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Chapter 27: Problem 32

In a hypothetical nuclear-fusion reactor, two deuterium nuclei combine or "fuse" to form one helium nucleus. The mass of a deuterium nucleus, expressed in atomic mass units (u), is \(2.0136 \mathrm{u}\) that of a helium nucleus is 4.0015 u. \(\left(1 \mathrm{u}=1.661 \times 10^{-27} \mathrm{~kg} .\right)\) (a) How much energy is released when \(1.0 \mathrm{~kg}\) of deuterium undergoes fusion? (b) The annual consumption of electrical energy in the United States is on the order of \(1.0 \times 10^{19} \mathrm{~J}\). How much deuterium must react to produce this much energy?

Short Answer

Expert verified
(a) Energy released: \(5.74 \times 10^{14} \) J; (b) Deuterium needed: \(1.74 \times 10^4\) kg.

Step by step solution

01

Determine Mass Defect Per Fusion

The mass of two deuterium nuclei before fusion is \(2 \times 2.0136 \text{ u} = 4.0272 \text{ u}\). The mass of one helium nucleus after fusion is \(4.0015 \text{ u}\). The mass defect \(\Delta m\) is the difference: \[\Delta m = 4.0272 \text{ u} - 4.0015 \text{ u} = 0.0257 \text{ u}.\]
02

Convert Mass Defect to Kilograms

Convert the mass defect from atomic mass units to kilograms using the conversion factor. \[\Delta m_{\text{kg}} = 0.0257 \text{ u} \times 1.661 \times 10^{-27} \text{ kg/u} = 4.27017 \times 10^{-29} \text{ kg}.\]
03

Calculate Energy Released Per Fusion

Use Einstein's equation \(E = mc^2\) to calculate the energy per fusion reaction. \[E = 4.27017 \times 10^{-29} \text{ kg} \times (3 \times 10^8 \text{ m/s})^2 = 3.84315 \times 10^{-12} \text{ J}.\]
04

Calculate Number of Deuterium Nuclei in 1.0 kg

Find the number of deuterium nuclei in 1.0 kg. Each deuterium nucleus has a mass of \(2.0136 \times 1.661 \times 10^{-27} \text{ kg} = 3.344 \times 10^{-27} \text{ kg}\). Dividing, we get: \[N = \frac{1.0 \text{ kg}}{3.344 \times 10^{-27} \text{ kg}} = 2.99 \times 10^{26}.\]
05

Calculate Total Energy Released from 1.0 kg Deuterium

Since each fusion releases \(3.84315 \times 10^{-12} \text{ J}\), the total energy released by 1.0 kg of deuterium is \[E_{\text{total}} = \frac{1}{2} \times 2.99 \times 10^{26} \times 3.84315 \times 10^{-12} \text{ J} = 5.74 \times 10^{14} \text{ J}.\]
06

Calculate Mass of Deuterium for Desired Energy

To produce \(1.0 \times 10^{19} \text{ J}\), calculate the required mass of deuterium. Using the earlier energy release calculation, \[m_{\text{needed}} = \frac{1.0 \times 10^{19} \text{ J}}{5.74 \times 10^{14} \text{ J/kg}} \approx 1.74 \times 10^4 \text{ kg}.\]

