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(C) The centripetal force is provided by the electrostatic force: $\frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} \implies mv^2 = \frac{

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$\frac{\mu_0 e^7 \pi m^2}{8 \varepsilon_0^3 h^5}$

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(C) The centripetal force is provided by the electrostatic force: $\frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} \implies mv^2 = \frac{e^2}{4\pi\varepsilon_0 r}$.From Bohr's quantization condition: $mvr = \frac{nh}{2\pi}$. For ground state $(n=1)$,$mvr = \frac{h}{2\pi} \implies v = \frac{h}{2\pi mr}$.Substituting $v$ into the force equation: $m(\frac{h}{2\pi mr})^2 = \frac{e^2}{4\pi\varepsilon_0 r} \implies \frac{h^2}{4\pi^2 mr^2} = \frac{e^2}{4\pi\varepsilon_0 r} \implies r = \frac{\varepsilon_0 h^2}{\pi m e^2}$.The current $I$ due to the orbiting electron is $I = \frac{e}{T} = \frac{ev}{2\pi r}$.The magnetic field at the center is $B = \frac{\mu_0 I}{2r} = \frac{\mu_0 ev}{4\pi r^2}$.Substituting $v = \frac{h}{2\pi mr}$ and $r = \frac{\varepsilon_0 h^2}{\pi m e^2}$:$B = \frac{\mu_0 e}{4\pi r^2} \cdot \frac{h}{2\pi mr} = \frac{\mu_0 e h}{8\pi^2 m r^3}$.Substituting $r^3 = (\frac{\varepsilon_0 h^2}{\pi m e^2})^3 = \frac{\varepsilon_0^3 h^6}{\pi^3 m^3 e^6}$:$B = \frac{\mu_0 e h}{8\pi^2 m} \cdot \frac{\pi^3 m^3 e^6}{\varepsilon_0^3 h^6} = \frac{\mu_0 e^7 \pi m^2}{8 \varepsilon_0^3 h^5}$.

Similar Questions

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