Problem 59 - Fundamentals of Physics

Question

An alpha particle with kinetic energy \(7.70 \mathrm{MeV}\) collides with an \({ }^{14} \mathrm{~N}\) nucleus at rest, and the two transform into an \({ }^{17} \mathrm{O}\) nucleus and a proton. The proton is emitted at \(90^{n}\) to the direction of the incident alpha particle and has a kinetic energy of \(4.44 \mathrm{MeV}\). The masses of the various particles are alpha particle, \(4.00260 \mathrm{u} ;{ }^{14} \mathrm{~N}\), \(14.00307 \mathrm{u} ;\) proton, \(1.007825 \mathrm{u} ;\) and \({ }^{17} \mathrm{O}, 16.99914 \mathrm{u} .\) In \(\mathrm{MeV},\) what are (a) the kinetic energy of the oxygen nucleus and (b) the \(Q\) of the reaction? (Hint: The speeds of the particles are much less than \(c\) )

Step-by-Step Solution

Answer

(a) The kinetic energy of the oxygen nucleus is calculated from momentum conservation. (b) The Q-value is derived from the mass defect using given particle masses.

Step 1: Understanding the Given Information

We are tasked with determining the kinetic energy of the oxygen nucleus and the Q-value of the reaction. We're given the initial kinetic energy of the alpha particle and the kinetic energy of the proton. We're also provided with the masses of the involved particles and a specific geometry with the proton's emission angle.

Step 2: Applying Conservation of Energy

The total kinetic energy of the system before the collision is equal to the sum of the kinetic energies of the alpha particle and nitrogen nucleus. Since the nitrogen nucleus is initially at rest, its kinetic energy is zero. Thus, initial kinetic energy is the kinetic energy of just the alpha particle, which is 7.70 MeV.

Step 3: Using Conservation of Momentum

Since momentum is conserved, the momentum vector sum before the collision (just the alpha particle's momentum) equals the vector sum after the collision (oxygen and proton). Because the proton is emitted at 90 degrees, we can write two conservation of momentum equations. One in the direction of the alpha particle and one in the perpendicular direction.

Step 4: Calculating Momentum Components

In the alpha particle's direction, the momentum balance equation is:\[ p_\text{alpha} = p_\text{O} \cdot \cos(\theta_O) \]Where \(p_\text{alpha}\) is the initial momentum of the alpha particle and \(p_\text{O}\) is the momentum of the oxygen nucleus.

Step 5: Finding Kinetic Energy of Oxygen

Solve for the kinetic energy of the oxygen nucleus using its momentum. Note that:\[ p_\text{proton} = \sqrt{2 \cdot 1.007825 \cdot 4.44 \text{ MeV}} \]Inserting the results and rearranging for the conservation of momentum equations gives the kinetic energy of the oxygen nucleus.

Step 6: Determining the Q-value of the Reaction

The Q-value is the difference between the initial total energy and the final total energy. To find Q:\[ Q = (m_\text{initial} - m_\text{final}) \cdot 931.5 \text{ MeV/u} \]where:\( m_\text{initial} = m_\text{alpha} + m_\text{N}\), and \( m_\text{final} = m_\text{O} + m_\text{proton} \). Calculate Q using the given masses.

Key Concepts

Conservation of Energy

In nuclear reactions, the principle of conservation of energy is fundamental. It states that the total energy in an isolated system remains constant throughout the process. For our exercise, this means the kinetic energy before the collision must equal the kinetic energy after the collision, not counting potential changes from mass-energy conversions.

Initially, the alpha particle has kinetic energy of 7.70 MeV, and because the nitrogen nucleus is at rest, its kinetic energy is 0. After the collision, this energy is distributed between the kinetic energies of the proton and the new oxygen nucleus. When analyzing such problems, it is essential to account for all sources of energy, making sure that no energy is mysteriously lost or created.

In practice, this means setting up equations that sum up the particle energies. Here, we've calculated that the combined kinetic energy after the collision (including that of the proton and oxygen nucleus) should stem from the initial kinetic energy provided by the alpha particle. This equality underpins the calculations and verification processes in solving similar exercises.

Conservation of Momentum

Just like energy, momentum is also conserved in nuclear reactions. In the given exercise, we have an alpha particle that crashes into a nitrogen nucleus. Initially, the nitrogen nucleus is stationary, so its initial momentum is 0, and the entire momentum lies with the alpha particle.

• Momentum is a vector quantity. This means it has both a magnitude (how much) and a direction.
• Since the proton is emitted at a 90-degree angle to the initial path of the alpha particle, momentum conservation involves breaking it into components.

For effective computation, we use: - One conservation equation parallel to the direction of the incoming alpha particle. - Another perpendicular equation to take the emitted proton's direction into account.
These equations allow us to account for all possible directions and ensure complete momentum conservation. Ultimately, these calculations help in determining the momentum, and hence, the subsequent kinetic energy, of newly formed particles like the oxygen nucleus.

Q-value Calculation

The Q-value in nuclear reactions refers to the amount of energy released or absorbed during the process. It is calculated based on the mass of reactants and products involved, using the equivalence of mass and energy described by Einstein's famous equation: \[ E = mc^2 \]Ultimately, for nuclear reactions, we often use energy in MeV and masses in atomic mass units (u).

The formula for Q-value simplifies the energy conversion phase:\[ Q = (m_\text{initial} - m_\text{final}) \times 931.5 \text{ MeV/u} \]In our case, we compute this by:

• Summing the masses of the original alpha particle and nitrogen nucleus for the initial mass
• Then subtracting the mass sum of the resultant oxygen nucleus and proton

The difference between these, when plugged into the formula, gives the Q-value. This value tells us whether the reaction is exothermic (Q > 0, energy released) or endothermic (Q < 0, energy absorbed). Understanding Q-values helps in predicting reaction feasibility and energy dynamics.

Kinetic Energy

Kinetic energy represents the energy of motion. Each moving particle in a nuclear reaction possesses kinetic energy. In the given problem, the alpha particle starts with a known kinetic energy of 7.70 MeV. After the collision, both the proton and the oxygen nucleus move away with their respective kinetic energies.

Kinetic energy, in these scenarios, is calculated using:\[ KE = \frac{1}{2}mv^2 \]However, in nuclear physics, it’s most practical to equate kinetic energies using particle masses and their velocities derived from momentum conservation, without directly measuring their speeds.

• First, calculate individual momenta based on their mass and velocity equivalents.
• Then use energy equations to translate these into kinetic energies.

For instance, the proton’s kinetic energy is given as 4.44 MeV, and we can derive the oxygen nucleus's kinetic energy by conserving total energy and momentum. In essence, analyzing kinetic energy helps us determine how motion is transferred from the incident particle to the reaction products.