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Step-by-Step Solution
Step 1: Write down the potential energy and the force
The given central potential is
$U(r) = \frac{1}{2} k r^2$.
The force $F$ acting on the particle is the negative gradient of the potential:
$F = - \frac{\partial U}{\partial r} = - \frac{\partial}{\partial r} \left( \frac{1}{2} k r^2 \right) = - k r.$
Step 2: Apply the condition for circular motion
For a particle of mass $m$ moving on a circular path of radius $r$, the centripetal force is
$\frac{m v^2}{r}$.
Equating this to the (magnitude of) the force from Step 1 gives:
$\frac{m v^2}{r} = k r.$
Rearranging, we get:
$v^2 = k \, r^2 \quad \Rightarrow \quad v \propto r.
Step 3: Use Bohr’s quantization condition
Bohr’s quantization condition for orbital angular momentum states that:
$m \, v \, r = \frac{n h}{2\pi}, \quad n = 1, 2, 3, \dots
Since from Step 2 we have $v \propto r$, we substitute $v$ in the above condition:
$m \, (\text{constant} \cdot r) \, r = \frac{n h}{2\pi}.
It follows that
$r_n \propto \sqrt{n}.$
Step 4: Determine the energy levels
The total energy $E_n$ for a particle in this potential is the sum of its kinetic and potential energies.
For a circular orbit under central force, one finds that the total energy is related to the potential energy.
Specifically, for a harmonic-like potential $U(r) = \frac{1}{2} k r^2$, one can show:
$E = \frac{1}{2} m v^2 + U(r).
$
Using the relation for a harmonic oscillator orbit, it turns out that (often shown more rigorously in advanced treatments):
$E_n \propto k \, r_n^2.
Since $r_n^2 \propto n$, we get:
$E_n \propto n.
Conclusion
Hence, according to Bohr’s quantization conditions, the radius of the $n$-th orbit scales as
$r_n \propto \sqrt{n}$,
and the energy of the $n$-th state scales as
$E_n \propto n.$
