Problem 3 - Physics

Question

Multiple-Concept Example 4 provides useful background for this problem. Point \(\mathrm{A}\) is at a potential of \(+250 \mathrm{~V}\), and point \(B\) is at a potential of \(-150 \mathrm{~V}\). An \(\alpha\) -particle is a helium nucleus that contains two protons and two neutrons; the neutrons are electrically neutral. An \(\alpha\) -particle starts from rest at \(A\) and accelerates toward \(B\). When the \(\alpha\) -particle arrives at \(B\), what kinetic energy (in electron volts) does it have?

Step-by-Step Solution

Answer

The alpha-particle's kinetic energy at point B is 800 eV.

Step 1: Determine the Charge of the Alpha-Particle

An alpha-particle consists of 2 protons and 2 neutrons. Since neutrons are neutral, we only consider the protons. Each proton carries a charge of \( +e \), where \( e = 1.6 \times 10^{-19} \) coulombs. Therefore, the total charge \( q \) of the alpha-particle is \( q = 2 \times e = 3.2 \times 10^{-19} \) C.

Step 2: Calculate the Potential Difference

The potential difference \( \Delta V \) between points A and B is given by:\[ \Delta V = V_B - V_A = (-150 - 250)\, \text{V} = -400\, \text{V} \]

Step 3: Use Energy Conservation Principle

When the alpha-particle moves from A to B, its change in electric potential energy becomes its kinetic energy because it starts from rest. This change is given by \( \Delta U = q \Delta V \). Thus, the kinetic energy \( K \) at B is:\[ K = q \Delta V = (3.2 \times 10^{-19} \, \text{C}) \times (-400 \, \text{V}) \]

Step 4: Calculate Kinetic Energy in Joules

Substitute the values to find the kinetic energy \( K \):\[ K = 3.2 \times 10^{-19} \, \text{C} \times (-400 \, \text{V}) = -1.28 \times 10^{-16} \, \text{J} \]Since we are interested in the magnitude, we consider \( 1.28 \times 10^{-16} \, \text{J} \).

Step 5: Convert Kinetic Energy to Electron Volts

To convert from joules to electron volts (eV), use the conversion factor where 1 eV = \( 1.6 \times 10^{-19} \) J. Thus:\[ K = \frac{1.28 \times 10^{-16} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} = 800 \, \text{eV} \]

Step 6: Conclusion

The kinetic energy of the alpha-particle when it arrives at point B is 800 eV.

Key Concepts

Alpha Particle

An alpha particle, denoted by the Greek letter \( \alpha \), is a type of particle characterized by its composition. It consists of two protons and two neutrons. This combination is essentially the same as a helium nucleus.

Due to the presence of two protons, an alpha particle carries a positive charge. Neutrons do not contribute to the charge as they are neutrally charged. Hence, the total charge of an alpha particle is determined solely by the protons.

Each proton carries a positive charge equivalent to an electron but with opposite sign, which is \(+e\), where \(e\) is approximately \(1.6 \times 10^{-19}\) coulombs. Consequently, the charge of an alpha particle is twice that of a single proton, equaling \(3.2 \times 10^{-19}\) c.

Energy Conservation

The conservation of energy is a fundamental principle in physics. It states that the total energy of an isolated system remains constant over time. In simpler terms, energy can neither be created nor destroyed; it can only be transferred from one form to another.

When discussing an alpha particle moving from point A to point B, we focus on energy transformation. Initially, the alpha particle possesses electric potential energy due to its position in the electric field.

• As it moves from rest at a higher potential point (A) to a lower potential point (B), this potential energy transforms into kinetic energy.
• Since it starts from rest, the energy it gains manifests entirely as kinetic energy upon reaching point B.

The kinetic energy acquired by the particle can be calculated using the change in potential energy as per the equation:\[ \Delta U = q \Delta V \]where \( \Delta U \) represents the change in potential energy, \( q \) the charge of the particle, and \( \Delta V \) the potential difference.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. For the alpha particle moving in an electric field, kinetic energy (\( K \)) is acquired by converting electric potential energy.

The formula to calculate kinetic energy resulting from an electric potential difference is:\[ K = q \Delta V \]This formula highlights that the kinetic energy depends directly on the charge of the particle \( q \) and the potential difference \( \Delta V \).

From the given exercise, we see that as the alpha particle moves from point A to point B, the total kinetic energy it gains is due to the potential difference between these points. For an alpha particle, it ends up being significant because of its relatively large charge.

Potential Difference

Potential difference, often referred to as voltage, indicates the difference in electric potential between two points in an electric field. It represents the work done per unit charge to move a charge between these points.

In this exercise, point A has a potential of \( +250 \) volts, while point B has a potential of \( -150 \) volts. The potential difference \( \Delta V \) is calculated as:\[ \Delta V = V_B - V_A \]Plugging in the values:\[ \Delta V = -150 - 250 = -400 \text{ volts} \]

A positive or negative potential difference signifies the direction of energy transfer. Here, a negative value indicates that as the alpha particle moves from A to B, it loses potential energy, which is converted into kinetic energy, increasing the particle's velocity.