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 \(90^{\circ}\) 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, \(4.00260 \mathrm{u} ;{ }^{14} \mathrm{~N}\). \(14.00307\) 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 3.26 MeV. (b) The Q of the reaction is -0.02 MeV.

Step 1: Identify Known Quantities

We know the kinetic energy of the incident alpha particle is \(7.70\, \mathrm{MeV}\) and the kinetic energy of the proton is \(4.44\, \mathrm{MeV}\). The masses of the particles are given as: alpha particle = \(4.00260 \mathrm{u}\), \(^{14} \mathrm{N}\) = \(14.00307\, \mathrm{u}\), proton = \(1.007825\, \mathrm{u}\), and \(^{17} \mathrm{O}\) = \(16.99914\, \mathrm{u}\).

Step 2: Conserve Momentum

Since the proton is emitted at \(90^{\circ}\), the alpha particle and \(^{14}\mathrm{N}\) nucleus initially move along the same line. By conservation of momentum for two particles in one direction, the initial momentum of the system equals the final momentum parallel to the alpha particle's direction.- Initial momentum, \( p_i = m_{\alpha}v_{\alpha} \)- Final momentum consists of the momentum of \(^{17}\mathrm{O}\) and the proton, which is zero in the alpha direction because they separate at a right angle.To find the initial speed of the alpha particle, we can calculate using its kinetic energy:\[ K_{\alpha} = \frac{1}{2} m_{\alpha} v_{\alpha}^2 \]Thus, the momentum of the alpha particle is \(p_{\alpha} = m_{\alpha}v_{\alpha}\).

Step 3: Solve for Momentum Components

For this right-angle scenario:- The momentum of the proton \(p_p = \sqrt{2 m_p K_p}\) is perpendicular to \(v_{\alpha}\).- The oxygen's momentum \(p_O\) can be calculated using the law of conservation of momentum, where the momentum component of \(^{17} \mathrm{O}\) will match that of the proton because they move perpendicular to each other: \[ p_O = \sqrt{p_{\alpha}^2 - p_p^2} \]

Step 4: Calculate Kinetic Energy of Oxygen

The kinetic energy of the oxygen nucleus \(K_O\) can be derived using the calculated momentum:\[ K_O = \frac{p_O^2}{2 m_O} \]Calculate \(p_O\) using the momentum components derived above, substitute it back here to find the kinetic energy of the oxygen nucleus.

Step 5: Calculate Q Value of Reaction

The Q value for the reaction is the net change in the system's energy, calculated from:\[ Q = (m_{\alpha} + m_{N} - m_{O} - m_p)c^2 \]Since energies are provided in \(\mathrm{MeV}\) and using mass-energy conversion,\[ Q = (\text{initial total rest mass energy} - \text{final total rest mass energy}) \].

Step 6: Plug in Values and Solve

Compute the kinetic energy of \(^{17}\mathrm{O}\) and the Q-value using the above formulas and given data. For the particle masses:- Convert these energies using: \(1 \mathrm{u} = 931.5 \mathrm{MeV/c^2}\).Solve to find the kinetic energy of \(^{17} \mathrm{O}\) and substitute into the Q-value expression.

Key Concepts

Alpha Particle

An alpha particle is a type of ionizing radiation ejected by certain radioactive materials. It consists of two protons and two neutrons, making it identical to the nucleus of a helium atom. These particles are positively charged and have substantial mass compared to other forms of radiation, such as beta particles and gamma rays. Alpha particles do not travel far because they are heavy and can be stopped by a sheet of paper or skin. However, they cause significant damage if they enter the body, due to their potential to ionize atoms in living cells.
In nuclear reactions like the one described in the exercise, an alpha particle can be used to initiate a transformation. In this case, the particle collides with a nitrogen nucleus to produce an oxygen nucleus and a proton. Understanding the properties of alpha particles, such as their energy and momentum, is essential for analyzing the outcomes of such nuclear reactions.

Momentum Conservation

In physics, the principle of momentum conservation states that within a closed system, the total momentum remains constant if no external forces act on the system. Momentum is a product of a particle's mass and velocity, expressed as \( p = mv \).
During the alpha particle and nitrogen nucleus collision, the system's momentum before and after the interaction must remain equal. Initially, the alpha particle carries a certain momentum towards the nitrogen nucleus, which is at rest. After the collision, the momentum is distributed between the new oxygen nucleus and the emitted proton. Even though particles may separate at right angles, as seen with the proton being ejected at \(90^{\circ}\), momentum must still be conserved overall.
Calculations often involve breaking down the momentum into components: one in the direction of the incident particle and the other perpendicular to it. By using the equation for momentum, you can solve for unknowns, such as the kinematic properties of the newly formed particles.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. It's given by the formula \( K = \frac{1}{2} mv^2 \), where \(m\) is the mass and \(v\) is the velocity of the object.
In the context of the exercise, we start by knowing the kinetic energy of the alpha particle and the emitted proton. This energy is crucial for determining the outcomes of the reaction, such as how the energy is distributed among the new particles. The energy initially supplied by the alpha particle is partly transferred to the proton and partly to the oxygen nucleus, with any remaining energy contributing to other energy forms such as binding energy.
By applying conservation of energy and momentum, you can predict the motion and energy of the transformed oxygen nucleus post-collision. This requires understanding both the kinetic energy of the moving objects and using the mass-energy equivalence to handle any interconversion between the mass and energy.

Q Value

The Q value of a nuclear reaction represents the net change in kinetic energy of the system due to the reaction. It's determined using the difference in mass-energy of the products and reactants and is expressed in energy units, typically MeV.
The equation \( Q = (m_{\text{initial}} - m_{\text{final}}) c^2 \) provides a way to calculate this value, where \(m_{\text{initial}}\) and \(m_{\text{final}}\) are the total rest masses of the initial and final-state particles, respectively. If the Q value is positive, the reaction releases energy and is exothermic; however, if it's negative, the reaction requires energy input and is endothermic.
In our exercise, the Q value tells us how much energy from the rest mass is converted into kinetic energy during the collision. This insight helps physicists understand how effective such reactions might be for applications like nuclear energy generation or particle collision experiments.