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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 \({ }_{+}{ }^{0} e\) 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 $$ \begin{aligned} \text { Product mass } &=\left(\text { Mass of }_{2}^{4} \text { He nucleus }\right)+2 m_{e} \\ &=\left(4.002604 \mathrm{u}-2 m_{e}\right)+2 m_{e} \\ &=4.002604 \mathrm{u} \end{aligned} $$ The mass loss is therefore $$ (\text { Reactant mass })-(\text { Product mass })=\left(4.0313 \mathrm{u}-4 m_{e}\right)-4.0026 \mathrm{u} $$ 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 $$ \begin{aligned} \text { Mass loss } / \mathrm{kg} &=(0.0265 \mathrm{u})\left(6.02 \times 10^{26} / 4\right)=3.99 \times 10^{24} \mathrm{u} \\ &=\left(3.99 \times 10^{24} \mathrm{u}\right)\left(1.66 \times 10^{-27} \mathrm{~kg} / \mathrm{u}\right)=0.00663 \mathrm{~kg} \end{aligned} $$ 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

Calculate the reactant mass

The reactant mass consists of 4 hydrogen atoms, which yields the equation \[ \text{Reactant Mass} = (4)(1.007825 \, \text{u}) - 4 \times m_{e} = 4.031300 \, \text{u} - 4 \times m_{e} \] where \( m_{e} \) is the electron mass.

02

Calculate the product mass

Product mass includes the helium nucleus and 2 electrons (positrons), resulting in: \[ \text{Product Mass} = (4.002604 \, \text{u} - 2 \times m_{e}) + 2 \times m_{e} = 4.002604 \, \text{u} \] because the electron mass adds back what was subtracted.

03

Determine mass loss per fusion reaction

Subtract the product mass from the reactant mass to find the mass loss: \[ \Delta m = \left( 4.031300 \, \text{u} - 4 \times m_{e} \right) - 4.002604 \, \text{u} = 0.026596 \, \text{u} \] and with \( m_{e} = 0.000549 \, \text{u} \), the mass loss simplifies to 0.0265 u.

04

Calculate total mass loss for 1 kg of hydrogen

Calculate the total mass loss by converting the number of hydrogen atoms in 1 kg and the loss per reaction:\[ \frac{6.02 \times 10^{26} \text{ atoms}}{4} \times 0.0265 \, \text{u} = 3.99 \times 10^{24} \, \text{u} \] Convert to kilograms: \[ 3.99 \times 10^{24} \, \text{u} \times 1.66 \times 10^{-27} \, \text{kg/u} = 0.00663 \, \text{kg} \]

05

Convert mass loss to energy using Einstein's equation

Use the formula \( E = \Delta m \cdot c^2 \): \[ \Delta E = 0.00663 \, \text{kg} \times (2.998 \times 10^8 \, \text{m/s})^2 = 5.96 \times 10^{14} \, \text{J} \] This value represents the energy released.

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Key Concepts

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

Mass-Energy Equivalence

The principle of mass-energy equivalence is a cornerstone in understanding nuclear fusion processes. This concept, introduced by Albert Einstein's famous equation \(E=mc^2\), tells us that mass can be converted into energy and vice versa. In this equation, \(E\) represents energy, \(m\) stands for mass, and \(c\) is the speed of light. When thinking about nuclear reactions such as fusion, the idea is that a small amount of mass is converted into a large amount of energy due to the massive value of \(c^2\) (the speed of light squared).
In the fusion reactions occurring within the sun, mass from hydrogen nuclei (protons) is converted into energy, which is why our sun shines so brightly. This mass-energy conversion is what fuels the sun and provides the thermal and light energy that supports life on Earth.

Hydrogen Fusion

Hydrogen fusion is the process where hydrogen atoms combine to form helium, releasing energy in the process. Inside the sun, this occurs under extreme heat and pressure, allowing hydrogen nuclei to overcome repulsive forces and fuse together. The primary reaction involved is when four hydrogen nuclei, or protons, come together to eventually form a helium nucleus.
During this process, not only is a helium nucleus formed, but positrons and neutrinos are also emitted. The emitted energy as a result of this fusion process is what powers the sun and, by extension, contributes massively to the solar energy reaching Earth. It’s important to understand the mechanics of this reaction to grasp how energy is made available in fusion. The transformation is accompanied by a faint alteration in mass, which becomes the source of energy.

Energy Calculation

To calculate the energy produced in a fusion reaction, several steps are involved. Initially, we must find the difference in mass between the reactants and the products. This difference, known as mass loss, is essential because it represents the mass that is converted into energy. The mass loss per fusion reaction is calculated by subtracting the final mass of helium and associated particles from the initial mass of the hydrogen atoms.
The energy calculation uses Einstein's mass-energy equivalence formula \(E=mc^2\). In this formula, the mass loss \(\Delta m\) is converted into energy. For a practical example, if you have 1kg of hydrogen undergoing fusion, the mass loss translates into a significant amount of energy due to the conversion factor of \(c^2\). This results in a large sum of energy, illustrating why nuclear fusion is such a powerful energy source.

Helium Production

The end product of hydrogen fusion in this context is helium, specifically the isotope \(^4He\), which comprises two protons and two neutrons. This step by step transformation from four hydrogen nuclei to one helium nucleus is not only crucial for energy release but also a primary method of helium production in stars.
The process is made possible within stars thanks to high temperatures and pressures. These conditions facilitate the overcoming of electrostatic forces that typically keep similarly charged protons apart. The synthesis of helium from hydrogen showcases stellar nucleosynthesis, where elements are formed in the cores of stars.

• Helium production signifies a transition from hydrogen to heavier elements, which is foundational to understanding stellar life cycles.
• This process indirectly highlights how heavier elements beyond helium are created through other nuclear reactions in more powerful stars.

Understanding helium production serves to grasp the lifecycle of stars and the creation of elements throughout the universe.