Understanding Electron Energy Near the Nucleus

When an electron is near the nucleus, it experiences a strong attractive force. This is because the electron has a negative charge and the nucleus has a positive charge (due to protons). The strength of this attractive force increases as the distance between the electron and the nucleus decreases.

Potential Energy of an Electron Near the Nucleus

The potential energy of an electron in the electric field of a nucleus is given by the formula:

\( U = \frac{k q_1 q_2}{r} \)

Where:

• \( U \) is the potential energy.
• \( k \) is Coulomb's constant.
• \( q_1 \) is the charge of the electron (\(-e\), which is negative).
• \( q_2 \) is the charge of the nucleus (\(+Ze\), which is positive, where \(Z\) is the atomic number).
• \( r \) is the distance between the electron and the nucleus.

Substituting the charges, the potential energy becomes:

\( U = \frac{k (-e) (+Ze)}{r} = -\frac{kZe^2}{r} \)

Since \(k\), \(Z\), \(e\), and \(r\) are all positive values, the potential energy \(U\) is always negative in an attractive system like an electron and nucleus.

Now, let's consider what happens when the electron is near the nucleus. Being near the nucleus means the distance \(r\) is small. In the formula \( U = -\frac{kZe^2}{r} \), if \(r\) is small, the value of \(\frac{kZe^2}{r}\) is large. Since the potential energy is negative, a large positive value multiplied by -1 results in a large negative value.

A large negative value represents a low potential energy compared to values closer to zero (which corresponds to larger distances or infinite separation, where potential energy is typically considered zero).

Therefore, an electron near the nucleus, experiencing strong attraction due to the small distance, has a low potential energy.

Considering Kinetic Energy

While the question focuses on potential energy, it's worth noting the relationship with kinetic energy in a bound system like an atom. For an electron in orbit around a nucleus, the total energy is the sum of kinetic and potential energy (\(E = K + U\)). For a bound electron, the total energy \(E\) is negative. In many simple atomic models (like the Bohr model or for inverse square forces), the average kinetic energy is related to the average potential energy. Specifically, for circular orbits, \( K = -U/2 \). If potential energy \(U\) is large and negative (low), the kinetic energy \(K\) is large and positive (high). So, an electron near the nucleus is often moving faster, thus having high kinetic energy. However, the question specifically asks what the electron "has" near the nucleus, and "Low potential energy" is a direct consequence of its position in the attractive electric field.

Evaluating the Options

Based on our understanding of electrostatic potential energy between an electron and a nucleus:

• Low kinetic energy: Not necessarily true; kinetic energy is often high near the nucleus in a bound state.
• Low potential energy: This is correct. The potential energy \(U = -\frac{kZe^2}{r}\) is large and negative (low) when \(r\) is small.
• High kinetic energy: Possible, but the potential energy is directly determined by position and charge interaction.
• High potential energy: This would mean the potential energy is less negative or positive, which happens at larger distances or for repulsive forces.

The strong attraction implies a position where the potential energy in the attractive field is minimized (most negative), which is near the nucleus.

The final answer is Low potential energy.