Problem 44 - Fundamentals of Physics

Question

In the reaction \(\mathrm{p}+{ }^{19} \mathrm{~F} \rightarrow \alpha+{ }^{16} \mathrm{O}\), the masses are $$ \begin{array}{ll} m(\mathrm{p})=1.007825 \mathrm{u}, & m(\alpha)=4.002603 \mathrm{u}, \\ m(\mathrm{~F})=18.998405 \mathrm{u}, & m(\mathrm{O})=15.994915 \mathrm{u} . \end{array} $$ Calculate the \(Q\) of the reaction from these data.

Step-by-Step Solution

Answer

The \(Q\) value of the reaction is 8.115 MeV.

Step 1: Understand the Reaction Equation

The given reaction is \( \mathrm{p}+{ }^{19} \mathrm{F} \rightarrow \alpha+{ }^{16} \mathrm{O} \). This means a proton \((\mathrm{p})\) and a fluorine nucleus \(({}^{19}\mathrm{F})\) react to produce an alpha particle \((\alpha)\) and an oxygen nucleus \(({}^{16}\mathrm{O})\).

Step 2: Understand Q Value

The \(Q\) value of a reaction is the energy released or absorbed during the reaction. It can be calculated by the formula \( Q = (m_\text{initial} - m_\text{final}) \times 931.5 \frac{\text{MeV}}{\text{u}} \), where \(m_\text{initial}\) and \(m_\text{final}\) are the total masses of reactants and products, respectively.

Step 3: Calculate Initial Mass

Add the masses of the reactants: \[m_{\text{initial}} = m(\mathrm{p}) + m({}^{19}\mathrm{F}) = 1.007825\, \text{u} + 18.998405\, \text{u} = 20.006230\, \text{u}.\]

Step 4: Calculate Final Mass

Add the masses of the products: \[m_{\text{final}} = m(\alpha) + m({}^{16}\mathrm{O}) = 4.002603\, \text{u} + 15.994915\, \text{u} = 19.997518\, \text{u}.\]

Step 5: Compute the Mass Difference

Calculate the mass difference, which is \( m_{\text{initial}} - m_{\text{final}} \):\[\Delta m = 20.006230\, \text{u} - 19.997518\, \text{u} = 0.008712\, \text{u}.\]

Step 6: Calculate Energy Released (Q value)

Use the mass difference to calculate the \(Q\) value:\[Q = \Delta m \times 931.5 \frac{\text{MeV}}{\text{u}} = 0.008712\, \text{u} \times 931.5 \frac{\text{MeV}}{\text{u}} = 8.115\, \text{MeV}.\]

Step 7: Conclusion

The \(Q\) value of the reaction, which is the energy released, is \(8.115\) MeV.

Key Concepts

Mass-Energy Equivalence

The concept of mass-energy equivalence is fundamental in understanding nuclear reactions. This principle is most famously summarized by Einstein's equation \( E = mc^2 \). It implies that mass can be converted into energy and vice versa. In nuclear physics, when a reaction occurs, the mass of the reactants usually differs from the mass of the products. This mass difference is known as the mass defect and is directly related to the energy released or absorbed in the reaction.

In nuclear reactions, even a small mass change can result in significant energy changes due to the multiplication factor \(c^2\), where \(c\) is the speed of light. This conversion is why nuclear fission and fusion release tremendous amounts of energy. Nuclear energetics often focus on calculating these energy changes to understand and predict the behavior of nuclear materials.

Understanding this equivalence is crucial for understanding the energy dynamics in nuclear reactions. It's why nuclear reactions can power everything from stars to nuclear reactors on Earth.

Q Value Calculation

The \(Q\) value of a nuclear reaction is a way to quantify the energy change during the reaction. It tells us whether energy is released or absorbed. You can calculate the \(Q\) value using the formula \( Q = (m_\text{initial} - m_\text{final}) \times 931.5 \frac{\text{MeV}}{\text{u}} \). In this formula, \(m_\text{initial}\) and \(m_\text{final}\) are the sums of the masses of the initial and final particles involved in the reaction.

To determine the \(Q\) value, you first calculate the total mass of reactants and the total mass of products. The mass difference reveals how much mass was converted to energy.

• If the \(Q\) value is positive, it means energy is released (exothermic reaction), which is a typical characteristic of many natural nuclear reactions.
• If the \(Q\) value is negative, energy is absorbed (endothermic reaction), indicating that energy must be supplied for the reaction to occur.

This concept is vital for applications like nuclear power generation, where maximizing the energy released is desired for efficiency.

Nuclear Physics

Nuclear physics focuses on the components and behavior of atomic nuclei. It explores how particles like protons and neutrons interact within an atom. Nuclear reactions study these interactions, including how they can transform one element into another and release or absorb energy.

A fundamental aspect of nuclear physics is understanding the forces that hold the nucleus together, as well as the conditions under which a nucleus may become unstable and decay. It's these reactions that are harnessed in nuclear energy applications and in medical applications like radiotherapy.

• Elementary particles like protons, neutrons, and electrons are studied extensively in this field.
• It explores the energy levels within a nucleus that lead to reactions like fission, fusion, and radioactive decay.

Nuclear physics discoveries have had profound impacts on technology and our understanding of the universe, from the Big Bang to the development of modern medical imaging techniques.