A particle of mass $m$ moves in a circular orbit in a central potential field $\mathrm{U}(\mathrm{r})=\frac{1}{2} \mathrm{kr}^{2}$. If Bohr's quantization conditions are applied, radii of possible orbitls and energy levels vary with quantum number $n$ as:
Correct Option: 1
(1) Let force of attraction towards the centre is $\mathrm{F}$, then
$\mathrm{F}=\frac{\mathrm{dU}}{\mathrm{dr}}=\mathrm{kr}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}=$ centripetal force
$\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$ [Bohr's Quantization rule]
so, $\frac{\mathrm{m}^{2} \mathrm{v}^{2}}{\mathrm{~m}}=\mathrm{kr}^{2} \Rightarrow\left(\frac{\mathrm{nh}}{2 \pi \mathrm{r}}\right)^{2} \frac{1}{\mathrm{~m}}=\mathrm{kr}^{2}$
$\Rightarrow \mathrm{r}^{2} \propto \mathrm{n}$
$\Rightarrow \quad r \propto \sqrt{n}$
Also $\mathrm{E}=\mathrm{K} . \mathrm{E}+$ P.E
$E=\frac{1}{2} \mathrm{kr}^{2}+\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{kr}^{2}+\frac{1}{2} \times\left(\mathrm{kr}^{2}\right)$
$=\mathrm{kr}^{2} \propto \mathrm{n}\left[\mathrm{as} \mathrm{k}\right.$ is constant $\left.\mathrm{and} \mathrm{r}^{2} \propto \mathrm{n}\right]$
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