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JEE Advanced 2020Paper 1
Exam
JEE Advanced
Year
2020
Paper
Paper 1
Subject
Physics
Chapter
Atomic Physics

JEE Advanced 2020 · Paper 1 · Physics · Atomic Physics

A particle of mass m moves in circular orbits with potential energy...

A particle of mass mm moves in circular orbits with potential energy U(r)=FrU\left(r\right) = Fr, where FF is a positive constant and rr is its distance from the origin. Its energies are calculated using the Bohr model. If the radius of the particle's orbit is denoted by RR and its speed and energy are denoted by vv and EE respectively, then for the nth{n}^{th} orbit (here hh is the Planck's constant)

Options

  1. AR∝n1/3R ∝ {n}^{1 / 3} and v∝n2/3v ∝ {n}^{2 / 3}
  2. BR∝n2/3R ∝ {n}^{2 / 3} and v∝n1/3v ∝ {n}^{1 / 3}
  3. CE=32(n2h2F24π2m)1/3E = \dfrac{3}{2}{\left(\dfrac{{n}^{2}{h}^{2}{F}^{2}}{4{\pi}^{2}m}\right)}^{1 / 3}
  4. DE=2(n2h2F24π2m)1/3E = 2{\left(\dfrac{{n}^{2}{h}^{2}{F}^{2}}{4{\pi}^{2}m}\right)}^{1 / 3}

Correct answer

B. R∝n2/3R ∝ {n}^{2 / 3} and v∝n1/3v ∝ {n}^{1 / 3}

Step-by-step solution

U=F⋅rU = F \cdot r

Fc=−dUdr=−F{F}_{c} = - \dfrac{dU}{dr} = - F

mv2r=F\dfrac{m{v}^{2}}{r} = F

mur=nh2πmur = \dfrac{nh}{2\pi}

⇒(mur2)mu(4r)=n2h24π2⋅F⇒ \dfrac{\left(mu{r}^{2}\right)}{mu\left(4r\right)} = \dfrac{{n}^{2}{h}^{2}}{4{\pi}^{2} \cdot F}

⇒mr3=n2h24π2F⇒ m{r}^{3} = \dfrac{{n}^{2}{h}^{2}}{4{\pi}^{2}F}

⇒r=(n2h24π2mF)1/3⇒ r = {\left(\dfrac{{n}^{2}{h}^{2}}{4{\pi}^{2}mF}\right)}^{1 / 3}

⇒r∝n2/3⇒ r ∝ {n}^{2 / 3}

E=U+K=F⋅r+1/2mv2E = U + K = F \cdot r + 1 / 2m{v}^{2}

=F⋅r+1/2F⋅r= F \cdot r + 1 / 2F \cdot r

E=3/2F⋅rE = 3 / 2F \cdot r

Put value of rr,

E=32Fr(n2h24π2mF)1/3E = \dfrac{3}{2}Fr{\left(\dfrac{{n}^{2}{h}^{2}}{4{\pi}^{2}mF}\right)}^{1 / 3}

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