F=maF=maF=ma
2q24πϵ0r2=mv2r→m2v2=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}4πϵ0r22q2=mrv2→m2v2=4πϵ0r2mq2
According to Bohr's model of the atom
mvr=nh2πmvr=n\frac{h}{2\pi}mvr=n2πh. In our case n=1n=1n=1
2mq24πϵ0r=h24π2r2→r=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}4πϵ0r2mq2=4π2r2h2→r=2πmq2ϵ0h2
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}v=2πmrh=2πmϵ0h22πhmq2=ϵ0hq2
So, we have
v=ωr→ω=vr=q2ϵ0h⋅2πmq2ϵ0h2=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}v=ωr→ω=rv=ϵ0hq2⋅ϵ0h22πmq2=ϵ02h32πmq4 Answer.
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