
  & \Rightarrow {v_n} \propto \dfrac{n}{{{n^2}}} \cr
  & \Rightarrow {v_n} \propto \dfrac{1}{n} \cr} $
Angular velocity of ${n^{th}}$ orbit is given by ${\omega _n} = \dfrac{{{v_n}}}{{{r_n}}}$
But ${v_n} \propto \dfrac{1}{n}$ and ${r_n} \propto {n^2}$ so
$\eqalign{
  & {\omega _n} = \dfrac{{{v_n}}}{{{r_n}}} \cr
  & \Rightarrow {\omega _n} \propto \dfrac{1}{n} \times \dfrac{1}{{{n^2}}} \cr
  & \Rightarrow {\omega _n} \propto \dfrac{1}{{{n^3}}} \cr} $
So if $\omega $ be the angular velocity of an electron and $v$ be its speed in ${n^{th}}$ orbit of hydrogen atom. Then,
$\omega \propto \dfrac{1}{{{n^3}}}$ and $v \propto \dfrac{1}{n}$

So, the correct answer is “Option C”.

Note: Note that according to Bohr’s model of an atom an orbit has fixed energy. So as long as an electron is revolving in a particular orbit, it will not emit or absorb energy. By absorbing energy it will make a transition to higher orbit and by losing or emitting energy it will jump to lower orbit.