JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Modern Physics

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

V (r) = F r

, 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^{th}

orbit (here h is the Planck's constant)-

(A) $R \propto n^{1/3}$ and $v \propto n^{2/3}$
(B) $R \propto n^{2/3}$ and $v \propto n^{1/3}$ ✓ Correct
(C) $E = \frac{3}{2} (\frac{n ^{2} h ^{2} F ^{2}}{4 π ^{2} m})^{1/3}$ ✓ Correct
(D) $E = 2 (\frac{n ^{2} h ^{2} F ^{2}}{4 π ^{2} m})^{1/3}$

Answer: (B), (C)

Solution

For U = Fr, the force is F = -dU/dr = -F (constant magnitude). For circular motion: F = mv²/R. Using Bohr quantization: mvR = nh/(2π). From these: F = m/(R) × (n²h²)/(4π²m²R²), giving R ∝ n^(2/3). Then v = (nh)/(2πmR) ∝ n^(1/3). The total energy E = (1/2)mv² + FR = (1/2)mv² + mv² = (3/2)mv². Substituting the expressions for v and R: E = (3/2)[(n²h²F²)/(4π²m)]^(1/3).

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · One or more correct options

Practise more: all JEE Advanced PYQs on Modern Physics →

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Modern Physics

Solution

More JEE Advanced questions on Modern Physics