/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 In the reaction \(\mathrm{p}+{ }... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The Q-value of the reaction is approximately 8.1 MeV.

Step by step solution

01

Understanding the Problem

In this step, we recognize that we are asked to calculate the Q-value of a nuclear reaction. Q-value represents the net energy change of the reaction and is calculated based on the difference in total mass of reactants and products.
02

Write the Reaction Components

Identify the reactants and products of the given reaction: \( \text{Reactants: p + }^{19} \text{F}\), \( \text{Products: } \alpha + {}^{16}\text{O} \).
03

Calculate Total Mass of Reactants

Sum the masses of the proton and fluorine: \[ m(\text{Reactants}) = m(p) + m(\text{F}) = 1.007825 \text{ u} + 18.998405 \text{ u} = 20.006230 \text{ u} \].
04

Calculate Total Mass of Products

Sum the masses of the alpha particle and oxygen: \[ m(\text{Products}) = m(\alpha) + m(\text{O}) = 4.002603 \text{ u} + 15.994915 \text{ u} = 19.997518 \text{ u} \].
05

Calculate the Q-value

Use the formula \(Q = (m(\text{Reactants}) - m(\text{Products})) \times 931.5 \, \text{MeV/u}\), substituting in the masses we calculated: \[ Q = (20.006230 \text{ u} - 19.997518 \text{ u}) \times 931.5 \, \text{MeV/u} \approx 8.1 \, \text{MeV} \].

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Key Concepts

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

Mass-Energy Equivalence
Mass-Energy Equivalence is a fundamental principle in physics, articulated by Einstein through his famous equation: \(E = mc^2\). This equation states that mass can be converted into energy and vice versa, with the speed of light squared \(c^2\) acting as the conversion factor. This means every particle with mass inherently has an energy equivalence.When applying this in nuclear reactions, the energy change \(Q\) is directly related to the mass difference between the reactants and the products. If a reaction releases energy, the products have less mass than the reactants. Thus, calculating the Q-value involves finding this mass difference and then converting that difference into energy units using 931.5 \(\text{MeV/u}\). This conversion factor simplifies the mass-energy computations, making it easier to express energy changes in the desired unit of MeV.
Alpha Particle
The alpha particle is a crucial component in many nuclear reactions. It consists of 2 protons and 2 neutrons, identical to the nucleus of the helium atom, and is denoted as \(\alpha\).Alpha particles are characterized by their relatively large mass, which amounts to approximately \(4.002603\ \text{u}\). They carry a positive charge, caused by their two protons, and are a common product of nuclear decay processes.In the reaction described, the transformation includes an alpha particle being one of the products, making it essential to know its mass accurately. Understanding the role of alpha particles helps in discerning the dynamics of nuclear reactions involving heavy nuclei.
Proton
Protons are fundamental particles that make up the nucleus of an atom, carrying a positive charge and a mass of around \(1.007825\ \text{u}\). In nuclear reactions, protons can engage in many transformation processes, such as turning into more complex nuclei or as projectiles that initiate reactions.In the given nuclear reaction, a proton is one of the reactants, colliding with a fluorine nucleus. Its conversion into an alpha particle and an oxygen nucleus is part of the transformation, demonstrating how protons contribute to the creation of new elements and isotopes through nuclear interactions.
Energy Conservation in Nuclear Reactions
Energy Conservation in Nuclear Reactions is a pivotal concept, ensuring that energy is neither created nor destroyed in the process of a reaction. Instead, it is transformed from one form to another while obeying this fundamental principle. In nuclear reactions, this principle is represented by the Q-value calculation. Here, the energy equivalence of the mass difference between reactants and products directly shows the energy that's absorbed or released. Positive Q-values indicate energy release, marking the reaction as exothermic. The concept of energy conservation reassures that the total energy before the reaction equals the total after, including kinetic energy changes and energy stored in newly formed particles. This understanding ensures accurate predictions of energy output from any nuclear process.

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Most popular questions from this chapter

Observer \(S\) reports that an event occurred on the \(x\) axis of his reference frame at \(x=3.00 \times 10^{8} \mathrm{~m}\) at time \(t=1.50 \mathrm{~s}\). Observer \(S^{\prime}\) and her frame are moving in the positive direction of the \(x\) axis at a speed of \(0.400 c\). Further, \(x=x^{\prime}=0\) at \(t=t^{\prime}=0\). What are the (a) spatial and (b) temporal coordinate of the event according to \(S^{\prime} ?\) If \(S^{\prime}\) were, instead, moving in the negative direction of the \(x\) axis, what would be the (c) spatial and (d) temporal coordinate of the event according to \(S^{\prime}\) ?

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