Problem 20 - College Physics

Question

(a) An electron is to be accelerated from \(3.00 \times 10^{6} \mathrm{m} / \mathrm{s}\right.\) to \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) . Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) to a halt?

Step-by-Step Solution

Answer

(a) 173.5 V, (b) 182.4 V

Step 1: Understand the Problem

We have two parts to the problem. In (a), an electron is being accelerated, and we need to find the potential difference required for this acceleration. In (b), we are slowing the electron down to a stop and must again find the potential difference needed. Both parts involve using kinetic energy changes.

Step 2: Write Down Known Values

The initial velocity of the electron, \( v_i = 3.00 \times 10^{6} \, \mathrm{m/s} \), and the final velocity \( v_f = 8.00 \times 10^{6} \, \mathrm{m/s} \) for part (a). For part (b), the initial velocity \( v_i = 8.00 \times 10^{6} \, \mathrm{m/s} \) and it is slowed to a halt, so \( v_f = 0 \, \mathrm{m/s} \). Also, remember the charge of an electron \( e = 1.60 \times 10^{-19} \, \mathrm{C} \) and its mass \( m = 9.11 \times 10^{-31} \, \mathrm{kg} \).

Step 3: Use Kinetic Energy Formula

The change in kinetic energy \( \Delta KE \) is equal to the work done by the electric field, which is also \( e \Delta V \), where \( \Delta V \) is the potential difference. So, \( \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) = e \Delta V \).

Step 4: Solve for Potential Difference in Part (a)

Substitute the values for part (a) into the kinetic energy formula: \[ \frac{1}{2} (9.11 \times 10^{-31} \mathrm{kg}) ((8.00 \times 10^{6})^2 - (3.00 \times 10^{6})^2) = (1.60 \times 10^{-19} \mathrm{C}) \Delta V \]. Solve for \( \Delta V \) to find the potential difference required for acceleration.

Step 5: Solve for Potential Difference in Part (b)

Part (b) asks for the potential difference to stop the electron. So, the change in kinetic energy will be all the initial kinetic energy because final velocity is zero. Use \( \frac{1}{2} (9.11 \times 10^{-31}) (8.00 \times 10^{6})^2 = (1.60 \times 10^{-19}) \Delta V \) and solve for \( \Delta V \).

Step 6: Calculate the Results

For part (a), \( \Delta V \approx 173.5 \, \mathrm{V} \). For part (b), \( \Delta V \approx 182.4 \, \mathrm{V} \). These are the potential differences required for acceleration and stopping, respectively.

Key Concepts

Kinetic Energy

The kinetic energy of an object is the energy it possesses due to its motion. For a particle like an electron, kinetic energy can be expressed using the formula:

• \( KE = \frac{1}{2} m v^2 \)

where \( m \) is the mass of the electron and \( v \) is its velocity. This equation shows that kinetic energy depends on both the mass and the square of the velocity of the particle.
In the context of electron acceleration in a potential difference, we need to account for the change in the electron's kinetic energy as it speeds up or slows down. The change in kinetic energy, \( \Delta KE \), is used to find how much work is done by the electric field on the electron.

• For acceleration: \( \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) \)

This formula calculates the energy required to change the electron’s velocity from \( v_i \) (initial) to \( v_f \) (final).
Understanding this concept is critical for using potential difference to modify the electron's speed.

Potential Difference

Potential difference, often referred to as voltage, is the measure of the work done per unit charge to move a charge between two points in an electric field. In the case of an electron moving in an electric field, this potential difference is responsible for accelerating or decelerating the electron.

• It can be represented by the symbol \( \Delta V \)

The relationship between potential difference and kinetic energy change is crucial. The work done by the potential difference on an electron can be expressed as:

• \( \Delta KE = e \Delta V \)

where \( e \) is the electron charge. This equation is derived from the principle of energy conservation, equating the change in kinetic energy to the work done by the electric field.
Hence, knowing how to calculate the potential difference is essential for determining the electron's speed change.

Electron Charge

The electron charge is a fundamental constant representing the quantity of charge carried by a single electron. It is denoted by \( e \) and has a value:

• \( e = 1.60 \times 10^{-19} \, \text{Coulombs} \)

This charge plays a pivotal role in calculating the work done on an electron when it moves through a potential difference. The work done is related to both the electron charge and the potential difference, used in the formula:

• \( \Delta KE = e \Delta V \)

Knowing the electron charge allows us to determine how much energy is required or released when an electron changes speed. It is a fundamental aspect of any calculation involving electron motion in electric fields.

Electron Mass

The mass of an electron is another fundamental physical constant needed in calculations involving kinetic energy. The mass is represented by \( m \) and is:

• \( m = 9.11 \times 10^{-31} \, \text{kg} \)

This extremely small value implies that even small potential differences can result in significant speed changes for electrons.
In kinetic energy calculations, the mass is directly used in the expression:

• \( KE = \frac{1}{2} m v^2 \)

As mass is a scalar quantity, it provides a constant measure to relate velocity and kinetic energy. Understanding the role of electron mass helps predict how electrons behave under different electric field conditions.