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)

Question

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) (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=(3/2)((n2 h2 F2/4 π2 m))1 / 3 (D) E=2((n2 h2 F2/4 π2 m))1 / 3

Tardigrade Admin · Accepted Answer

F =-( dV / dr ) ( mv 2/ R )= F mvr =( nh /2 π) Solving above equations R 3=( n 2 h 2/4 π2 m ) v propto n 1 / 3

$R \propto n^{1/3}$ and $v \propto n^{2/3}$

$R \propto n^{2/3}$ and $v \propto n^{1/3}$

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

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

$F = - \frac{d V}{d r}$
$\frac{m v ^{2}}{R} = F$
$m v r = \frac{nh}{2 π}$
Solving above equations
$R^{3} = \frac{n ^{2} h ^{2}}{4 π ^{2} m}$
$v \propto n^{1/3}$

R∝n1/3 and v∝n2/3

R∝n2/3 and v∝n1/3

E=23​(4π2mn2h2F2​)1/3

E=2(4π2mn2h2F2​)1/3

F=−drdV​ Rmv2​=F mvr=2πnh​ Solving above equations R3=4π2mn2h2​ v∝n1/3

$R \propto n^{1/3}$ and $v \propto n^{2/3}$

$R \propto n^{2/3}$ and $v \propto n^{1/3}$

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

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

$F = - \frac{d V}{d r}$
$\frac{m v ^{2}}{R} = F$
$m v r = \frac{nh}{2 π}$
Solving above equations
$R^{3} = \frac{n ^{2} h ^{2}}{4 π ^{2} m}$
$v \propto n^{1/3}$