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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
  • JEE Mains 2013
  • Hard
  • MCQ

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the \(n^{th}\) orbital will therefore be proportional to

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Answer & Solution

Correct Answer

(C) \(n^{1/2}\)

Step-by-step Solution

Detailed explanation

According to the question,

\(2 \pi r=n \lambda=\frac{n h}{p}=\frac{n h}{m v}\)

or \(\operatorname{mvr}=\frac{\mathrm{nh}}{2 \pi}\) or \(\mathrm{mv}=\frac{\mathrm{nh}}{2 \pi \mathrm{r}}\)

\(\mathrm{F}=\mathrm{qv}_{\mathrm{B}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\)

or, \(\quad \mathrm{q}_{B}=\frac{\mathrm{mv}}{\mathrm{r}}=\frac{\mathrm{nh}}{2 \pi \mathrm{r} \cdot \mathrm{r}}\)

or, \(\quad \mathrm{r}^{2}=\frac{\mathrm{nh}}{2 \pi \mathrm{qB}}\)

or, \(\quad r=\sqrt{\frac{n h}{2 \pi q B}}\)

i.e., \(r\propto n^{1/2}\)

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