Question

A particle of mass $m$ moves in circular orbits with potential energy $V\left(r\right)=Fr$, 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 $n^{th}$orbit (here $h$ is the Planck’s constant)

• A. $R\alpha 1^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}andv\alpha n^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$
• B. $R\alpha n^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}andv\alpha n^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$
• C. $E=\frac{3}{2}{\left(\frac{n^2{h}^2{F}^2}{4{\mathrm{\pi }}^2\mathrm{m}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• D. $E=2{\left(\frac{n^2{h}^2{F}^2}{4{\mathrm{\pi }}^2\mathrm{m}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Answer 1

Answer: (B), (C)

$E=\frac{3}{2}{\left(\frac{n^2{h}^2{F}^2}{4{\mathrm{\pi }}^2\mathrm{m}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Radius and velocity

Potential Energy, $V\left(r\right)=Fr$

$Force=-\frac{dV}{dr}=-F$

Equating to the magnitude of the centripetal force,

$F=\frac{m{v}^2}{R}$

We also know that

$mvR=\frac{nh}{2\mathrm{\pi }}$

which can be rearranged to

$v=\frac{nh}{2\mathrm{\pi mR}}$

$R=\frac{nh}{2\mathrm{\pi mv}}$

substituting $v$ in $F$

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

rearranging,

$R={\left(\frac{n^2{h}^2}{4{\mathrm{\pi }}^2\mathrm{mF}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

and using $R$ in $v$

$v={\left(\frac{nhF}{2{\mathrm{\pi m}}^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

So option $\left(B\right)$ is correct.

Energy

$$\begin{gathered} E=\frac{1}{2}m{v}^2+V \\ =\frac{1}{2}m{v}^2+FR \end{gathered}$$

Substituting $v$ and $R$

$E=\frac{3}{2}{\left(\frac{n^2{h}^2{F}^2}{4{\mathrm{\pi }}^2\mathrm{m}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

This means option $\left(C\right)$is correct.

Hence, the correct options are $\left(B\right)$ and $\left(C\right)$.