Problem 103 - University Physics with Modern Physics

Question

Antineutrino. In beta decay, a nucleus emits an elec- tron. A \(^{210} \mathrm{Bi}\) (bismuth) nucleus at rest undergoes beta decay to \(^{210} \mathrm{Po}\) (polonium). Suppose the emitted electron moves to the right with a momentum of \(5.60 \times 10^{-22} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .\) The \(^{210} \mathrm{Po}\) nucleus, with mass \(3.50 \times 10^{-25} \mathrm{kg}\) , recoils to the left at a speed of \(1.14 \times 10^{3} \mathrm{m} / \mathrm{s}\) . Momentum conservation requires that a second particle, called an antineutrino, must also be emitted. Calculate the magnitude and direction of the momentum of the antineutrino that is emitted in this decay.

Step-by-Step Solution

Answer

The antineutrino's momentum is \(1.61 \times 10^{-22} \; \text{kg} \cdot \text{m/s}\) to the right.

Step 1: Identify Given Values

In the problem, we are given the following values: momentum of the electron, \( p_e = 5.60 \times 10^{-22} \; \text{kg} \cdot \text{m/s} \), speed of the Po nucleus, \( v_{Po} = 1.14 \times 10^{3} \; \text{m/s} \), and mass of the Po nucleus, \( m_{Po} = 3.50 \times 10^{-25} \; \text{kg} \). We need to calculate the momentum of the antineutrino.

Step 2: Calculate Initial Momentum

Since the initial nucleus is at rest, its momentum is 0. Therefore, the total initial momentum of the system is \( P_{i} = 0 \; \text{kg} \cdot \text{m/s} \).

Step 3: Determine Total Final Momentum

In beta decay, the principle of conservation of momentum states that the total initial momentum should equal the total final momentum. Hence, the total final momentum is also \( 0 \; \text{kg} \cdot \text{m/s} \).

Step 4: Calculate Momentum of Polonium Nucleus

The momentum of the recoiling polonium nucleus is calculated using the formula \( p_{Po} = m_{Po} \times v_{Po} \). Substituting the values, \( p_{Po} = 3.50 \times 10^{-25} \; \text{kg} \times 1.14 \times 10^{3} \; \text{m/s} = 3.99 \times 10^{-22} \; \text{kg} \cdot \text{m/s} \).

Step 5: Apply Conservation of Momentum

Applying the conservation of momentum: \( 0 = p_{e} - p_{Po} - p_{antineutrino} \), where \( p_{antineutrino} \) is the momentum of the antineutrino. Rearrange the equation to find \( p_{antineutrino} = p_{e} - p_{Po} \).

Step 6: Substitute Values to Find Antineutrino Momentum

Substitute the known values \( p_{e} = 5.60 \times 10^{-22} \; \text{kg} \cdot \text{m/s} \) and \( p_{Po} = 3.99 \times 10^{-22} \; \text{kg} \cdot \text{m/s} \) into the equation: \( p_{antineutrino} = 5.60 \times 10^{-22} - 3.99 \times 10^{-22} = 1.61 \times 10^{-22} \; \text{kg} \cdot \text{m/s} \).

Step 7: Determine the Direction of Antineutrino Momentum

Since the system's initial momentum was zero, the direction of the antineutrino's momentum must be to the right, opposite to that of the recoiling polonium nucleus, which is to the left. Thus, the direction is to the right.

Key Concepts

Momentum Conservation in Physics

Momentum conservation is a core principle in physics stating that the total momentum of a closed system remains constant if no external forces act upon it. This concept is crucial in understanding interactions such as collisions or, in our case, beta decay processes in nuclear physics. When a system is initially at rest, like our bismuth nucleus, the total initial momentum is zero. Any movement or emission of particles within the system should result in a total final momentum of zero as well. This means the sum of the momenta of the emitted particles and the remaining ones must cancel each other out in both magnitude and direction.

To understand momentum conservation better, think of it like a tug-of-war where both teams pull with equal strength and no one moves. If one team suddenly pulls, the other team is either pulled toward them or must exert an equal force to maintain competition balance. In our beta decay, the polonium nucleus and the electron are akin to the teams, while the antineutrino ensures balance by providing the necessary counteract.

The Beta Decay Process

Beta decay is a fascinating process in nuclear physics where an unstable nucleus releases an electron, transforming into a different element. During this decay, a neutron in the nucleus changes into a proton, thereby emitting a beta particle (electron) and an antineutrino. This transformation alters the element—like changing bismuth to polonium in our case.

The beta decay process is vital for understanding radioactive decay and nuclear stability. As the neutron turns into a proton, the atomic number of the nucleus increases by one, which changes the identity of the atom without altering the nucleon number, as the emitted electron (beta particle) and antineutrino are not part of the nucleus's structure. This decay proceeds under the weak nuclear force, one of the four fundamental forces in nature, showcasing its critical role in governing nuclear reactions.

Fundamentals of Nuclear Physics

Nuclear physics is the field that studies atomic nuclei and their interactions. It encompasses concepts like radioactivity, nuclear decay, and interactions within the atomic nucleus. These interactions are mediated by fundamental forces, primarily the strong and weak nuclear forces. Understanding such concepts is pivotal because nuclear reactions—like beta decay—are not only natural processes but also the foundation for technologies like nuclear energy and medical imaging. In nuclear reactions, energy and momentum conservation principles are always adhered to, helping scientists and engineers predict the behavior of nuclear materials. These predictions allow us to harness nuclear processes for various applications, making nuclear physics an essential area of study.

Physics Problem-Solving Approaches

Physics problem-solving often involves following systematic steps to clarify complex scenarios. It starts with identifying given values and understanding the context of a problem. For instance, knowing the initial conditions and properties like mass, velocity, and momentum were crucial in solving the given exercise about antineutrinos.

• Identify the known and unknown variables.
• Apply relevant physical principles, like momentum conservation.
• Perform calculations step-by-step, substituting known values into equations.
• Check your solution for consistency with initial assumptions and known theories.

This structured approach not only leads to correct solutions but also enhances comprehension of how theoretical concepts apply to real-world scenarios, allowing students to tackle a wide range of scientific problems confidently.

Nuclear Reactions and Their Implications

Nuclear reactions such as beta decay impact both scientific perspectives and practical applications. When a nuclear reaction occurs, it signifies a change in the nucleus's structure, releasing energy and often emitting particles. These changes can lead to the production of energy or the transformation of elements. Understanding nuclear reactions aids in various fields:

• Energy: Fission and fusion reactions are the basis of nuclear power.
• Medicine: Radioactive isotopes are used in imaging and cancer treatments.
• Astronomy: Stellar processes, like those fueling our sun, are nuclear reactions at a grand scale.

Grasping the implications of these reactions helps harness nuclear energy efficiently and safely while expanding scientific knowledge about the universe.