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(C) Given the force $F = \frac{k}{r^4}$.For circular motion,the centripetal force is provided by this force:$\frac{mv^2}{r} = \frac{k}{r^4} \implies v

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$n^6$

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(C) Given the force $F = \frac{k}{r^4}$.For circular motion,the centripetal force is provided by this force:$\frac{mv^2}{r} = \frac{k}{r^4} \implies v^2 = \frac{k}{mr^3}$.According to Bohr's quantization condition:$mvr = \frac{nh}{2\pi} \implies v = \frac{nh}{2\pi mr}$.Squaring both sides:$v^2 = \frac{n^2h^2}{4\pi^2m^2r^2}$.Equating the two expressions for $v^2$:$\frac{k}{mr^3} = \frac{n^2h^2}{4\pi^2m^2r^2} \implies r = \frac{4\pi^2m k}{n^2h^2} \propto \frac{1}{n^2}$.The total energy $E$ is the sum of kinetic and potential energy. Since the force $F \propto r^{-4}$,the potential energy $U = -\int F dr = -\int \frac{k}{r^4} dr = -\frac{k}{3r^3}$.The kinetic energy $K = \frac{1}{2}mv^2 = \frac{1}{2}m(\frac{k}{mr^3}) = \frac{k}{2r^3}$.Total energy $E = K + U = \frac{k}{2r^3} - \frac{k}{3r^3} = \frac{k}{6r^3}$.Since $r \propto n^{-2}$,then $r^3 \propto n^{-6}$.Therefore,$E \propto \frac{1}{n^{-6}} \implies E \propto n^6$.

If in Bohr's atomic model,it is assumed that the force between the electron and the proton varies inversely as $r^4$,the energy of the system will be proportional to

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