Search any question & find its solution
Question:
Answered & Verified by Expert
A particle of mass $\mathrm{m}$ moves around the origin in a potential $\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{r}^{2}$, where $\mathrm{r}$ is the distance from the origin. Applying the Bohr model in this case, the radius of the particle in its $\mathrm{n}^{\text {th }}$ orbit in terms of $\mathrm{a}=\sqrt{\mathrm{h} /(2 \pi \mathrm{m} \omega)}$ is
Options:
Solution:
1567 Upvotes
Verified Answer
The correct answer is:
$a \sqrt{n}$
$\mathrm{mvr}=\mathrm{n} \frac{\mathrm{h}}{2 \pi}$
$\mathrm{v}=\mathrm{r} \omega$
$\mathrm{mr}^{2} \omega=\mathrm{n} \frac{\mathrm{h}}{2 \pi}$
$\mathrm{r}=\sqrt{\frac{\mathrm{nh}^{2}}{2 \pi \mathrm{m} \omega}}$
$\mathrm{r}=\sqrt{\mathrm{n}} \sqrt{\frac{\mathrm{h}^{2}}{2 \pi \mathrm{m} \omega}}$
$\mathrm{r}=\sqrt{\mathrm{n}} \cdot \mathrm{a}$
$\mathrm{v}=\mathrm{r} \omega$
$\mathrm{mr}^{2} \omega=\mathrm{n} \frac{\mathrm{h}}{2 \pi}$
$\mathrm{r}=\sqrt{\frac{\mathrm{nh}^{2}}{2 \pi \mathrm{m} \omega}}$
$\mathrm{r}=\sqrt{\mathrm{n}} \sqrt{\frac{\mathrm{h}^{2}}{2 \pi \mathrm{m} \omega}}$
$\mathrm{r}=\sqrt{\mathrm{n}} \cdot \mathrm{a}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.