Question

An electron of charge $e$ is moving round the nucleus of a hydrogen atom in a circular orbit of radius $r$. The Coulomb's force $F$ between the two is:

• A. $-\frac{2k{e}^2}{r^3}\overrightarrow{r}$
• B. $\frac{k{e}^2}{r^2}\overrightarrow{r}$
• C. $\frac{-k{e}^2}{r^3}\overrightarrow{r}$
• D. $\frac{-k{e}^2}{r^2}\overrightarrow{r}$

Answer 1

The correct option is C $\frac{-k{e}^2}{r^3}\overrightarrow{r}$

Using Coulomb's law, the force between two charges is given by,
$\overrightarrow{F}=\frac{k{q}_1{q}_2}{r^3}\overrightarrow{r}\dots \dots \left(i\right)$
where $r=$ distance between charges
$\overrightarrow{r}=$ Position vector of one charge w.r.t another one.
Here, $q_1=-e$ (charge of electron)
$q_2=+Ze$ (charge of nucleus)
i.e. $q_2=+e$ (for hydrogen atom $Z=1$)

From eq. $\left(i\right)$, the force between electron and nucleus is
$\overrightarrow{F}=\frac{k(-e)\left(e\right)}{r^3}\overrightarrow{r}$
$\therefore \overrightarrow{F}=\frac{-k{e}^2}{r^3}\overrightarrow{r}$

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: Magnitude of electric force between electron and nucleus is,}\\ F=\frac{k\left(e\right)\left(Ze\right)}{r^2}=\frac{kZ{e}^2}{r^2}\end{array}$