Problem 13 - Chemistry The Central Science In Si Units

Question

(a) What is the electrostatic potential energy (in joules) between an electron and a proton that are separated by \(230 \mathrm{pm} ?\) (b) What is the change in potential energy if the distance separating the electron and proton is increased to \(1.0 \mathrm{nm} ?\) (c) Does the potential energy of the two particles increase or decrease when the distance is increased to \(1.0 \mathrm{nm} ?\)

Step-by-Step Solution

Answer

Potential energy decreases when distance increases to 1.0 nm.

Step 1: Understanding Coulomb's Law for potential energy

The electrostatic potential energy between two point charges can be calculated using Coulomb's law for potential energy: \[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]where \(U\) is the potential energy, \(k\) is Coulomb's constant \((8.988 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2)\), \(q_1\) and \(q_2\) are the charges of the particles (in coulombs), and \(r\) is the separation distance (in meters). The charge of an electron \(q_e\) is \(-1.602 \times 10^{-19}\, \text{C}\), and proton \(q_p\) is \(+1.602 \times 10^{-19}\, \text{C}\). Conversion of distance is necessary from picometers to meters.

Step 2: Calculate the potential energy at 230 pm

Given \(r = 230\, \text{pm} = 230 \times 10^{-12} \text{ m}\). Substitute the values into the potential energy formula:\[ U = \frac{(8.988 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2) \times (-1.602 \times 10^{-19}\, \text{C}) \times (1.602 \times 10^{-19}\, \text{C})}{230 \times 10^{-12} \text{ m}} \]Calculate \(U\) to find the potential energy at 230 pm.

Step 3: Calculate the potential energy at 1.0 nm

Convert 1.0 nm to meters, \(r = 1.0 \text{ nm} = 1.0 \times 10^{-9} \text{ m}\). Substitute this distance into the potential energy formula:\[ U' = \frac{(8.988 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2) \times (-1.602 \times 10^{-19}\, \text{C}) \times (1.602 \times 10^{-19}\, \text{C})}{1.0 \times 10^{-9} \text{ m}} \]Calculate \(U'\) to determine the potential energy at 1.0 nm.

Step 4: Determine the change in potential energy

To find the change in potential energy when the distance changes from 230 pm to 1.0 nm, calculate:\[ \Delta U = U' - U \]Subtract the potential energy at 230 pm from that at 1.0 nm to find the change.

Step 5: Analyze the change in potential energy

If \(\Delta U\) is positive, the potential energy increased. If it's negative, the potential energy decreased. This step explains the physical significance of the change, showing that potential energy decreases as the distance between attractive charges increases.

Key Concepts

Coulomb's Law

Coulomb's law is a fundamental principle used to calculate the electrostatic potential energy between two point charges. This law helps us understand how two charged particles interact with each other. The formula for electrostatic potential energy is given by:\[U = \frac{k \cdot q_1 \cdot q_2}{r} \]Where:

• \( U \) is the potential energy in joules.
• \( k \) is Coulomb's constant, approximately \( 8.988 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2 \).
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges (in coulombs).
• \( r \) is the separation distance between the charges (in meters).

This equation tells us how strongly two charges of opposite signs (like a proton and an electron) are attracted to each other based on their magnitude and the distance between them.

Point charges

In electrostatics, we often model charged particles as 'point charges'. This simply means we consider the charges to be concentrated at a single point in space. This is a useful simplification because it allows us to calculate electric forces and potential energies without being concerned about the shape or size of the objects involved.Point charges can have either positive or negative values. In our example, the proton carries a positive charge of \(+1.602 \times 10^{-19}\, \text{C} \), and the electron carries a negative charge of \(-1.602 \times 10^{-19}\, \text{C} \). Due to their opposite charges, they attract each other. This attraction can be quantified using Coulomb's law. Using these simplified models of particle charges helps tremendously when solving many problems in physics and chemistry.

Separation distance

The separation distance between two point charges significantly influences the electrostatic potential energy. The distance, denoted by \( r \) in Coulomb's law, is crucial because potential energy is inversely proportional to it. In simple terms, as the distance between two charges increases:

• The denominator in the formula gets larger, meaning the overall potential energy \,\( U \) becomes smaller.
• This results in a decrease in the attractive forces between oppositely charged particles.

Here's how it applies to the exercise: When a proton and an electron are separated by \(230 \mathrm{pm} \) (picometers, which is \(230 \times 10^{-12} \text{m}\)), the potential energy is calculated using that specific distance. If you increase the separation to \(1.0 \mathrm{nm} \) (nanometers, which is \(1.0 \times 10^{-9} \text{m}\)), the potential energy must be recalculated by substituting the new distance into the formula.

Change in potential energy

The concept of change in potential energy is central to understanding how energy transforms as two charged particles move closer together or further apart.We calculate the change in potential energy, \( \Delta U \), by computing the difference between the initial and final potential energies:\[\Delta U = U' - U \]Where:

• \( U \) is the initial potential energy at the starting distance.
• \( U' \) is the potential energy at the new, changed distance.

If \( \Delta U \) turns out to be negative, it indicates that the potential energy decreased, meaning the particles are more stable further apart, which is often the case for opposite charges. For the given example, when you increase the separation between an electron and a proton from \(230 \mathrm{pm} \) to \(1.0 \mathrm{nm} \), you should notice a decrease in potential energy. This decrease signals that the system of charges requires less energy when the two are further apart, reinforcing the concept that energy decreases with increased distance in such setups.