Answer to Question #103796 in Atomic and Nuclear Physics for sridhar

Question #103796
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

Answer: (2πmq^4)/(€²h^3)
1
Expert's answer
2020-02-26T10:24:52-0500

F=maF=ma


2q24πϵ0r2=mv2rm2v2=2mq24πϵ0r\frac{2q^2}{4\pi \epsilon_0r^2}=m\frac{v^2}{r}\to m^2v^2=\frac{2mq^2}{4\pi \epsilon_0r}


According to Bohr's model of the atom


mvr=nh2πmvr=n\frac{h}{2\pi}. In our case n=1n=1


2mq24πϵ0r=h24π2r2r=ϵ0h22πmq2\frac{2mq^2}{4\pi \epsilon_0r}=\frac{h^2}{4\pi^2r^2}\to r=\frac{\epsilon_0h^2}{2\pi mq^2}


v=h2πmr=2πhmq22πmϵ0h2=q2ϵ0hv=\frac{h}{2\pi mr}=\frac{2\pi hmq^2}{2\pi m\epsilon_0h^2}=\frac{q^2}{\epsilon_0h}


So, we have


v=ωrω=vr=q2ϵ0h2πmq2ϵ0h2=2πmq4ϵ02h3v=\omega r\to \omega=\frac{v}{r}=\frac{q^2}{\epsilon_0h}\cdot \frac{2\pi mq^2}{\epsilon_0h^2}=\frac{2\pi mq^4}{\epsilon^2_0h^3} Answer.









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