An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius `r`. The coulomb force `vec(F)` between the two is (where `k=(1)/(4piepsilon_(0)))`

Options

A. `k(e^(2))/(r^(3))r`
B. `-k(e^(2))/(r^(3))r`
C. `k(e^(2))/(r^(2))r`
D. `-k(e^(2))/(r^(3))hatr`

Answer

To find the Coulomb force between the electron and the nucleus of a hydrogen atom in vector form, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges**:
- The charge of the proton (nucleus) is \( +e \).
- The charge of the electron is \( -e \).

2. **Coulomb's Law**:
- The magnitude of the electrostatic force \( F \) between two point charges is given by Coulomb's law:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
- Here, \( k = \frac{1}{4\pi\epsilon_0} \), \( q_1 = +e \), and \( q_2 = -e \).

3. **Substituting the Charges**:
- Substituting the values of the charges into the equation:
\[ F = k \frac{|-e \cdot e|}{r^2} = k \frac{e^2}{r^2} \]

4. **Direction of the Force**:
- Since the force is attractive (the electron is attracted to the proton), we need to include the direction. The force vector points from the electron towards the nucleus.
- We can express this direction using a unit vector \( \hat{r} \), which points from the electron to the nucleus.

5. **Vector Form of the Force**:
- The vector form of the Coulomb force \( \vec{F} \) can be written as:
\[ \vec{F} = -k \frac{e^2}{r^2} \hat{r} \]
- Here, the negative sign indicates that the force is attractive, pulling the electron towards the nucleus.

6. **Final Expression**:
- Therefore, the final expression for the Coulomb force in vector form is:
\[ \vec{F} = -k \frac{e^2}{r^2} \hat{r} \]