Question

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

• 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, $F_c$
$\frac{k{q}_1{q}_2}{r^2}=mr{\omega }^2\dots (i)$
Given, $q_1=2q,q_2=q$ and $k=\frac{1}{4\pi {\epsilon }_0}$
Putting thesc valucs in Eq. (i), we get
$\frac{1}{4\pi {\epsilon }_0}\frac{q(2q)}{r^2}=mr{\omega }^2\dots (ii)$
According to the Bohr's model,
$mvr=\frac{nh}{2\pi }$
$\Rightarrow m{r}^2\omega =\frac{nh}{2\pi }\dots (iii)$
$\left[\because \omega =\frac{v}{r}\right]$
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
$\frac{2\pi m{q}^4}{{\epsilon }_0^2{h}^3}$.