Question

In a system, a particle $A$ of mass $m$ and charge $-2 q$ is moving in the nearest orbit around a very heavy particle $B$ having charge $+q$. Assuming Bohr's model of the atom to be applicable to this system, the orbital angular velocity of the particle $A$ is

• A. $\frac{2\pi m^{2} q^{2}}{\in_{0 }h^{4} }$
• B. $\frac{3 \pi m^{3} q^{2}}{\in_{0}^3 h^{2} }$
• C. $\frac{2\pi m q^{4}}{\in_{0}^2 h^{3} }$
• D. $\frac{5 \pi m^{2} q^{3}}{\in_{0}^3 h^{2} }$

Answer 1

Answer: (C)

In a system, a particle $A$ moving around a particle $B$ in a circular path then applied the centripetal force directed and center of the circular path.
So, the electric force, $F_{e}=$ centripetal force, $F_{c}$
$\frac{k q_{1} q_{2}}{r^{2}}=m r \omega^{2}\,\,\,\,\,\,\,\,\,\dots(i)$
Given, $ q_{1}=2 q, q_{2}=q$ and $k=\frac{1}{4 \pi \varepsilon_{0}}$
Putting thesc valucs in Eq. (i), we get
$\frac{1}{4 \pi \varepsilon_{0}} \frac{q(2 q)}{r^{2}}=m r \omega^{2}\,\,\,\,\,\,\,\dots(ii)$
According to the Bohr's model,
$m v r=\frac{n h}{2 \pi} $
$\Rightarrow m r^{2} \omega=\frac{n h}{2 \pi} \,\,\,\,\,\,\,\ldots( iii )$
$\left[\because \omega=\frac{v}{r}\right]$
From Eqs. (ii) and (iii), we get
$\omega=\frac{2 \pi m q^{4}}{c_{0}^{2} h^{3}} $
$[\because n =1]$
Hence, the orbital angular velocity of the particle $A$ is
$\frac{2 \pi \,m q^{4}}{\varepsilon_{0}^{2} \,h^{3}}$.