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JEE Main 201912 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Morning Shift
Subject
Physics
Chapter
Atomic Physics

JEE Main 2019 · 12 January · Morning Shift · Physics · Atomic Physics

A particle of mass m moves in a circular orbit in a central...

A particle of mass mm moves in a circular orbit in a central potential field U(r)=12kr2.U\left(r\right) = \dfrac{1}{2}k{r}^{2} . If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number nn as:

Options

  1. Arn∝n2,En∝1n2{r}_{n} ∝ {n}^{2}, {E}_{n} ∝ \dfrac{1}{{n}^{2}}
  2. Brn∝n,En∝n{r}_{n} ∝ \sqrt{n}, {E}_{n} ∝ n
  3. Crn∝n,En∝n{r}_{n} ∝ n, {E}_{n} ∝ n
  4. Drn∝n,En∝1n{r}_{n} ∝ \sqrt{n}, {E}_{n} ∝ \dfrac{1}{n}

Correct answer

B. rn∝n,En∝n{r}_{n} ∝ \sqrt{n}, {E}_{n} ∝ n

Step-by-step solution

Fr=−dUdr=−kr{F}_{r} = \dfrac{- dU}{dr} = - kr

For circular motion

∣Fr∣=kr=mv2r⇒kr2=mv2| {F}_{r} | = kr = \dfrac{m{v}^{2}}{r} ⇒ k{r}^{2} = m{v}^{2} ........(i)

Bohr's quantization ⇒mvr=nh2π⇒ mvr = \dfrac{nh}{2\pi} ....(ii)

From (i) and (ii)

m2v2m=kr2\dfrac{{m}^{2}{v}^{2}}{m} = k{r}^{2}

⇒1m(nh2\pir)2=kr2⇒n2h24π2mk=r4⇒ \dfrac{1}{m}{\left(\dfrac{nh}{2\pir}\right)}^{2} = k{r}^{2} ⇒ \dfrac{{n}^{2}{h}^{2}}{4{\pi}^{2}mk} = {r}^{4}

⇒r=(h24π2mk)1/4n1/2⇒ r = {\left(\dfrac{{h}^{2}}{4{\pi}^{2}mk}\right)}^{1 / 4}{n}^{1 / 2}

r∝nr ∝ \sqrt{n}

From equation (i) U∝nU ∝ \sqrt{n}

KE=12mv2;KE = \dfrac{1}{2}m{v}^{2};PE=12kr2PE = \dfrac{1}{2}k{r}^{2}

E=K+U=12mv2+12kr2=kr2∝nE = K + U = \dfrac{1}{2}m{v}^{2} + \dfrac{1}{2}k{r}^{2} = k{r}^{2} ∝ n
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