A particle of mass $m$ moves in a circular orbit in a central potential field
Question:

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:

  1. (1) $\mathrm{r}_{\mathrm{n}} \propto \sqrt{\mathrm{n}}, \mathrm{E}_{\mathrm{n}} \propto \mathrm{n}$

  2. (2) $r_{n} \propto \sqrt{n}, E_{n} \propto \frac{1}{n}$

  3. (3) $\mathrm{r}_{\mathrm{n}} \propto \mathrm{n}, \mathrm{E}_{\mathrm{n}} \propto \mathrm{n}$

  4. (4) $\mathrm{r}_{\mathrm{n}} \propto \mathrm{n}^{2}, \mathrm{E}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}^{2}}$


Correct Option: 1

Solution:

(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 $\left.\mathrm{and} \mathrm{r}^{2} \propto \mathrm{n}\right]$

Administrator

Leave a comment

Please enter comment.
Please enter your name.