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
Pick an option to check your answer.
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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