A particle of mass m moves in a circular orbit in a central potential field $$U(r) = \frac{1}{2}kr^2$$. If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number n as:

We have a particle of mass $$m$$ moving in a central potential field $$U(r)=\tfrac12 k r^2$$. For a circular orbit of radius $$r$$ the necessary centripetal force must be supplied entirely by the central force obtained from the potential.

First, we write the radial force that comes from the potential. By definition

$$F_r \;=\;-\frac{dU}{dr}.$$

Because $$U(r)=\tfrac12 k r^2,$$ we get

$$F_r \;=\;-\frac{d}{dr}\!\Bigl(\tfrac12 k r^2\Bigr) \;=\; -k r.$$

In a circular orbit the magnitude of this inward force equals the required centripetal force $$\dfrac{m v^2}{r}$$, so

$$\frac{m v^2}{r}\;=\;k r.$$

Multiplying both sides by $$r$$ gives

$$m v^2 \;=\;k r^2,$$

and hence

$$v^2 \;=\;\frac{k}{m}\,r^2 \quad\Longrightarrow\quad v \;=\;\sqrt{\frac{k}{m}}\;r.$$

Now we invoke Bohr’s quantization condition for angular momentum, which states

$$L \;=\;m v r \;=\;n\hbar,$$

where $$n=1,2,3,\dots$$ is the principal quantum number and $$\hbar$$ is the reduced Planck constant.

Substituting the expression for $$v$$ into $$L$$, we obtain

$$L \;=\;m(\sqrt{\tfrac{k}{m}}\,r)\,r \;=\;\sqrt{m k}\;r^2.$$

Setting this equal to $$n\hbar$$ gives

$$\sqrt{m k}\;r^2 \;=\;n\hbar.$$

Solving for $$r^2$$ we find

$$r^2 \;=\;\frac{n\hbar}{\sqrt{m k}}.$$

Taking the square root on both sides,

$$r \;\propto\;\sqrt{n}.$$

Hence the radius of the allowed orbitals increases as the square root of the quantum number $$n$$.

Next, we calculate the total mechanical energy $$E$$ of the particle. The kinetic energy is

$$K \;=\;\tfrac12 m v^2,$$

and the potential energy is

$$U \;=\;\tfrac12 k r^2.$$

Using $$v^2=\dfrac{k}{m}r^2,$$ we get

$$K \;=\;\tfrac12 m\Bigl(\tfrac{k}{m}r^2\Bigr) \;=\;\tfrac12 k r^2.$$

Therefore the total energy becomes

$$E \;=\;K + U \;=\;\tfrac12 k r^2 + \tfrac12 k r^2 \;=\;k r^2.$$

Because we have already shown that $$r^2 \propto n,$$ it follows directly that

$$E \;\propto\;n.$$

Summarizing our results,

$$r_n \propto \sqrt{n}, \qquad E_n \propto n.$$

Comparing with the given options, this corresponds exactly to Option B.

Hence, the correct answer is Option B.