Question

An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The Coulomb force $ \vec{F} $ between the two is

• A. $ k\frac{{{e}^{2}}}{{{r}^{3}}}\vec{r} $
• B. $ -k\frac{{{e}^{2}}}{{{r}^{3}}}\vec{r} $
• C. $ -k\frac{{{e}^{2}}}{{{r}^{2}}}\hat{r} $
• D. $ -k\frac{{{e}^{2}}}{{{r}^{3}}}\hat{r} $ (where $ k=\frac{1}{4\pi {{\varepsilon }_{0}}} $ )

Answer 1

Answer: (B)

Let charges on an electron and hydrogen nucleus are $q_{1}$ and $q_{2}$.
The Coulomb's force between them at a distance $r$ is,
$\vec{ F }=-\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} \hat{ r }$
Putting, $\frac{1}{4 \pi \varepsilon_{0}}=k$ (given)
$\vec{ F }=-k \frac{q_{1} q_{2}}{r^{2}} \hat{ r }$
Since, the nucleus of hydrogen atom has one proton, so charge on nucleus is $e$
ie, $q_{2}=e$ also $q_{1}=e$ for electron.
So, $\vec{ F }=-k \frac{ e \cdot e}{r^{2}} \hat{ r }=-k \frac{e^{2}}{r^{2}} \hat{ r }$
but $\hat{ r }=\frac{\vec{ r }}{|\vec{ r }|}=\frac{\vec{ r }}{r}$
Hence, $\vec{ F }=-k \frac{e^{2}}{r^{2}} \cdot \frac{\vec{ r }}{r}=-k \frac{e^{2}}{r^{3}} \cdot \vec{ r }$
Note : Negative sign in the expression for Coulomb's force shows that force between electron and hydrogen nucleus is of attraction.