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Question: Answered & Verified by Expert
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
PhysicsAtomic PhysicsAP EAMCETAP EAMCET 2019 (20 Apr Shift 1)
Options:
  • A \(\frac{2 \pi m^2 q^2}{\varepsilon_0 h^4}\)
  • B \(\frac{3 \pi m^3 q^2}{\varepsilon_0^3 h^2}\)
  • C \(\frac{2 \pi m q^4}{\varepsilon_0^2 h^3}\)
  • D \(\frac{5 \pi m^2 q^3}{\varepsilon_0^3 h^2}\)
Solution:
2407 Upvotes Verified Answer
The correct answer is: \(\frac{2 \pi m q^4}{\varepsilon_0^2 h^3}\)
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 \(\frac{2 \pi m q^4}{\varepsilon_0^2 h^3}\).

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