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JEE Main 201322 AprilOnline
Exam
JEE Main
Year
2013
Date
22 April
Mode
Online
Subject
Physics
Chapter
Atomic Physics

JEE Main 2013 · 22 April · Online · Physics · Atomic Physics

Orbits of a particle moving in a circle are such that the perimeter...

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 nth n^{\text{th }} orbital will therefore be proportional to :

Options

  1. An2n^2
  2. Bnn
  3. Cn1/2n^{1 / 2}
  4. Dn1/4n^{1 / 4}

Correct answer

C. n1/2n^{1 / 2}

Step-by-step solution

According to the question, 2πr=nλ=nhp=nhmv or mvr=nh2π or mv=nh2πrF=qvB=mv2r or, qB=mvr=nh2πr⋅r or, r2=nh2πqB or, r=nh2πqB i.e., r∝\mathrmn1/2\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 \mathrmn^{1 / 2} \end{aligned}
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