Q144 / 160 · AP EAMCET 2023 (17 May Shift 2) · Physics

An electron of charge ' $\mathrm{e}$ ' is moving round the nucleus of a hydrogen atom in a circular orbit of radius ' $r$ '. The coulomb force $\overrightarrow{\mathrm{F}}$ between the two is $\left(\right.$ here $\left.\mathrm{K}=\frac{1}{4 \pi \varepsilon_0}\right)$

  • A
    $-\mathrm{K} \frac{\mathrm{e}^2}{\mathrm{r}^3} \hat{\mathrm{r}}$
  • B
    $\mathrm{K} \frac{\mathrm{e}^2}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}$
  • Correct answer
    C
    $-\mathrm{K} \frac{\mathrm{e}^2}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}$
  • D
    $\mathrm{K} \frac{\mathrm{e}^2}{\mathrm{r}^2} \overrightarrow{\mathrm{r}}$

Solution

Using Coulomb's law, the force between two charges is given by $\overrightarrow{\mathrm{F}}=\frac{\mathrm{k} \mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}$ Here, charge of an electron, $q_1=-e$ Charge of the nucleus, $\mathrm{q}_2=\mathrm{e}$ $\overrightarrow{\mathrm{F}}=\frac{\mathrm{k}(-\mathrm{e}) \mathrm{e}}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}=-\frac{\mathrm{ke}^2}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}$

Hence, the correct answer is option: 'C'.

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