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Question

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

Vedclass · Accepted Answer

$r_n = a_0 n^2 - \beta$

Vedclass · Answer

(C) The centripetal force is provided by the electrostatic force: $\frac{mv^2}{r} = \frac{e^2}{4\pi \varepsilon_0} \left( \frac{1}{r^2} + \frac{\beta}{r^3} \right)$.From Bohr's quantization condition,$mvr = \frac{nh}{2\pi}$,so $v = \frac{nh}{2\pi mr}$.Substituting $v$ into the force equation: $\frac{m}{r} \left( \frac{nh}{2\pi mr} \right)^2 = \frac{e^2}{4\pi \varepsilon_0} \left( \frac{r + \beta}{r^3} \right)$.Simplifying: $\frac{n^2 h^2}{4\pi^2 m r^3} = \frac{e^2}{4\pi \varepsilon_0} \left( \frac{r + \beta}{r^3} \right)$.Rearranging terms: $\frac{n^2 h^2 \varepsilon_0}{\pi m e^2} = r + \beta$.Given $a_0 = \frac{\varepsilon_0 h^2}{m \pi e^2}$,we get $a_0 n^2 = r + \beta$.Therefore,$r_n = a_0 n^2 - \beta$.

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