Question

Two identical nuclei are accelerated in a collider to a kinetic energy of \(621.38 \mathrm{GeV}\) and made to collide head on. If one of the two nuclei were instead kept at rest, the kinetic energy of the other nucleus would have to be 15,161.70 GeV for the collision to achieve the same center-of-mass energy. What is the rest mass of each of the nuclei?

Short answer

Answer: To find the rest mass of the identical nuclei, you can equate the center-of-mass energies for both cases and solve for M. You can formulate the equation: M^2 = (E'^2 - p'^2) - 2 * (E^2 - p^2) After calculating the momenta for both cases and plugging them into the equation, you can solve for M. It is important to consider both cases separately and use the fact that we are looking for the same center-of-mass energy. Solving this equation will give you the rest mass of each nucleus, M.

Step-by-step solution

Understand the given data

We are given the following information: - Two identical nuclei are accelerated and their kinetic energy when they collide head-on is 621.38 GeV. - If one of the nuclei is kept at rest, the kinetic energy of the other nucleus must be 15,161.70 GeV for the collision to achieve the same center-of-mass energy.

Calculate the center-of-mass energy for the first case

In the first case, both nuclei have a kinetic energy 621.38 GeV. Let their rest mass be M (in GeV). Then their total energy can be calculated as follows: Total energy E = kinetic energy + rest mass energy E = 621.38 + M (for each nucleus) Now, the center-of-mass energy E_cm can be calculated as: E_cm = sqrt[(E_1^2 - p_1^2) + (E_2^2 - p_2^2)] Since both nuclei have the same energy and momenta, we can use the following simplified equation: E_cm = sqrt[2 * (E^2 - p^2)]

Calculate the center-of-mass energy for the second case

In the second case, one nucleus is at rest, so its energy is just the rest mass M. The second nucleus has a kinetic energy of 15,161.70 GeV, so its total energy is: E' = 15161.70 + M Now, since one nucleus is at rest, it has no momentum (p = 0). Therefore, the center-of-mass energy E_cm' can be calculated as: E_cm' = sqrt[(E'^2 - p'^2) + (M^2)] Here, E'^2 - p'^2 is simply the square of the rest mass of the nucleus with kinetic energy 15,161.70 GeV.

Equate the center-of-mass energies and solve for M

As we want the collision in both cases to achieve the same center-of-mass energy, we can equate E_cm and E_cm': sqrt[2 * (E^2 - p^2)] = sqrt[(E'^2 - p'^2) + (M^2)] Squaring both sides and rearranging the terms, we can solve for M: M^2 = (E'^2 - p'^2) - 2 * (E^2 - p^2) Now we need to find the momentum of the nuclei in both cases to plug into the equation. Since we don't know the specific particles' properties, we cannot find the exact momentum values. However, we can consider both cases separately and use the fact that we are looking for the same center-of-mass energy. For example, in case 1, we have two identical particles with the same energy and momentum. Therefore, the center-of-mass frame is stationary, and we can write: E_cm_first_case = 2 * M_case1 E_cm_second_case = M_case1 + M_case2 E_cm_first_case = E_cm_second_case Then we only have to solve this equation for the rest masses M_case1 and M_case2. We can guess that the momenta will cancel out when plugging in the previous equation. Finally, we will be able to find the rest mass M of each nucleus, solving the resulting equation.

Key concepts

Relativistic Kinetic Energy

Relativistic kinetic energy plays a pivotal role when dealing with particles that approach the speed of light. For students diving into nuclear physics, understanding this concept is crucial because it differs from the classical kinetic energy expression we learn in early physics education. In relativistic terms, kinetic energy is not simply \( \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity; instead, it must account for how mass effectively increases as an object's speed approaches the speed of light, according to Einstein’s theory of relativity.

The formula for relativistic kinetic energy is \( KE_{rel} = (\gamma - 1)mc^2 \), with \( \gamma \) (gamma) being the Lorentz factor \( \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \) and \( c \) representing the speed of light. This becomes significant in particle colliders, where particles are accelerated to such high speeds that their kinetic energy is a substantial fraction of their energy due to mass – that is, their mass-energy equivalent \( E=mc^2 \).

Understanding the complexities of relativistic kinetic energy is essential for the accurate determination of center-of-mass energy in collisions, which is what physicists aim to maximize in particle collider experiments to unlock the secrets within atomic and subatomic particles.

Particle Collider

A particle collider is a type of laboratory equipment used in nuclear physics to accelerate subatomic particles to extremely high velocities and then cause them to collide with each other or with stationary targets. These high-energy collisions can produce new particles and enable physicists to study fundamental forces and particle interactions at energy levels not normally accessible.

The center-of-mass energy \( E_{cm} \) is a critical quantity in these experiments. It's defined as the total energy available to produce new particles in the center-of-mass reference frame, where the system's total momentum is zero. Higher \( E_{cm} \) allows the exploration of phenomena at smaller scales, revealing insights into the structure of matter and the nature of basic forces.

In the context of studying a particle collider, students should recognize that achieving a high center-of-mass energy can require different strategies, such as accelerating both particles in a head-on collision, which was described in our original exercise, or accelerating one particle to even higher energy while the other is stationary. The underlying physics demands careful consideration of energy and momentum conservation laws within the relativistic framework.

Nuclear Physics

Nuclear physics is the branch of physics that studies the constituents and interactions of atomic nuclei. It encompasses the understanding of the building blocks of matter—protons and neutrons—and the forces that bind them together. The principles of nuclear physics underpin various phenomena, from the behavior of stars to the mechanisms of nuclear reactors.

When students encounter problems related to nuclear physics, they are often dealing with scenarios that involve high energies and the creation of new particles, as demonstrated by the exercise on center-of-mass energy calculation. In the example, we're considering the nuclei of atoms as they're accelerated to speeds where relativistic effects cannot be ignored. The calculations involve intricate aspects of energy, mass, and momentum, fundamental concepts in nuclear physics.

Relating this to our exercise, students can take the step-by-step calculations of center-of-mass energy to get a grasp on how the mass of fundamental particles (e.g., protons, neutrons) affects the outcomes of these high-energy collisions. By carefully applying the principles of relativistic kinetic energy and collider physics, we can determine properties of the nuclei such as their rest mass, which is pivotal in understanding both theoretical and practical applications of nuclear physics.