Atoms & Nuclei

JEE Main level, single correct, medium

85 of 239

Question 85 of 239

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 :

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Answer: (C) n1/2n^{1 / 2}

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∝n1/2\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 \mathrm{n}^{1 / 2} \end{aligned}

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