Question

A particle of mass m moves in a circular orbit in a central potential field $U\left( r \right) = \dfrac{1}{2}k{r^2}$. If Bohr’s quantization condition is applied, radii of possible orbitals and energy level vary with quantum number $n$ as:
A. ${r_n} \propto {n^2},{E_n} \propto \dfrac{1}{{{n^2}}}$
B. ${r_n} \propto \sqrt n ,{E_n} \propto \dfrac{1}{n}$
C. ${r_n} \propto n,{E_n} \propto n$
D. \[{r_n} \propto \sqrt n ,{E_n} \propto n\]

Answer 1

Hint: In this, we first start by finding the force due to the electric field that is \[F = \dfrac{{dU(r)}}{{dr}}\] then equate it to the centrifugal force that is $F = \dfrac{{m{v_n}^2}}{{{r_n}}}$ which gives us ${r_n} \propto \sqrt n $. Then we for the energy we find the total energy as \[{E_n} = \dfrac{1}{2}k{r_n}^2 + \dfrac{1}{2}m{v_n}^2\] which will give us \[{E_n} \propto n\].

Complete Step-by-Step solution:
Given the potential field in which the particle is moving is $U\left( r \right) = \dfrac{1}{2}k{r_n}^2$
Now we will write the Bohr’s Quantization condition for the allowed state of an electron is
$m{v_n}{r_n} = \dfrac{{nh}}{{2\pi }}$-------------------------- (1)
Where $n = 1,2,3....$
$m$is the mass which is taken constant
${v_n}$ is the velocity of the allowed state
${r_n}$ is the orbital radii of the allowed state
$h$is the planck's constant
Now we need to find the relation between ${r_n}$ and $n$ but, we cannot write it directly as ${v_n}$ is a variable so we need to find how ${v_n}$ is varying with the ${r_n}$.
 For that what we will do is that we first found the force due to the electric field using the potential energy expression given in the question that is
\[F = \dfrac{{dU(r)}}{{dr}} = \dfrac{{d\left( {\dfrac{1}{2}k{r^2}} \right)}}{{dx}}\]
$ \Rightarrow F = Kr$--------------------------- (2)
We also know that this force is going to be canceled out by the centrifugal force which is the force exerted on a body due to its motion in a circular path that is
$F = \dfrac{{m{v_n}^2}}{{{r_n}}}$----------------------------- (3)
Now equating equation (2) and (3)
We get
$k{r_n} = \dfrac{{m{v_n}^2}}{{{r_n}}}$
$ \Rightarrow {r_n}^2 = \dfrac{{m{v_n}^2}}{k}$
\[ \Rightarrow {v_n} \propto {r_n}\]
$ \Rightarrow {v_n} = C{r_n}$----------------------------- (4)
Where, C is a constant now Substituting (4) in equation (1), we get
$m(C{r_n}){r_n} = \dfrac{{nh}}{{2\pi }}$
$ \Rightarrow n \propto {r_n}^2$
$ \Rightarrow {r_n} \propto \sqrt n $
Now we find the total energy ${E_n}$ that is the sum of potential energy which is the energy occupied by the particle due to its height and kinetic energy which is the energy occupied by the particle due to its motion.
\[{E_n} = {E_{PE}} + {E_{KE}}\]
\[ \Rightarrow {E_n} = \dfrac{1}{2}k{r_n}^2 + \dfrac{1}{2}m{v_n}^2\]------------------------ (5)
Now again substituting (4) in (5) we will get
\[ \Rightarrow {E_n} = \dfrac{1}{2}k{r_n}^2 + \dfrac{1}{2}m{C^2}{r_n}^2\]
\[ \Rightarrow {E_n} \propto {r_n}^2 \propto n\]
Hence option D is correct that is ${r_n} \propto \sqrt n ,{E_n} \propto n$

Note: For these types of questions we need to know about different types of forces like force due to electric field, Centrifugal, and centripetal forces. We also need to know about their expression and about Bohr’s law. Coming to energies we need to have a clear understanding of the potential energy and kinetic energies and their expression.