Q: A particle of mass m moves in circular orbits with potential energy V(r) = F r , where F is positive constant and r is its distance from origin . Its energy are calculated using the Bohr model . If 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}$

(c) $ E = \frac{3}{2} (\frac{n^2 h^2 F^2}{4\pi^2 m})^{1/3}$

(d) $ E = 2 (\frac{n^2 h^2 F^2}{4\pi^2 m})^{1/3}$

**Click to See Solution : **

$F = \frac{m v^2}{r} $ …(i)

$m v r = \frac{n h}{2 \pi}$ ….(ii)

From (i) & (ii)

$F = \frac{m}{r} (\frac{n h}{2 \pi m r})^2$

$F = \frac{m n^2 h^2}{4 \pi^2 m^2 r^3}$

$F = \frac{ n^2 h^2}{4 \pi^2 m r^3}$

Since F = constant

$r^3 \propto n^2 $

$r \propto n^{2/3} $

$v = \frac{n h}{2 \pi m r} $

$ v \propto \frac{n}{n^{2/3}} $

$ v \propto n^{1/3} $

$ K.E = \frac{1}{2}m v^2 = \frac{F r}{2} $

Total Energy = P.E + K.E

$T.E = F r + \frac{F r}{2}$

$T.E = \frac{3 F r}{2}$

$T.E = E = \frac{3F}{2} \times (\frac{n^2 h^2}{4 \pi^2 r n F})^{1/3} $

$E = \frac{3}{2} (\frac{n^2 h^2 F^2}{4 \pi^2 m})^{1/3}$