Question

If potential energy between a proton and an electron is given by $|U| = {{k{e^2}} \mathord{\left/ {\vphantom {{k{e^2}} {2{R^3}}}} \right.} {2{R^3}}}$, where $e$ is the charge of electron and $R$ is the radius of the atom, then, the radius of Bohr’s orbit is given by ($h$ = Planck’s constant, $k$ = constant)
A. $\dfrac{{k{e^2}m}}{{{h^2}}}$
B. $\dfrac{{6{\pi ^2}}}{{{n^2}}}\dfrac{{k{e^2}m}}{{{h^2}}}$
C. $\dfrac{{2\pi }}{n}\dfrac{{k{e^2}m}}{{{h^2}}}$
D. $\dfrac{{4{\pi ^2}k{e^2}m}}{{{n^2}{h^2}}}$

Answer 1

Hint
We need to first find the derivative of the given potential energy, it gives the force. By comparing this force with the centrifugal force and equating it in the Bohr’s angular momentum equation and rearranging the terms we can obtain the expression in terms of the radius of an atom.
$\Rightarrow F = - \dfrac{{dU}}{{dR}}$
where $F$ is the force
$U$ is the potential energy and $R$ is the radius.
$\Rightarrow mvr = \dfrac{{nh}}{{2\pi }}$
where $m$ is mass and $v$ and $r$ are the velocity and the radius of the particle.

Complete step by step answer
The potential energy at a distance R is given by the equation,
$\Rightarrow U = \dfrac{{k{e^2}}}{{2{R^3}}}$
This potential energy is basically the internal energy of the particle. So the force can be represented in terms of potential energy as the derivative of the potential energy like,
$\Rightarrow F = - \dfrac{{dU}}{{dR}}$
Now by substituting the equation of the potential energy given in the above equation we get,
$\Rightarrow F = - \dfrac{d}{{dR}}\left( {\dfrac{{k{e^2}}}{{2{R^3}}}} \right)$
Now, the constants will come outside the derivative so we will have,
$\Rightarrow F = - \dfrac{{k{e^2}}}{2}\dfrac{d}{{dR}}\left( {\dfrac{1}{{{R^3}}}} \right)$
On calculating the derivative of the above function we get
$\Rightarrow F = - \dfrac{{k{e^2}}}{2}\left( { - \dfrac{3}{{{R^4}}}} \right)$
This can be rearranged as,
$\Rightarrow F = \dfrac{{3k{e^2}}}{{2{R^4}}}$…….(1)
This derivational force of the potential energy is the cause of the centrifugal acceleration of the system.
The centrifugal acceleration of the system is given by the formula as follows.
$\Rightarrow F = \dfrac{{m{v^2}}}{R}$…… (2)
Thus, upon comparing the equations (1) and (2), it is clear that the LHS part of both the equations is the same, that is, the force. So, we can equate the RHS as,
$\Rightarrow \dfrac{{3k{e^2}}}{{2{R^4}}} = \dfrac{{m{v^2}}}{R}$…… (3)
The angular momentum of the electron as per the Bohr’s model is given by the following equation,
$\Rightarrow mvr = \dfrac{{nh}}{{2\pi }}$
Rearranging the terms of the above equation we can obtain the expression in terms of the velocity of the electron as,
$\Rightarrow v = \dfrac{{nh}}{{2\pi mr}}$
Now we can substitute this equation in the RHS of equation (3). And hence we get,
$\Rightarrow \dfrac{{3k{e^2}}}{{2{R^4}}} = \dfrac{m}{R}{\left( {\dfrac{{nh}}{{2\pi mR}}} \right)^2}$
On opening the bracket we get,
$\Rightarrow \dfrac{{3k{e^2}}}{{2{R^4}}} = \dfrac{{m{n^2}{h^2}}}{{4{\pi ^2}{m^2}{R^3}}}$
We can cancel the $2{R^3}$ from the denominator of both LHS and RHS. Further on cancelling the $m$ from the RHS we get,
$\Rightarrow \dfrac{{3k{e^2}}}{R} = \dfrac{{{n^2}{h^2}}}{{2{\pi ^2}m}}$
Now we take the $R$ from the LHS to the RHS and the rest terms to the LHS we get,
$\Rightarrow R = \dfrac{{6k{e^2}{\pi ^2}m}}{{{n^2}{h^2}}}$
Therefore, the radius of the Bohr’s atom can be represented in the terms of the expression $\Rightarrow R = \dfrac{{6k{e^2}{\pi ^2}m}}{{{n^2}{h^2}}}$.
Thus, the option (B) is correct.

Note
In atomic physics, the Bohr’s model is used to depict the model of an atom containing a positively charged nucleus in the center and the electrons revolving around it. This model is quite similar to that of the solar system. The energy in the orbits is discrete and the electrons lose or gain energy by jumping from or to different levels.