Question

A particle of mass $m$ moves in circular orbits with potential energy $V\left(r\right)$=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. $R\propto n^{\frac{1}{3}}\text{and}V\propto n^{\frac{2}{3}}$
• B. $R\propto n^{\frac{2}{3}}\text{and}V\propto n^{\frac{1}{3}}$
• C. $E=\frac{3}{2}{\left(\frac{n^2{h}^2{F}^2}{4{\pi }^{2}m}\right)}^{\frac{1}{3}}$
• D. $E=2{\left(\frac{n^2{h}^2{F}^2}{4{\pi }^{2}m}\right)}^{\frac{1}{3}}$

Answer 1

Answer: (B), (C)

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

$U$ = $Fr$
[Using $U$ = Potential energy and $v$ = velocity, to avoid confusion between their symbols]
$\Rightarrow \text{Force}=\frac{-dU}{dr}=-F$
$\Rightarrow$Magnitude of force = Constant = $F$
$\Rightarrow F=\frac{m{v}^2}{R}.......\left(1\right)$
$\Rightarrow mvR=\frac{nh}{2\pi }.......\left(2\right)$
$\Rightarrow F=\frac{m}{R}\times \frac{n^2{h}^2}{4{\pi }^2}\times \frac{1}{m^2{R}^2}$
$\Rightarrow R={\left(\frac{n^2{h}^2}{4{\pi }^{2}mF}\right)}^{\frac{1}{3}}.......\left(3\right)$

From $\left(2\right)$
$v=\frac{nh}{2\pi mR}$
$\Rightarrow v=\frac{nh}{2\pi m}{\left(\frac{4{\pi }^{2}mF}{n^2{h}^2}\right)}^{\frac{1}{3}}$
$\Rightarrow v=\frac{n^{\frac{1}{3}}{h}^{\frac{1}{3}}{F}^{\frac{1}{3}}}{2^{\frac{1}{3}}{\pi }^{\frac{1}{3}}{m}^{\frac{2}{3}}}.......\left(4\right)$
Hence, (B) is correct

Using
$\Rightarrow E=\frac{1}{2}m{v}^2+U=\frac{1}{2}m{v}^2+FR$
$\Rightarrow E=\frac{1}{2}m\left(\frac{n^{\frac{2}{3}}{h}^{\frac{2}{3}}{F}^{\frac{2}{3}}}{2^{\frac{2}{3}}{\pi }^{\frac{2}{3}}{m}^{\frac{2}{3}}}\right)+F\times {\left(\frac{n^2{h}^2{F}^2}{4{\pi }^{2}mF}\right)}^{\frac{1}{3}}$
$\Rightarrow E={\left(\frac{n^2{h}^2{F}^2}{4{\pi }^{2}mF}\right)}^{\frac{1}{3}}[\frac{1}{2}+1]$
$\therefore E=\frac{3}{2}{\left(\frac{n^2{h}^2{F}^2}{4{\pi }^{2}m}\right)}^{\frac{1}{3}}$