JEE Advanced 2020 · Paper 1 · Physics Q7

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PhysicsMultiple correct+4 / -2Question 7 of 54

A particle of mass $m$ moves in circular orbits with potential energy $V(r) = Fr$ , where $F$ is a positive constant and $r$ 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 $R$ and its speed and energy are denoted by $v$ and $E$ , respectively, then for the $n^{\text{th}}$ orbit (here $h$ is the Planck's constant)

R \propto n^{1/3}

and

v \propto n^{2/3}

R \propto n^{2/3}

and

v \propto n^{1/3}

E = \dfrac32\left(\dfrac{n^2h^2F^2}{4\pi^2 m}\right)^{1/3}

E = 2\left(\dfrac{n^2h^2F^2}{4\pi^2 m}\right)^{1/3}

Answer and worked solution

Official answer: B, C

$V=Fr$ gives a constant attractive force $F$ .

Orbit and quantisation.
$\frac{mv^2}R=F,\qquad mvR=\frac{nh}{2\pi}.$
Eliminating $v$ :
$\frac{n^2h^2}{4\pi^2mR^3}=F\implies R=\left(\frac{n^2h^2}{4\pi^2mF}\right)^{1/3}\propto n^{2/3},\qquad v=\frac{nh}{2\pi mR}\propto n^{1/3}.$
So (B) is true and (A) is false.

Energy.
$E=\tfrac12mv^2+FR=\tfrac12FR+FR=\tfrac32FR=\frac32\left(\frac{n^2h^2F^2}{4\pi^2m}\right)^{1/3}.$
So (C) is true and (D) is false.

(B), (C)

Topics:Bohr model and hydrogen spectrum Circular motion dynamics

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JEE Advanced 2020 · Paper 1, Physics question 7