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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

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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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