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JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2019
- Hard
- MCQ
A particle of mass \(m\) moves in a circular orbit in a central potential field \(U\left( r \right) = \frac{1}{2}k{r^2}\). If Bohr’s quantization conditions are applied, radii of possible orbits and energy levels vary with quantum number \(n\) as
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \({r_n} \propto \sqrt n ,{E_n} \propto n\)
Step-by-step Solution
Detailed explanation
\(\mathrm{U}=\frac{1}{2} \mathrm{kr}^{2}\)
Force, \(F=-\frac{d U}{d r}=-k r\)
For circular motion \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{kr}\) .... \((i)\)
And \({ mvr }=\frac{n h}{2 \pi}\) .... \((ii)\)
\({ \Rightarrow {r^2} = \frac{{{\text{nh}}}}{{2\pi \sqrt {{\text{km}}} }}}\)
\({ \Rightarrow {\text{r}} \propto \sqrt {\text{n}} }\)
Total energy, \(E=k+U\)
\( = \frac{1}{2}m{v^2} + \frac{1}{2}k{r^2}\)
\( = \frac{1}{2}k{r^2} + \frac{1}{2}k{r^2}\) [From equation \((i)\)]
\(E = k{r^2}\)
\( \Rightarrow \,\,E\, \propto \,\,n\)
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