/*! 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 9 Much of the research on controll... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 9

Much of the research on controlled fusion focuses on the problem of how to contain the reacting material. Magnetic fields appear to be the most promising mode of containment. Why is containment such a problem? Why must one resort to magnetic fields for containment?

Short Answer

Expert verified
In short, containment in controlled fusion is crucial due to the high temperatures involved and the need to maintain plasma stability. Conventional materials cannot withstand such temperatures, and any contact between the plasma and reactor walls leads to energy loss and halting of the fusion process. Magnetic fields are a promising solution for containment, as they can confine the charged plasma particles away from the reactor walls while maintaining stability, ensuring an appropriate environment for fusion reactions to occur.

Step by step solution

01

Controlled Fusion and Containment

Controlled fusion is a process where hydrogen nuclei are combined at very high temperatures to form helium nuclei, releasing a substantial amount of energy in the process. Containment is essential because the reacting materials reach extremely high temperatures, and if not properly contained, they can cause severe damage to the reactor or the environment.
02

Challenges of Containment

The primary challenge of containment in controlled fusion is the extremely high temperatures required for the reaction to occur, which can reach millions of degrees Celsius. Such temperatures are too high for any material to withstand. If the reacting materials come into contact with the walls of the reactor, they can cause the reactor to fail and potentially lead to accidents. Another issue is that the plasma, consisting of ions and free electrons, must be maintained in a stable state within the reactor. Plasma stability is crucial, as any interaction with the reactor walls will result in a loss of energy, leading to a reduction in temperature and the halting of the fusion process.
03

Magnetic Fields for Containment

Magnetic confinement is the most promising solution for containment because it can maintain the required high temperatures by keeping the reacting materials away from the reactor walls. The plasma inside the reactor can be confined with magnetic fields because the charged particles follow the magnetic field lines. This means that by creating a magnetic field with a specific shape, the plasma can be kept away from the walls and in a stable state within the reactor. Magnetic confinement can be achieved in various ways, such as using a toroidal-shaped magnetic field called a tokamak or a linear-shaped magnetic field called a stellarator. Both of these designs aim to create a stable magnetic field that can confine the plasma long enough for fusion reactions to occur.
04

Conclusion

In summary, containment in controlled fusion is a critical challenge due to the extremely high temperatures and maintaining plasma stability. Magnetic fields provide a promising solution to containment as they can keep reacting materials away from the reactor walls by confining the charged plasma particles along the magnetic field lines, ensuring a stable environment for the fusion reactions to take place.

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 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The easiest fusion reaction to initiate is $${ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} \mathrm{n}$$ Calculate the energy released per \({ }_{2}^{4} \mathrm{He}\) nucleus produced and per mole of \({ }_{2}^{4}\) He produced. The atomic masses are \({ }_{1}^{2} \mathrm{H}, 2.01410 ;{ }_{1}^{3} \mathrm{H}\), \(3.01605\); and \({ }_{2}^{4}\) He, \(4.00260\). The masses of the electron and neutron are \(5.4858 \times 10^{-4}\) amu and \(1.00866\) amu, respectively.

During the research that led to production of the two atomic bombs used against Japan in World War II, different mechanisms for obtaining a supercritical mass of fissionable material were investigated. In one type of bomb, a "gun" shot one piece of fissionable material into a cavity containing another piece of fissionable material. In the second type of bomb, the fissionable material was surrounded with a high explosive that, when detonated, compressed the fissionable material into a smaller volume. Discuss what is meant by critical mass, and explain why the ability to achieve a critical mass is essential to sustaining a nuclear reaction.

To determine the \(K_{\mathrm{sp}}\) value of \(\mathrm{Hg}_{2} \mathrm{I}_{2}\), a chemist obtained a solid sample of \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) in which some of the iodine is present as radioactive \({ }^{131} \mathrm{I}\). The count rate of the \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) sample is \(5.0 \times 10^{11}\) counts per minute per mole of \(I\). An excess amount of \(\mathrm{Hg}_{2} \mathrm{I}_{2}(s)\) is placed into some water, and the solid is allowed to come to equilibrium with its respective ions. A \(150.0-\mathrm{mL}\) sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute. From this information, calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) $$\mathrm{Hg}_{2} \mathrm{I}_{2}(s) \rightleftharpoons \mathrm{Hg}_{2}^{2+}(a q)+2 \mathrm{I}^{-}(a q) \quad K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}$$

The only stable isotope of fluorine is fluorine- \(19 .\) Predict possible modes of decay for fluorine- 21 , fluorine- 18 , and fluorine- 17 .

Using the kinetic molecular theory (Section 5.6), calculate the root mean square velocity and the average kinetic energy of \({ }_{1}^{2} \mathrm{H}\) nuclei at a temperature of \(4 \times 10^{7} \mathrm{~K}\). (See Exercise 46 for the appropriate mass values.)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.