Question:
An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The Coulomb force F between the two is
An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The Coulomb force F between the two is
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\(k = \frac{1}{4\pi\epsilon_0}\). Negative sign indicates attractive force.
Updated On: Apr 23, 2026
- \(\frac{ke^2}{r^3} \mathbf{r}\)
- \(-\frac{ke^2}{r^3} \mathbf{r}\)
- \(\frac{ke^2}{r^2} \mathbf{r}\)
- \(-\frac{ke^2}{r^2} \mathbf{r}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Coulomb force is attractive, directed from electron to nucleus. \(\mathbf{F} = \frac{k q_1 q_2}{r^2} \hat{r}\). Here \(q_1 = +e\), \(q_2 = -e\).
Step 2: Detailed Explanation:
Magnitude \(F = \frac{k e^2}{r^2}\). Direction: towards nucleus, i.e., opposite to position vector \(\mathbf{r}\) (if origin at nucleus).
\(\mathbf{F} = -\frac{k e^2}{r^2} \hat{r}\). But \(\hat{r} = \frac{\mathbf{r}}{r}\), so \(\mathbf{F} = -\frac{k e^2}{r^3} \mathbf{r}\).
Step 3: Final Answer:
Thus, \(\mathbf{F} = -\frac{ke^2}{r^3} \mathbf{r}\).
Coulomb force is attractive, directed from electron to nucleus. \(\mathbf{F} = \frac{k q_1 q_2}{r^2} \hat{r}\). Here \(q_1 = +e\), \(q_2 = -e\).
Step 2: Detailed Explanation:
Magnitude \(F = \frac{k e^2}{r^2}\). Direction: towards nucleus, i.e., opposite to position vector \(\mathbf{r}\) (if origin at nucleus).
\(\mathbf{F} = -\frac{k e^2}{r^2} \hat{r}\). But \(\hat{r} = \frac{\mathbf{r}}{r}\), so \(\mathbf{F} = -\frac{k e^2}{r^3} \mathbf{r}\).
Step 3: Final Answer:
Thus, \(\mathbf{F} = -\frac{ke^2}{r^3} \mathbf{r}\).
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