In a system, a particle \(A\) of mass \(m\) and charge \(-2 q\) is moving in the nearest orbit around a very…
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
\(\frac{2 \pi m^2 q^2}{\varepsilon_0 h^4}\)
\(\frac{3 \pi m^3 q^2}{\varepsilon_0^3 h^2}\)
\(\frac{2 \pi m q^4}{\varepsilon_0^2 h^3}\)
\(\frac{5 \pi m^2 q^3}{\varepsilon_0^3 h^2}\)
Solution
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, \(F_c\)
\(\frac{k q_1 q_2}{r^2}=m r \omega^2\) ...(i)
Given, \(\quad q_1=2 q, q_2=q\) and \(k=\frac{1}{4 \pi \varepsilon_0}\)
Putting these values in Eq. (i), we get
\(\frac{1}{4 \pi \varepsilon_0} \frac{q(2 q)}{r^2}=m r \omega^2\) ...(ii)
According to the Bohr's model,
\(m v r=\frac{n h}{2 \pi} \Rightarrow m r^2 \omega=\frac{n h}{2 \pi} \quad \ldots \text { (iii) }\left[\because \omega=\frac{v}{r}\right]\)
(iii) \(\left[\because \omega=\frac{v}{r}\right]\)
From Eqs. (ii) and (iii), we get
\(\omega=\frac{2 \pi m q^4}{\varepsilon_0^2 h^3} \quad[\because \mathrm{n}=1]\)
Hence, the orbital angular velocity of the particle \(A\) is \(\frac{2 \pi m q^4}{\varepsilon_0^2 h^3}\).