91Ó°ÊÓ

The following fusion reaction takes place in the Sun and furnishes much of its energy: $$ 4_{1}^{1} \mathrm{H} \rightarrow 4{ }_{2}^{4} \mathrm{He}+2_{+1}^{0} e+\text { energy } $$ where \(_{+1}^{0} e\) is a positron electron. How much energy is released as \(1.00 \mathrm{~kg}\) of hydrogen is consumed? The masses of \({ }^{1} \mathrm{H},{ }^{4} \mathrm{He}\), and \({ }_{+1}^{0} e\) are, respectively, 1.007825, 4.002604, and \(0.000549 \mathrm{u}\), where atomic electrons are included in the first two values. Ignoring the electron binding energy, the mass of the reactants, 4 protons, is 4 times the atomic mass of hydrogen \(\left({ }^{1} \mathrm{H}\right)\), less the mass of 4 electrons: $$ \begin{aligned} \text { Reactant Mass } &=(4)(1.007825 \mathrm{u})-4 m_{e} \\ &=4.031300 \mathrm{u}-4 m_{e} \end{aligned} $$ where \(m_{e}\) is the mass of the electron (or positron). The reaction products have a combined mass $$ \begin{aligned} \text { Product mass } &=\left(\text { Mass of }{ }_{2}^{4} \mathrm{He} \text { nucleus }\right)+2 m_{e} \\ &=\left(4.002604 \mathrm{u}-2 m_{e}\right)+2 m_{e} \\ &=4.002604 \mathrm{u} \end{aligned} $$ The mass loss is therefore (Reactant mass) \(-\) (Product mass) \(=\left(4.0313 \mathrm{u}-4 m_{e}\right)-4.0026 \mathrm{u}\) Substituting \(m_{e}=0.000549\) u gives the mass loss as \(0.0265 \mathrm{u}\). But \(1.00 \mathrm{~kg}\) of \({ }^{1} \mathrm{H}\) contains \(6.02 \times 10^{26}\) atoms. For each four atoms that undergo fusion, \(0.0265 \mathrm{u}\) is lost. The mass lost when \(1.00 \mathrm{~kg}\) undergoes fusion is therefore $$ \begin{aligned} \text { Mass } \operatorname{los} / / \mathrm{kg} &=(0.0265 \mathrm{u})\left(6.02 \times 10^{26} / 4\right)=3.99 \times 10^{24} \mathrm{u} \\\ &=\left(3.99 \times 10^{24} \mathrm{u}\right)\left(1.66 \times 10^{-27} \mathrm{~kg} / \mathrm{u}\right)=0.00663 \mathrm{~kg} \end{aligned} $$ Then, from the Einstein relation, \(\Delta \mathrm{E}=(\Delta m) \mathrm{c}^{2}=(0.00663 \mathrm{~kg})\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)^{2}=5.96 \times 10^{14} \mathrm{~J}\)

Step by step solution

01

Determine Mass Difference per Reaction

Calculate the mass loss for one fusion reaction: \[\Delta m = \text{Reactant mass} - \text{Product mass}\]Substitute the values:\[\Delta m = \left(4.0313\, \mathrm{u} - 4 m_{e}\right) - 4.0026\, \mathrm{u} = 0.0265\, \mathrm{u}\]This shows the mass is reduced by 0.0265 atomic mass units (u) per reaction.

02

Convert Mass Loss to Kilograms

Convert the mass loss from atomic mass units to kilograms. We know that 1 atomic mass unit (u) is approximately \(1.66 \times 10^{-27}\) kg. Hence, the mass loss for each reaction in kg is:\[\Delta m = 0.0265\, \text{u} \times 1.66 \times 10^{-27}\, \text{kg/u} = 4.4039 \times 10^{-29}\, \text{kg}\]

03

Calculate Total Mass Loss for 1 kg of Hydrogen

There are \(6.02 \times 10^{26}\) hydrogen atoms in 1 kg. Since the reaction involves 4 hydrogen atoms, the number of reactions is:\[\frac{6.02 \times 10^{26}}{4} = 1.505 \times 10^{26}\, \text{reactions}\]The total mass loss is:\[\text{Total mass loss} = 1.505 \times 10^{26} \times 4.4039 \times 10^{-29}\, \text{kg} = 0.00663\, \text{kg}\]

04

Calculate Energy Released using Einstein's Equation

Use Einstein's relation \(\Delta E = (\Delta m) c^2\) to calculate the energy released. The speed of light \(c\) is \(2.998 \times 10^8\, \text{m/s}^2\).\[\Delta E = 0.00663\, \text{kg} \times (2.998 \times 10^8\, \text{m/s})^2\]\[\Delta E = 5.957 \times 10^{14}\, \text{J}\]Hence, this is the energy released by converting 1 kg of hydrogen to helium.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 of physics, articulated by Albert Einstein in his famous equation, \( E = mc^2 \). This equation shows that mass (\( m \)) and energy (\( E \)) are interchangeable; in other words, mass can be converted into energy and vice versa. In the context of nuclear fusion, this principle is crucial because it explains how energy is released during fusion reactions. When light nuclei, such as hydrogen, combine to form a heavier nucleus, the mass of the resulting nuclei is less than the sum of the masses of the separate particles. This lost mass is converted into energy, which is released during the reaction.

• This energy release is the driving force behind the energy production in stars, including our Sun.
• The energy produced can be calculated by determining the mass lost and applying the mass-energy equivalence equation.

Understanding \( E = mc^2 \) helps in grasping why even a tiny amount of mass can release an enormous amount of energy, emphasizing the enormous potential of nuclear fusion power.

Atomic Mass Units

An atomic mass unit (amu) is a unit of mass used to express atomic and molecular weights. It is defined as one twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state.Using atomic mass units simplifies complex nuclear calculations. In nuclear reactions, the masses of individual particles and nuclei are often very small, and expressing them in atomic mass units allows for easier handling and understanding.

• In the context of nuclear fusion, the mass of reactants and products is often measured in atomic mass units.
• The slight difference in mass before and after the reaction (mass defect) directly corresponds to the amount of energy released.

When these mass differences are translated into kilograms (using the conversion \( 1 \, \text{amu} = 1.66 \times 10^{-27} \, \text{kg} \)), it's possible to calculate the total energy released using Einstein’s equation. This move from atomic to macroscopic scales is essential for practical fusion calculations.

Hydrogen Fusion

Hydrogen fusion is the process that powers stars, including our Sun, through the fusion of hydrogen nuclei to form helium. Fusion in the Sun primarily involves the conversion of four hydrogen nuclei into a single helium atom. This process occurs under immense heat and pressure in the Sun's core, allowing the hydrogen nuclei to overcome their natural repulsion to each other. Here’s how hydrogen fusion happens:

• Four protons (hydrogen nuclei) are needed to start the process.
• The protons merge through a series of steps to form helium, positrons, and energy is released.
• The net loss of mass in this process is what gets converted into energy, illustrating mass-energy equivalence.

Hydrogen fusion is considered a potential energy source on Earth because it produces vast amounts of energy with minimal, non-radioactive waste. However, replicating the necessary conditions on Earth is highly challenging and currently the focus of considerable research into clean and sustainable energy sources.