An early model for an atom considered to have a positively charged point nucleus of charge Ze surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field a distance r from the nucleus, where `r gt R` ?

Options

A. `(1)/(4pi epsilon_(0)) (Ze)/(r^(2))`
B. `(1)/(4pi epsilon_(0)) (Ze)/(R^(2))`
C. `(1)/(4pi epsilon_(0)) (Ze)/(r^(2)-R^(2))`
D. 0

Answer

To solve the problem, we need to find the electric field at a distance \( r \) from the nucleus of an atom, where \( r > R \) (the radius of the atom).

### Step-by-step Solution:

1. **Understanding the Charge Distribution**:
- The atom has a positively charged nucleus with charge \( Ze \) (where \( Z \) is the atomic number and \( e \) is the elementary charge).
- Surrounding the nucleus, there is a uniform density of negative charge that extends up to a radius \( R \).
- The atom as a whole is neutral, meaning the total positive charge from the nucleus is balanced by the total negative charge from the electrons.

2. **Applying Gauss's Law**:
- Gauss's Law states that the electric flux \( \Phi \) through a closed surface is equal to the charge enclosed \( Q_{\text{enc}} \) divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi = \oint E \cdot dS = \frac{Q_{\text{enc}}}{\epsilon_0} \]

3. **Choosing a Gaussian Surface**:
- We choose a spherical Gaussian surface of radius \( r \) such that \( r > R \). This means the surface is outside the region where the negative charge is distributed.

4. **Calculating the Enclosed Charge**:
- Inside the Gaussian surface, the charge from the nucleus is \( +Ze \).
- The total negative charge from the electrons is \( -Ze \) (since the atom is neutral).
- Therefore, the total charge enclosed by the Gaussian surface is:
\[ Q_{\text{enc}} = +Ze + (-Ze) = 0 \]

5. **Applying Gauss's Law**:
- Since the enclosed charge \( Q_{\text{enc}} = 0 \), we can substitute this into Gauss's Law:
\[ \oint E \cdot dS = \frac{0}{\epsilon_0} = 0 \]
- This implies that the electric field \( E \) at a distance \( r \) from the nucleus (where \( r > R \)) is zero.

6. **Conclusion**:
- Therefore, the electric field at a distance \( r \) from the nucleus, where \( r > R \), is:
\[ E = 0 \]

### Final Answer:
The electric field at a distance \( r \) from the nucleus, where \( r > R \), is \( E = 0 \).