Question

An old model of a hydrogen atom has the charge +e of the proton uniformly distributed over a sphere of radius${{a}_{0}}$, with the electron of charge –e and mass m at its center. What would then be the force on the electron if it were displaced from the center by a distance $r\le {{a}_{0}}$?

Answer 1

Hint : Firstly, you could recall the expression for force on an electron in a uniform electric field. Electric field to be substituted there could be derived using Gauss's law. Consider a sphere of radius r which will have the same volume charge density as the sphere of radius ${{a}_{0}}$ and thus get the charge enclosed in terms of the given quantities.
Formula used:
Gauss law,
$\oint{\overrightarrow{E}}\bullet \overrightarrow{da}=\dfrac{{{q}_{en}}}{{{\varepsilon }_{0}}}$
Complete answer:
In the question, we are given an old model of hydrogen atom where a charge of +e is distributed uniformly over a sphere of radius ${{a}_{0}}$with an electron of charge –e and mass m at the center. We are supposed to find the force on the electron if it was displaced from the centre by $r\le {{a}_{0}}$.
We know that the force on an electron in the vicinity of an electric field E is given by the product of the charge of the electron and the magnitude of the electric field. So,
$F=-eE$ ………………………………………. (1)
Since the given question deals with uniform distribution of charge, we could use the Gauss law to find the electric field. That is,
$\oint{\overrightarrow{E}}\bullet \overrightarrow{da}=\dfrac{{{q}_{en}}}{{{\varepsilon }_{0}}}$ ……………………………………….. (2)
Let us consider a sphere of radius $r\le {{a}_{0}}$ within or on the sphere of uniform charge distribution. Let ${{q}_{en}}$ be the charge enclosed in it, then, volume charge density of both spheres will be equal, that is,
$\dfrac{{{q}_{en}}}{\dfrac{4}{3}\pi {{r}^{3}}}=\dfrac{e}{\dfrac{4}{3}\pi {{a}_{0}}^{3}}$
$\Rightarrow {{q}_{en}}=e{{\left( \dfrac{r}{{{a}_{0}}} \right)}^{3}}$
So, (2) will become,
$E\left( 4\pi {{r}^{2}} \right)=\dfrac{e{{\left( \dfrac{r}{{{a}_{0}}} \right)}^{3}}}{{{\varepsilon }_{0}}}$
$\Rightarrow E=\dfrac{e}{4\pi {{\varepsilon }_{0}}{{a}_{0}}^{3}}r$
Substituting this in (1), we will get,
$F=-\dfrac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}{{a}_{0}}^{3}}r$
Hence, we found the force on the electron when the electron was displaced from the center by a distance$r\le {{a}_{0}}$to be,
$F=-\dfrac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}{{a}_{0}}^{3}}r$
Note :
You may have noticed the negative sign in the final expression of force. That negative sign indicates that the force on the electron is towards the centre, that is, attractive force. Gauss law could be stated as, ‘Electric flux out of any arbitrary closed surface is known to be proportional to the charge enclosed by the surface’.