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(C) For a particle $A$ moving in a circular orbit around a heavy particle $B$, the electrostatic force provides the necessary centripetal force.$F_e =

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$\frac{2 \pi m q^4}{\varepsilon_0^2 h^3}$

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(C) For a particle $A$ moving in a circular orbit around a heavy particle $B$, the electrostatic force provides the necessary centripetal force.$F_e = F_c$$\frac{1}{4 \pi \varepsilon_0} \frac{|q_A| |q_B|}{r^2} = m r \omega^2$Given $q_A = 2q$ and $q_B = q$, we have:$\frac{1}{4 \pi \varepsilon_0} \frac{2q^2}{r^2} = m r \omega^2 \implies r^3 = \frac{2q^2}{4 \pi \varepsilon_0 m \omega^2} = \frac{q^2}{2 \pi \varepsilon_0 m \omega^2} \quad \dots(i)$According to Bohr's quantization condition for the nearest orbit $(n=1)$:$m v r = \frac{h}{2 \pi} \implies m r^2 \omega = \frac{h}{2 \pi} \implies r^2 = \frac{h}{2 \pi m \omega} \quad \dots(ii)$Squaring (ii) and dividing by $(i)$ to eliminate $r$:$r^6 = \frac{h^2}{4 \pi^2 m^2 \omega^2}$ and $r^6 = \left(\frac{q^2}{2 \pi \varepsilon_0 m \omega^2}\right)^3 = \frac{q^6}{8 \pi^3 \varepsilon_0^3 m^3 \omega^6}$Equating the two expressions for $r^6$:$\frac{h^2}{4 \pi^2 m^2 \omega^2} = \frac{q^6}{8 \pi^3 \varepsilon_0^3 m^3 \omega^6}$$\omega^4 = \frac{q^6}{8 \pi^3 \varepsilon_0^3 m^3} \cdot \frac{4 \pi^2 m^2}{h^2} = \frac{q^6}{2 \pi \varepsilon_0^3 m h^2}$Solving for $\omega$, we obtain $\omega = \frac{2 \pi m q^4}{\varepsilon_0^2 h^3}$.

In a system, a particle $A$ of mass $m$ and charge $-2q$ 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

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