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Calculate \(\Delta \mathrm{E}\) for the proposed basis of an absolutely clean source of nuclear energy $$ { }^{11}{ }_{5} \mathrm{~B}+{ }_{1}^{1} \mathrm{H} \rightarrow{ }^{12}{ }_{6} \mathrm{C} \rightarrow 3{ }_{2}^{4}{ }_{2} \mathrm{He} $$ Atomic masses \(:{ }^{11} \mathrm{~B}=11.00931,{ }^{4} \mathrm{He}=4.00260,{ }^{1} \mathrm{H}=\) \(1.00783 .\)

Short Answer

Expert verified
The change in energy for the proposed basis of an absolutely clean source of nuclear energy is \(8.403 \times 10^{11}\) Joules.

Step by step solution

01

Find the mass defect for the first step

To find the mass defect for the first step, first, calculate the total mass before the reaction occurs (mass of Boron-11 and Hydrogen-1) and the mass after (mass of Carbon-12). The difference between the initial and final mass will be the mass defect. Mass before reaction: \(m_{B_{11}} + m_{H_{1}} = 11.00931 + 1.00783 = 12.01714\) Now, we must find the mass of Carbon-12 by adding together the mass of the three resulting Helium-4 particles. \(3\times m_{He_{4}} = 3 * 4.00260 = 12.00780\) Mass defect in the first step: \(\Delta m_{1} = 12.01714 - 12.00780 = 0.00934\)
02

Find the energy change for the first step

Now, we can use the mass defect to find the energy change for the reaction using the equation \(\Delta E = \Delta m \times c^2\). We will use a value of \(c\) of \(3 \times 10^8\) m/s for the speed of light. \( \Delta E_{1} = \Delta m_{1} \times c^2 = 0.00934 \times (3 \times 10^8)^2 = 8.403 \times 10^{11}\) J
03

Find the mass defect for the second step

In this step, we need to find the mass defect for when Carbon-12 splits into three Helium-4 particles. We have already calculated the mass of Carbon-12 (12.00780), so we will use that as the initial mass. Mass defect in the second step: \(\Delta m_{2} = 0\) (Since the mass of Carbon-12 equals the combined mass of three Helium-4 particles)
04

Find the energy change for the second step

Since there is no mass defect in the second step, the energy change for the second step will also be zero. \( \Delta E_{2} = 0\)
05

Calculate the total energy change for the reaction

Now, we can find the total energy change for the entire reaction by adding the energy change from both steps. \( \Delta E_{total} = \Delta E_{1} + \Delta E_{2} = 8.403 \times 10^{11} + 0 = 8.403 \times 10^{11}\) J So, the change in energy for the proposed basis of absolutely clean source of nuclear energy is \(8.403 \times 10^{11}\) Joules.

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Key Concepts

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

Mass Defect
Mass defect is a critical concept in understanding nuclear energy. It refers to the difference in mass between the starting reactants and the final products in a nuclear reaction. In simpler terms, when an atom's nucleus is formed, a small amount of total mass is missing compared to the sum of individual parts. This missing mass is what we call mass defect.

To see how this works, let’s consider the first step of the reaction: Boron-11 ( ^{11}{ }_{5} ext{B}) and Hydrogen-1 ( ^{1}{ }_{1} ext{H}) combine to form an intermediate Carbon-12 ( ^{12}{ }_{6} ext{C}). We calculate the initial mass (before reaction) as the sum of Boron-11 and Hydrogen-1, which equals 12.01714 atomic mass units (amu). The mass of Carbon-12 is found to be 12.00780 amu.

The difference between these two masses (12.01714 - 12.00780) gives us the mass defect, Δm = 0.00934 amu. Although this mass defect seems tiny, it holds the key to understanding the immense energy released in nuclear reactions, as this mass converts to energy.
Energy Change
Energy change ( ΔE) in a nuclear reaction is directly related to the mass defect through Einstein's famous equation ΔE = Δm × c^2, where c is the speed of light. This equation tells us that the mass defect transforms into energy.

Using the calculated mass defect of 0.00934 amu in the first step, and knowing that the speed of light ( c) is approximately 3 × 10^8 m/s, we substitute these values into the equation. The energy change ΔE is calculated as 0.00934 × (3 × 10^8)^2, resulting in 8.403 × 10^{11} Joules.

This considerable amount of energy elucidates why nuclear reactions can be such potent energy sources. In the second step of our reaction, no mass defect means no change in energy, keeping the total energy change at 8.403 × 10^{11} Joules.
Nuclear Reaction
Nuclear reactions involve changes in an atom's nucleus, unlike chemical reactions that involve electrons. These reactions can release or absorb significant energy, primarily due to the mass defect.

In our example, a two-step nuclear reaction begins with Boron-11 and Hydrogen-1 nuclei combining to eventually form Carbon-12, and then transform into three Helium-4 particles. The reason such reactions are pivotal for energy production is due to the large amounts of energy released.

A step-by-step analysis of the reaction provides insights into nuclear energy mechanisms. Understanding these reactions involves:
  • Recognizing the initial and final states of involved nuclear particles
  • Calculating mass defects at every step
  • Computing energy change through mass-energy equivalency
In summary, nuclear reactions, with their associated mass defects and energy transformations, form the basis of nuclear energy production, offering immense potential as clean energy sources.

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Most popular questions from this chapter

Chromium exists in four isotopic forms. The atomic masses and percent occurrences of these isotopes are listed in the following table: $$ \begin{array}{|l|l|} \hline \text { Isotopic mass (amu) } & \underline{\text { Percent occurrence }} \\ \hline 50 & 4.31 \% \\ \hline 52 & 83.76 \% \\ \hline 53 & 9.55 \% \\ \hline 54 & 2.38 \% \\ \hline \end{array} $$ Calculate the average atomic mass of chromium.

The first step in the radioactive decay of \({ }^{238}{ }_{92} \mathrm{U}\) is \({ }^{238}{ }_{92} \mathrm{U}={ }^{234}{ }_{90} \mathrm{Th}+{ }_{2}^{4} \mathrm{He}\). Calculate the energy released in this reaction. The exact masses of \({ }^{238}{ }_{92} \mathrm{U},{ }^{234}{ }_{90} \mathrm{Th}\), and \({ }_{2}^{4} \mathrm{He}\) are \(238.0508,234.0437\) and \(4.0026 \mathrm{amu}\), respectively. \(1.0073 \mathrm{amu}=1.673 \times 10^{-24} \mathrm{~g}\)

Given that the masses of a proton, neutron, and electron are \(1.00728,1.00867\), and \(.000549\) amu, respectively, how much missing mass is there in \({ }^{19}{ }_{9} \mathrm{~F}\) (atomic weight = \(18.9984) ?\)

Chlorine is found in nature in two isotopic forms, one of atomic mass 35 amu and one of atomic mass 37 amu. The average atomic mass of chlorine is \(35.453\) amu. What is the percent with which each of these isotopes occurs in nature?

Complete the following nuclear equations. (a) \({ }_{7} \mathrm{~N}^{14}+{ }_{2} \mathrm{He}^{4}\) \(\rightarrow{ }_{8} \mathrm{O}^{17}+\ldots \ldots\) (b) \(_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4}\) \(\rightarrow{ }_{6} \mathrm{C}^{12}+\ldots \ldots\) (c) \({ }_{15} \mathrm{P}^{90} \rightarrow{ }_{14} \mathrm{Si}^{30}+\ldots \ldots\) (d) \({ }_{1} \mathrm{H}^{3} \quad \rightarrow{ }_{2} \mathrm{He}^{3}+\ldots \ldots\)

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