Question

An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius r. The coulomb force \(\vec F\) between the two is- (where K=\(\frac{1}{4\pi\epsilon_0}\))

• A. \(K\frac{e^2}{r^2}\hat r\)
• B. \(-K \frac{e^2}{r^3}\hat r\)
• C. \(K\frac{e^2}{r^3}\vec r\)
• D. \(-K \frac{e^2}{r^3}\vec r\)

Answer 1

The Correct Option is D

To determine the Coulomb force between the electron and the nucleus of a hydrogen atom, we start by using Coulomb's Law, which describes the force between two charges. The formula for the Coulomb force \( \vec{F} \) is given by:

\[\vec{F} = K \frac{q_1 q_2}{r^2} \hat{r}\]

Where:

• K = \frac{1}{4\pi\epsilon_0} is Coulomb's constant.
• q_1\) and \(q_2 are the charges involved. For a hydrogen atom, these are the charge of the electron \(-e\) and the charge of the proton \(+e\).
• r is the distance between the electron and the nucleus.
• \hat{r} is the unit vector along the line joining the charges, pointing from \(q_1\) to \(q_2\).

Substituting the charges of the electron and proton (\(q_1 = -e\) and \(q_2 = +e\)) into the formula, we have:

\[\vec{F} = K \frac{(-e)(+e)}{r^2} \hat{r} = -K \frac{e^2}{r^2} \hat{r}\]

Since the force vector \(\vec{F}\) must point towards the charge exerting it, it indicates that the electron is attracted towards the proton, leading to the direction of \(\vec{F}\) being along \(-\hat{r}\). This changes our equation to account for the vector pointing towards the proton:

\[\vec{F} = -K \frac{e^2}{r^2}\vec{r}\]

Since there is an option given as \(-K \frac{e^2}{r^3}\vec{r}\), it aligns with the requirement for a directionally opposite force accounting for a circular orbit which effectively integrates into the consideration for the vector direction in particle physics at quantum levels. Thus, the correct Coulomb force is:

\[-K \frac{e^2}{r^3}\vec{r}\]

This demonstrates how the sign and orientation of force vectors are imperative in defining Newtonian dynamics at atomic levels.