Question

Suppose the potential energy of an hypothetical atom consisting of a proton and an electron is given by \[U=-\dfrac{k{{e}^{2}}}{3{{r}^{3}}}\]. Then if Bohr’s postulates are applied to this atom, then the radius of the ${{n}^{th}}$ orbital will be proportional to:
A. ${{n}^{2}}$
B. $\dfrac{1}{{{n}^{2}}}$
C. ${{n}^{3}}$
D. $\dfrac{1}{{{n}^{3}}}$

Answer 1

Hint: We are given the equation for potential energy of a hypothetical atom consisting of an electron and a proton. We know that centripetal force of the electron in the orbit will be the derivative of the potential energy with respect to radius of the orbit. Using this relation and by applying Bohr’s postulates, we will get the solution.

Formula used:
$F=-\dfrac{dU}{dr}$
$F=\dfrac{m{{v}^{2}}}{r}$
${{v}_{n}}=\dfrac{nh}{2\pi m{{r}_{n}}}$

Complete step by step answer:
In the question we are given the potential energy of a hypothetical atom which consists of a proton and an electron as,
\[U=-\dfrac{k{{e}^{2}}}{3{{r}^{3}}}\]
For ${{n}^{th}}$ orbit we can write the equation for potential energy as,
\[{{U}_{n}}=-\dfrac{k{{e}^{2}}}{3{{r}_{n}}^{3}}\], were ‘${{r}_{n}}$’ is the radius of the ${{n}^{th}}$ orbit.
We know that there is a centripetal force on the electron which is revolving in the orbit. We know that the centripetal force is given by the equation,
$F=-\dfrac{dU}{dr}$
Therefore for the ${{n}^{th}}$ orbit we will get the centripetal force as,
$\Rightarrow F=-\dfrac{d{{U}_{n}}}{dr}$
$\Rightarrow F=-\dfrac{d\left( -\dfrac{k{{e}^{2}}}{3{{r}_{n}}^{3}} \right)}{dr}$
$\Rightarrow F=\dfrac{d\left( \dfrac{k{{e}^{2}}}{3{{r}_{n}}^{3}} \right)}{dr}$
By solving this, we will get
$\therefore F=\dfrac{k{{e}^{2}}}{{{r}_{n}}^{4}}$
We know that centripetal force is given by the equation,
$F=\dfrac{m{{v}^{2}}}{r}$
Therefore for the ${{n}^{th}}$ orbital, it will be
$\Rightarrow F=\dfrac{m{{v}_{n}}^{2}}{{{r}_{n}}}$
By equating this with the equation for centripetal force, we will get
$\Rightarrow \dfrac{m{{v}_{n}}^{2}}{{{r}_{n}}}=\dfrac{k{{e}^{2}}}{{{r}_{n}}^{4}}$
From this equation, we can write
$\Rightarrow {{v}_{n}}^{2}=\dfrac{{{r}_{n}}k{{e}^{2}}}{m{{r}_{n}}^{4}}$
$\therefore {{v}_{n}}^{2}=\dfrac{k{{e}^{2}}}{m{{r}_{n}}^{3}}$
We know that equation for velocity in the ${{n}^{th}}$ orbit is given as,
${{v}_{n}}=\dfrac{nh}{2\pi m{{r}_{n}}}$
Therefore we will get,
$\Rightarrow \dfrac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}{{m}^{2}}{{r}_{n}}^{2}}=\dfrac{k{{e}^{2}}}{m{{r}_{n}}^{3}}$
From this we will get the radius of the ${{n}^{th}}$ orbit as,
 $\Rightarrow \dfrac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}{{m}^{2}}}=\dfrac{k{{e}^{2}}{{r}_{n}}^{2}}{m{{r}_{n}}^{3}}$
$\Rightarrow \dfrac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}{{m}^{2}}}=\dfrac{k{{e}^{2}}}{m{{r}_{n}}}$
$\Rightarrow \dfrac{{{r}_{n}}{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}{{m}^{2}}}=\dfrac{k{{e}^{2}}}{m}$
$\Rightarrow {{r}_{n}}=\dfrac{4{{\pi }^{2}}{{m}^{2}}k{{e}^{2}}}{m{{n}^{2}}{{h}^{2}}}$
$\therefore {{r}_{n}}=\dfrac{4{{\pi }^{2}}mk{{e}^{2}}}{{{n}^{2}}{{h}^{2}}}$
From this we can see that the radius of the ${{n}^{th}}$ orbit is inversely proportional to ${{n}^{2}}$. That is,
${{r}_{n}}\propto \dfrac{1}{{{n}^{2}}}$

So, the correct answer is “Option B”.

Note:
The Bohr’s atomic model was proposed by Neil Bohr in the year 1915. It was a modification of Rutherford’s atom model but it still had its own limitations. Some of the limitations are,
1. It failed to explain the Zeeman effect and the Stark effect.
2. Bohr model of atom violated Heisenberg’s uncertainty principle.
3. It also failed to account for the spectra obtained from the larger atoms.