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

• A. \(\frac{ke^2}{r^3} \mathbf{r}\)
• B. \(-\frac{ke^2}{r^3} \mathbf{r}\)
• C. \(\frac{ke^2}{r^2} \mathbf{r}\)
• D. \(-\frac{ke^2}{r^2} \mathbf{r}\)

Hint:

\(k = \frac{1}{4\pi\epsilon_0}\). Negative sign indicates attractive force.

Answer 1

The Correct Option is B

To solve the problem of determining the Coulomb force acting on an electron moving around the nucleus of a hydrogen atom in a circular orbit, we begin by reviewing the concept of electrostatic forces.

The Coulomb force between two point charges is given by Coulomb's law:

\(F = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\)

where:

• \(F\) is the magnitude of the force between the charges.
• \(k\) is Coulomb's constant \((8.9875 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2)\).
• \(q_1\) and \(q_2\) are the magnitudes of the charges.
• \(r\) is the distance between the charges.

In the hydrogen atom, the charges are:

• \(q_1 = e\) (the charge of the electron, \(-1.6 \times 10^{-19} \, \text{C}\))
• \(q_2 = e\) (the charge of the proton in the nucleus, \(+1.6 \times 10^{-19} \, \text{C}\))

The vector form of the Coulomb force includes the direction of the force. Since the electron is negatively charged and the proton is positively charged, the force is attractive and directed towards the nucleus. Therefore, the vector form of the force is:

\(\mathbf{F} = -\frac{k \cdot e^2}{r^2} \cdot \mathbf{r}_{unit}\)

The vector \(\mathbf{r}_{unit}\) is the unit vector in the direction from the nucleus to the electron. The negative sign indicates the force is attractive.

However, the given answer option that captures both direction and the expression as described is:

\(-\frac{ke^2}{r^3} \mathbf{r}\)

where \(\mathbf{r}\) is a vector of magnitude \(r\) from the nucleus to the electron. The use of \(-\frac{ke^2}{r^3} \mathbf{r}\) effectively incorporates the direction of the force, showing it is an attractive (negative) force.

Thus, the correct answer is that the Coulomb force between the electron and the nucleus is given by the vectorial expression:

Correct Answer: \(-\frac{ke^2}{r^3} \mathbf{r}\)

This option is correct because it correctly represents the magnitude and direction of the Coulomb force in terms of vectors.