Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of…

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 $n^{\text {th }}$ orbital will therefore be proportional to :
  1. $n^2$
  2. $n$
  3. $n^{1 / 2}$
  4. $n^{1 / 4}$

Solution

According to the question, $ \begin{aligned} & 2 \pi \mathrm{r}=\mathrm{n} \lambda=\frac{\mathrm{nh}}{\mathrm{p}}=\frac{\mathrm{nh}}{\mathrm{mv}} \\ & \text { or } \mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \text { or } \mathrm{mv}=\frac{\mathrm{nh}}{2 \pi \mathrm{r}} \\ & \mathrm{F}=\mathrm{qv}_{\mathrm{B}}=\frac{\mathrm{mv}^2}{\mathrm{r}} \\ & \text { or, } \mathrm{q}_{\mathrm{B}}=\frac{\mathrm{mv}}{\mathrm{r}}=\frac{\mathrm{nh}}{2 \pi \mathrm{r} \cdot \mathrm{r}} \\ & \text { or, } \mathrm{r}^2=\frac{\mathrm{nh}}{2 \pi \mathrm{qB}} \\ & \text { or, } \mathrm{r}=\sqrt{\frac{\mathrm{nh}}{2 \pi \mathrm{qB}}} \\ & \text { i.e., } \mathrm{r} \propto \mathrm{n}^{1 / 2} \end{aligned} $

Asked in: JEE Main 2013 (22 Apr Online)

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