Problem 14 - Chemistry The Central Science

Question

(a) What is the electrostatic potential energy (in joules) between two protons that are separated by 62 pm? (b) What is the change in potential energy if the distance separating the two is increased to 1.0 nm? (c) Does the potential energy of the two particles increase or decrease when the distance is increased to 1.0 nm?

Step-by-Step Solution

Answer

(a) The electrostatic potential energy between two protons separated by 62 pm is approximately \( 3.664 × 10^{-19} J \). (b) The change in potential energy when the distance is increased to 1.0 nm is approximately \( -3.434 × 10^{-19} J \). (c) The potential energy of the two protons decreases when the distance is increased to 1.0 nm.

Step 1: Gather given information and convert units if necessary.

The distance between the protons is initially 62 pm. We need to first convert it to meters: 62 pm = \( 62 × 10^{-12} \) m Next, we need to find the charges of the protons. Each proton carries a positive charge equal to the elementary charge: \( q = 1.6 × 10^{-19} C \)

Step 2: Calculate the electrostatic potential energy at 62 pm (Part a).

Plug the values into the formula \( U = k \frac{q_{1}q_{2}}{r} \): \( U = (8.99 × 10^9 Nm^2/C^2) \frac{(1.6 × 10^{-19} C)^2}{62 × 10^{-12} m} \) Now, calculate the value of U: \( U ≈ 3.664 × 10^{-19} J \) So, the electrostatic potential energy between two protons separated by 62 pm is approximately \( 3.664 × 10^{-19} J \).

Step 3: Calculate the electrostatic potential energy at 1.0 nm (Part b).

First, convert 1.0 nm to meters: 1.0 nm = \( 1.0 × 10^{-9} \) m Now, use the formula \( U = k \frac{q_{1}q_{2}}{r} \) with the new distance: \( U' = (8.99 × 10^9 Nm^2/C^2) \frac{(1.6 × 10^{-19} C)^2}{1.0 × 10^{-9} m} \) Calculate the new value of U: \( U' ≈ 2.302 × 10^{-20} J \) So, the electrostatic potential energy between two protons separated by 1.0 nm is approximately \( 2.302 × 10^{-20} J \).

Step 4: Calculate the change in potential energy (Part b).

To find the change in potential energy, subtract the initial potential energy from the final potential energy: \( ΔU = U' - U = (2.302 × 10^{-20} J) - (3.664 × 10^{-19} J) ≈ -3.434 × 10^{-19} J \) So, the change in potential energy is approximately \( -3.434 × 10^{-19} J \).

Step 5: Determine whether the potential energy increases or decreases (Part c).

Since the change in potential energy is negative, it means that the potential energy of the two protons decreases when the distance between them is increased to 1.0 nm.

Key Concepts

Coulomb's Law

Understanding Coulomb's law is essential for grasping how electric charges interact. Simply put, this law states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This relationship can be represented by the equation:
\[\begin{equation} F = k \frac{q_1 q_2}{r^2} \end{equation}\]
where \( F \) is the electrostatic force, \( k \) is Coulomb's constant (approximately \(8.99 \times 10^9 Nm^2/C^2\)), \( q_1 \) and \( q_2 \) are the charged particles, and \( r \) is the distance between the centers of the two charges. Coulomb's law is crucial in calculating the electrostatic potential energy between charges as it essentially reveals how this energy varies with distance and charge magnitude.

Elementary Charge

The concept of elementary charge is fundamental in the study of electricity and electronics. It is the smallest unit of electric charge that is considered indivisible in the context of charge interactions. This charge is carried by a single proton or, conversely, a single electron (with a negative sign). The value of an elementary charge is approximately \(1.6 \times 10^{-19} \) coulombs (C). When we reference charges in physics problems like our electrostatic potential problem, we're often measuring them in multiples of this elementary charge, particularly when dealing with atomic or subatomic particles such as protons.

Energy Units Conversion

In physics, energy can be measured in various units, and often, conversions are necessary to ensure that calculations are consistent. The problem at hand uses joules (J), which is the standard unit of energy in the International System of Units (SI). However, when working with atomic and subatomic scales, incredibly small units such as electronvolts (eV) may also be used. An electronvolt is the amount of kinetic energy gained by a single electron when it accelerates through an electric potential difference of one volt. In converting between these units, we use the fact that one electronvolt is equal to approximately \(1.602 \times 10^{-19}\) joules. Knowing how to convert between these units allows you to interpret energy values across various branches of physics.

Potential Energy Change

The change in potential energy is a crucial concept when evaluating how systems evolve over time. In the context of electrostatics, when the distance between two charged particles increases, their potential energy changes. According to Coulomb's law, if the separation distance increases, the potential energy typically decreases, since the particles are subjected to weaker electrostatic forces. Conversely, if the distance decreases, the potential energy between the charges increases. Our textbook problem illustrates this principle: by increasing the distance between the protons, the energy stored in the system decreased, represented by a negative change in potential energy.