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Key Concepts

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

Mass Defect
When two deuterium nuclei fuse into a helium nucleus, there's a difference in mass before and after the fusion. This difference is called the "mass defect." This occurs because the total mass of the newly formed helium nucleus is slightly less than the mass of the two separate deuterium nuclei.
The original combined mass from the deuterium nuclei is 4.0272 atomic mass units, while the resulting helium nucleus is 4.0015 atomic mass units. The difference of 0.0257 atomic mass units is the mass defect.
In essence, the mass defect represents the mass that has been converted into energy during the fusion process. It's a pivotal concept as it demonstrates that energy is released in nuclear reactions, primarily due to the small change in mass.
Einstein's Equation E=mc^2
Einstein's equation, represented as \(E=mc^2\), connects mass with energy. This formula explains how tiny amounts of mass can be converted into vast amounts of energy.
In the context of nuclear fusion, the mass defect calculated from the fusion of deuterium nuclei can be converted to energy using this equation.
Here, "\(E\)" is the energy produced, "\(m\)" is the mass measured in kilograms, and "\(c\)" is the speed of light, approximately \(3 imes 10^8\) meters per second.
  • This equation is what helps us understand the energy yield from nuclear reactions.
  • By using \(E=mc^2\), the mass defect of 0.0257 atomic mass units turns into real energy output, allowing scientists to harness this energy in practical scenarios.
Deuterium Fusion
Deuterium fusion involves two deuterium nuclei coming together to form a single helium nucleus. Deuterium is a stable isotope of hydrogen, consisting of one proton and one neutron.
The fusion process beneficially releases energy, making it a potential source for generating power.
  • Each deuterium nucleus involved in the fusion process has a mass of approximately 2.0136 atomic mass units.
  • The fusion releases energy due to the mass defect. The difference is transformed into energy, making it a promising method for future energy solutions.
The reactions involved in deuterium fusion are essential in closed energy cycles, contributing to clean and efficient energy production.
Atomic Mass Units
The atomic mass unit (u) is a unit of mass used to express atomic and molecular weights. It is defined as one twelfth of the mass of a carbon-12 atom.
This unit is incredibly useful in calculations involving atomic-scale particles, such as atoms and molecules.
  • For nuclear fusion calculations, deuterium nuclei and the resulting helium nucleus are compared in terms of atomic mass units.
  • The conversion of atomic mass units to kilograms is important since practical applications, like energy calculations, require SI units.
Using atomic mass units allows scientists to easily measure and compare the minuscule masses of atoms and nuclei involved in reactions, thus facilitating the understanding and application of nuclear processes.

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Most popular questions from this chapter

You take a trip to Pluto and back (round trip 11.5 billion \(\mathrm{km}\) ), traveling at a constant speed (except for the turnaround at Pluto) of \(45,000 \mathrm{~km} / \mathrm{h}\). (a) How long does the trip take, in hours, from the point of view of a friend on earth? About how many years is this? (b) When you return, what will be the difference between the time on your atomic wristwatch and the time on your friend's? (Hint: Assume the distance and speed are highly precise, and carry a lot of significant digits in your calculation!)

A spacecraft flies away from the earth with a speed of \(0.6 c\) relative to the earth and then returns at the same speed. The spacecraft carries an atomic clock that has been carefully synchronized with an identical clock that remains at rest on earth. The spacecraft returns to its starting point 365 days ( 1 year) later, as measured by the clock that remained on earth. (a) Which clock records the proper time between the two events that correspond to leaving earth and then arriving at earth? (b) What is the difference in the elapsed times on the two clocks, measured in days? (c) Which clock, the one in the spacecraft or the one on earth, shows the shorter elapsed time?

Space travel? Travel to the stars requires hundreds or thousands of years, even at the speed of light. Some people have suggested that we can get around this difficulty by accelerating the rocket (and its astronauts) to very high speeds so that they will age less due to time dilation. The fly in this ointment is that it takes a great deal of energy to do this. Suppose you want to go to the immense red giant Betelgeuse, which is about 500 light-years away. You plan to travel at constant speed in a \(1000 \mathrm{~kg}\) rocket ship (a little over a ton), which, in reality, is far too small for this purpose. In each case that follows, calculate the time for the trip, as measured by people on earth and by astronauts in the rocket ship, the energy needed in joules, and the energy needed as a percent of U.S. yearly use (which is \(1.0 \times 10^{20} \mathrm{~J}\) ). For comparison, arrange your results in a table showing \(v_{\text {rocket }}, t_{\text {earth }}, t_{\text {rocket }}, E(\) in \(\mathbf{J}),\) and \(E\) (as \(\%\) of U.S. use). The rocket ship's speed is (a) \(0.50 c,\) (b) \(0.99 c\), and (c) \(0.9999 c\). On the basis of your results, does it seem likely that any government will invest in such high- speed space travel any time soon?

At what speed is the momentum of a particle three times as great as the result obtained from the nonrelativistic expression \(m v ?\)

An electron is acted upon by a force of \(5.00 \times 10^{-15} \mathrm{~N}\) due to an electric field. Find the acceleration this force produces in each case: (a) The electron's speed is \(1.00 \mathrm{~km} / \mathrm{s}\). (b) The electron's speed is \(2.50 \times 10^{8} \mathrm{~m} / \mathrm{s}\) and the force is parallel to the velocity.
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