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(C) The force $F$ is given by the negative gradient of the potential energy: $F = -\frac{dU}{dr} = -\frac{d}{dr}(U_{0}r^{4}) = -4U_{0}r^{3}$.For a cir

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$3$

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(C) The force $F$ is given by the negative gradient of the potential energy: $F = -\frac{dU}{dr} = -\frac{d}{dr}(U_{0}r^{4}) = -4U_{0}r^{3}$.For a circular orbit,the centripetal force is provided by this central force: $\frac{mv^{2}}{r} = 4U_{0}r^{3}$,which implies $v^{2} \propto r^{4}$,so $v \propto r^{2}$.According to Bohr's quantization condition,the angular momentum is quantized: $mvr = \frac{nh}{2\pi}$.Substituting $v \propto r^{2}$ into the quantization condition: $m(r^{2})r \propto n$,which simplifies to $r^{3} \propto n$.Therefore,$r \propto n^{1/3}$.Comparing this with $r_{n} \propto n^{1/\alpha}$,we get $\alpha = 3$.

$A$ particle of mass $m$ moves in a circular orbit in a central potential field $U(r) = U_{0}r^{4}$. If Bohr's quantization conditions are applied,the radii of possible orbitals $r_{n}$ vary with $n^{1/\alpha}$,where $\alpha$ is ....... .

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