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

Short Answer

Expert verified
The Q-value of the reaction is 8.119 MeV.

Step by step solution

01

Understand the Reaction

The given nuclear reaction is \( p + \ ^{19}F \rightarrow \ ^4He + \ ^{16}O \), where \( p \) is a proton, \( ^{19}F \) is fluorine, \( ^4He \) (represented as \( a \)) is an alpha particle, and \( ^{16}O \) is oxygen. The task is to find the energy released (\( Q \) value) during this reaction.
02

Write the Q-value Formula

The Q-value of a reaction is calculated using the formula: \[ Q = (m_{ ext{initial}} - m_{ ext{final}}) \times c^2 \] where \( m_{ ext{initial}} \) is the total mass of the reactants, and \( m_{ ext{final}} \) is the total mass of the products. For this reaction, the Q-value becomes: \[ Q = ((m(p) + m(F)) - (m(a) + m(O))) \times c^2 \] We know \( c^2 \) is converted into MeV using the conversion factor \( 1 \,u = 931.5 \, ext{MeV/c}^2 \).
03

Calculate Initial Mass

Calculate the total mass of the reactants (initial mass): \[ m_{ ext{initial}} = m(p) + m(F) = 1.007825 \, u + 18.998405 \, u = 20.006230 \, u \]
04

Calculate Final Mass

Calculate the total mass of the products (final mass): \[ m_{ ext{final}} = m(a) + m(O) = 4.002603 \, u + 15.994915 \, u = 19.997518 \, u \]
05

Subtract and Convert to Energy

Determine the mass difference and convert it to energy (MeV): \[ \Delta m = m_{ ext{initial}} - m_{ ext{final}} = 20.006230 \, u - 19.997518 \, u = 0.008712 \, u \] \[ Q = 0.008712 \, u \times 931.5 \, \text{MeV/u} = 8.119 \, \text{MeV} \]
06

Conclusion

The positive Q-value indicates that the reaction releases energy. The energy released in this reaction is 8.119 MeV.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Q-value calculation
In nuclear reactions, the Q-value provides essential information about the energy dynamics. It represents the amount of energy released or absorbed during the reaction. This is calculated using the fundamental principle of conservation of energy.

To determine the Q-value, you need the initial and final masses involved in the reaction. This involves all reactants and products, respectively. The formula employed is:

\[ Q = (m_{\text{initial}} - m_{\text{final}}) \times c^2 \]

Here, \(m_{\text{initial}}\) is the sum of the masses of the reactants, and \(m_{\text{final}}\) is the sum of the masses of the products. The speed of light squared \(c^2\) acts as a conversion factor from mass units to energy units. In practice, 1 unified atomic mass unit (u) is converted to 931.5 MeV to simplify the calculations in nuclear physics.
  • Positive Q-value: the reaction releases energy, indicating an exothermic process.
  • Negative Q-value: the reaction absorbs energy, indicating an endothermic process.
Understanding the Q-value helps in predicting the feasibility of a reaction and is crucial for applications like nuclear power and medical radioisotopes.
mass-energy equivalence
The concept of mass-energy equivalence is pivotal in understanding the energy changes in nuclear reactions. This principle, famously formulated by Albert Einstein as \( E = mc^2 \), shows that mass can be converted into energy and vice versa.

In nuclear reactions, tiny amounts of mass are converted into large amounts of energy because of the immense value of \( c^2 \), the speed of light in a vacuum squared. This conversion is the basis for both atomic power generation and the destructive power of nuclear weapons.
  • Mass Defect: In nuclear reactions, when the initial reactants fuse or break apart, the total mass of the products is often different from the reactants due to energy emission or absorption.
  • Energy Release: A small mass discrepancy can lead to significant energy release due to the conversion through \(931.5 \text{ MeV/u}\).
This principle underscores why nuclear reactions are so potent despite involving minuscule masses.
nuclear physics
Nuclear physics is the branch of physics that studies the components and behavior of the atomic nucleus. It aims to understand the strong nuclear forces that hold nucleons—protons and neutrons—together.

Nuclear reactions are the heart of nuclear physics. These reactions involve changes in an atom's nucleus and can release or absorb vast amounts of energy compared to chemical reactions. The fields of nuclear power, nuclear medicine, and particle physics are all grounded on principles derived from nuclear physics.
  • Radioactive Decay: This natural process involves the transformation of an unstable atomic nucleus into a more stable one over time, emitting energy in the form of radiation.
  • Nuclear Fission: A nuclear reaction where a heavy nucleus splits into lighter nuclei, often used in nuclear reactors and weapons.
  • Nuclear Fusion: A reaction where light nuclei combine to form a heavier nucleus, which powers the sun and holds potential for future energy solutions.
These concepts illustrate the fundamental interactions taking place at the nuclear level, driving innovations and knowledge in energy production and beyond.

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