AP EAMCET · PHYSICS · Electrostatics
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 \(\mathrm{K} \frac{\mathrm{e}^2}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}\)
- C \(-\mathrm{K} \frac{\mathrm{e}^2}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}\)
- D \(\mathrm{K} \frac{\mathrm{e}^2}{\mathrm{r}^2} \overrightarrow{\mathrm{r}}\)
Answer & Solution
Correct Answer
(C) \(-\mathrm{K} \frac{\mathrm{e}^2}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}\)
Step-by-step Solution
Detailed explanation
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, \(\mathrm{q}_2=\mathrm{e}\)…
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