/*! 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 61 A portion of the Sun's energy co... [FREE SOLUTION] | 91影视

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Chapter 21: Problem 61

A portion of the Sun's energy comes from the reaction $$ 4_{1}^{1} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+2_{1}^{0} \mathrm{e} $$ which requires a temperature of \(10^{6}\) to \(10^{7} \mathrm{~K}\). (a) Use the mass of the helium- 4 nucleus given in Table 21.7 to determine how much energy is released when the reaction is run with \(1 \mathrm{~mol}\) of hydrogen atoms. (b) Why is such a high temperature required?

Short Answer

Expert verified
(a) The energy released when the reaction is run with 1 mole of hydrogen atoms is \(2.58 \times 10^{12}\,\mathrm{J}\). (b) The high temperature is required for this reaction because, at such high temperatures, hydrogen nuclei (protons) obtain sufficient kinetic energy to overcome the electrostatic repulsion between them. This allows the protons to come close enough to each other for the strong nuclear force to act, which leads to nuclear fusion. The temperature should be in the range of \(10^6\) to \(10^7\) K, to provide the required kinetic energy for this reaction to occur.

Step by step solution

01

Determine the mass difference of the reaction

To find the energy released, we first need to determine the mass difference between the reactants and products. For this reaction, 4 hydrogen atoms combine to form a helium-4 nucleus and 2 positrons (e+). Therefore, the mass difference can be calculated using the equation: \[螖m = m_{reactants} - m_{products} \] From Table 21.7, we know the mass of helium-4 nucleus: \(4_{1}^{1} \mathrm{H} = 1.007825\,\mathrm{amu}\) \(_{2}^{4} \mathrm{He} = 4.001506\,\mathrm{amu}\) \(_{1}^{0} \mathrm{e} = 0.000548\,\mathrm{amu}\)
02

Calculate the mass difference

Now, we can calculate the mass difference using the masses of the reactants and products: \(螖m = 4(1.007825\,\mathrm{amu}) - (4.001506\,\mathrm{amu} + 2(0.000548\,\mathrm{amu}))\) \(螖m = 4.0313\,\mathrm{amu} - 4.002602\,\mathrm{amu}\) \(螖m = 0.028698\,\mathrm{amu}\)
03

Convert mass difference to energy

Now, we can convert the mass difference to energy using Einstein's mass-energy equivalence equation: \(E = 螖m \times c^2\) where \(c\) is the speed of light (\(2.9979 \times 10^8\,\mathrm{m/s}\)). First, we need to convert the mass difference from amu to kg. We know that 1 amu = \(1.6605 \times 10^{-27}\,\mathrm{kg}\): \(螖m = 0.028698\,\mathrm{amu} \times\frac{1.6605 \times 10^{-27}\,\mathrm{kg}}{1\,\mathrm{amu}} = 4.766 \times 10^{-29}\,\mathrm{kg}\) Now we can calculate the energy: \(E = (4.766 \times 10^{-29}\,\mathrm{kg}) \times (2.9979 \times 10^8\,\mathrm{m/s})^2\) \(E = 4.29 \times 10^{-12}\,\mathrm{J}\)
04

Find the energy per mole of hydrogen atoms

Now, we have found the energy released for the reaction on a per atom basis. We need to find the energy released for 1 mole of hydrogen atoms. We know that 1 mole has \(6.022 \times 10^{23}\) atoms. Therefore, we can find the energy for 1 mol of hydrogen atoms as follows: \(E_{mol} = E \times 6.022 \times 10^{23}\,\mathrm{atoms}\) \(E_{mol} = (4.29 \times 10^{-12}\,\mathrm{J}) \times (6.022 \times 10^{23}\,\mathrm{atoms})\) \(E_{mol} = 2.58 \times 10^{12}\,\mathrm{J}\) (a) The energy released when the reaction is run with 1 mole of hydrogen atoms is \(2.58 \times 10^{12}\,\mathrm{J}\).
05

Explain the high temperature requirement

(b) The high temperature is required for this reaction because, at such high temperatures, hydrogen nuclei (protons) obtain sufficient kinetic energy to overcome the electrostatic repulsion between them. This allows the protons to come close enough to each other for the strong nuclear force to act, which leads to nuclear fusion. The temperature should be in the range of \(10^6\) to \(10^7\) K, to provide the required kinetic energy for this reaction to occur.

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Key Concepts

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

Understanding Mass-Energy Equivalence
The notion behind mass-energy equivalence is elegantly summarized by Albert Einstein's famous equation,\(E = mc^2\). This tells us that mass (\

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Most popular questions from this chapter

Tests on human subjects in Boston in 1965 and \(1966,\) following the era of atomic bomb testing, revealed average quantities of about \(2 \mathrm{pCi}\) of plutonium radioactivity in the average person. How many disintegrations per second does this level of activity imply? If each alpha particle deposits \(8 \times 10^{-13} \mathrm{~J}\) of energy and if the average person weighs \(75 \mathrm{~kg}\), calculate the number of rads and rems of radiation in 1 yr from such a level of plutonium.

It has been suggested that strontium-90 (generated by nuclear testing) deposited in the hot desert will undergo radioactive decay more rapidly because it will be exposed to much higher average temperatures. (a) Is this a reasonable suggestion? (b) Does the process of radioactive decay have an activation energy, like the Arrhenius behavior of many chemical reactions (Section 14.5\() ?\) Discuss.

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The average energy released in the fission of a single uranium- 235 nucleus is about \(3 \times 10^{-11} \mathrm{~J}\). If the conversion of this energy to electricity in a nuclear power plant is \(40 \%\) efficient, what mass of uranium-235 undergoes fission in a year in a plant that produces 1000 megawatts? Recall that a watt is \(1 \mathrm{~J} / \mathrm{s}\)
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