Question

A proton (nucleus of the hydrogen atom) is being accelerated in a linear accelerator. In each stage of such an accelerator the proton is accelerated along a straight line at \(3.60 \times 10^{15} \mathrm{~m} / \mathrm{s}^{2}\). If a proton enters such a stage moving initially with a speed of \(2.40 \times 10^{7} \mathrm{~m} / \mathrm{s}\) and the stage is \(3.50 \mathrm{~cm}\) long, compute \((a)\) its speed at the end of the stage and \((b)\) the gain in kinetic energy resulting from the acceleration. The mass of the proton is \(1.67 \times 10^{-27} \mathrm{~kg} .\) Express the energy in electron-volts.

Short answer

The proton's speed at the end of the stage is \(3.00 \times 10^{7} \mathrm{~m/s}\) and the gain in kinetic energy is 552 eV.

Step-by-step solution

Identify Given Values

The given values are Initial velocity (\(v_i\)) = \(2.40 \times 10^{7} \mathrm{~m} / \mathrm{s}\), Acceleration (\(a\)) = \(3.60 \times 10^{15} \mathrm{~m} / \mathrm{s}^{2}\), distance (\(d\)) = 3.50 cm = \(0.035 m\) (converted from cm to m), Mass of proton (\(m\)) = \(1.67 \times 10^{-27} kg\).

Calculate Final Speed

The final velocity (\(v_f\)) can be found using the kinematic equation \(v_f^2 = v_i^2 + 2ad\). We substitute the values: \(v_f^2 = (2.40 \times 10^{7})^2 + 2 \times (3.60 \times 10^{15}) \times 0.035\). Solving, we get \(v_f = 3.00 \times 10^{7} \mathrm{~m/s}\).

Calculate Gain in Kinetic Energy (in Joules)

The gain in kinetic energy can be calculated as: \(ΔK = 1/2 m v_f^2 - 1/2 m v_i^2\). Substituting the given values: \(ΔK = 1/2 \times (1.67 \times 10^{-27}) \times (3.00 \times 10^{7})^2 - 1/2 \times (1.67 \times 10^{-27}) \times (2.40 \times 10^{7})^2\). Solving, we get \(ΔK = 8.84 \times 10^{-14} \mathrm{~J}\).

Convert Energy from Joules to Electron Volts

The energy \(E\) in electron volts is obtained from the energy \(ΔK\) in joules through the conversion: \(E = ΔK /(1.6 \times 10^{-19})\). Substituting the value, \(E = 8.84 \times 10^{-14} /(1.6 \times 10^{-19})\). Simplifying, we get \(E = 552 \mathrm{~eV}\).

Key concepts

Kinematic Equations

Kinematic equations are the cornerstone for understanding motion in physics. They help us to calculate and predict the position, velocity, and acceleration of an object over time. One such equation is \( v_f^2 = v_i^2 + 2ad \), where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( d \) is the distance covered.

In our proton acceleration problem, we know the initial velocity of the proton, the acceleration it undergoes in the accelerator, and the distance of the stage. By plugging these values into the kinematic formula, we determine the final speed of the proton after it has been accelerated. This method is indispensable in physics as it gives us a clear view of how the proton moves within the accelerator without demanding complex equipment to measure the final speed directly.

Proton Acceleration

Proton acceleration is a process central to the operation of a linear accelerator. Linear accelerators, or 'linacs,' are devices used in physics experiments and medical treatments to accelerate charged particles like protons along a linear path. The acceleration is typically uniform, meaning the proton experiences a constant increase in velocity as it moves through the accelerator.

This acceleration can be described using Newton's second law, \( F=ma \), where force \( F \) acting on the proton with mass \( m \) results in an acceleration \( a \). In our exercise, the proton increases its speed thanks to the electromagnetic fields within the linear accelerator stages, which push it forward, causing the proton to gain energy and velocity with each section it passes through.

Kinetic Energy Calculation

Kinetic energy is the energy that an object possesses due to its motion. It can be calculated with the formula \( K = 1/2 mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. When it comes to subatomic particles like protons, the kinetic energy quantifies how much work is needed to accelerate the proton from rest to a given velocity.

In the solution to our problem, we calculate the change in kinetic energy (\(\Delta K\)) by finding the difference between the final and initial kinetic energies. This reveals how much energy the proton has gained as a result of being accelerated through the linac stage. Understanding this energy gain is crucial in both experimental particle physics and in applications such as radiation therapy where precise energy control is necessary for treatment effectiveness.

Energy Conversion Electron Volts

The electron volt (eV) is a unit of energy commonly used in the field of atomic and particle physics to denote the amount of kinetic energy gained or lost by an electron when it moves through an electric potential difference of one volt. It is a convenient unit because the energies involved in subatomic processes are often very small, and using joules would result in cumbersome numbers. The conversion factor is 1 eV = 1.6 x 10^-19 joules.

In our proton acceleration problem, once we have the proton's gain in kinetic energy in joules, we convert it to electron volts to express the energy in a scale that is more meaningful in the context of particle physics. This unit of measurement provides a universal language for physicists worldwide, allowing them to compare and communicate their results effectively. The conversion process is straightforward and requires only the division of the energy quantity in joules by the charge of an electron in coulombs.