/*! 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 11 Is \(^{238} \mathrm{U}\) fission... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Is \(^{238} \mathrm{U}\) fissionable? Is it fissile? Explain the distinction.

Short Answer

Expert verified
Uranium-238 is fissionable, but it is not fissile. The distinction lies in the neutron energy required to cause fission. For \(^{238} U\), high energy neutrons are needed, which is characteristic of fissionable isotopes, however, it cannot undergo fission with low energy or thermal neutrons, a characteristic of fissile isotopes.

Step by step solution

01

Define Uranium-238

Uranium-238 (\(^{238} U\)) is the most common isotope of uranium found in nature, being considerably more abundant than uranium-235.
02

Determine if Uranium-238 is fissionable

\(^{238} U\) is fissionable. This means that uranium-238 can undergo fission, but it requires a high energy neutron to cause the fission.
03

Determine if Uranium-238 is fissile

\(^{238} U\) is not fissile. Fission in uranium-238 does not easily occur with a slow or thermal neutron. Instead, uranium-238 mostly absorbs the neutron and transforms into uranium-239.
04

Explanation of the distinction

The key difference between being fissionable and being fissile is in the neutron energy required to cause fission. Fissionable isotopes like uranium-238 need high energy neutrons. On the other hand, fissile isotopes can use thermal (slow) neutrons for fission. This is the distinction between the two terms.

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 91Ó°ÊÓ!

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

An NMR spectrometer is described as a " \(300-\) MHz instrument". meaning \(3.00 \times 10^{8} \mathrm{Hz}\) is the frequency supplied to its transmitter coil to flip the spin states of bare protons. What's the strength of its unperturbed magnetic field?

Nitrogen-13 is a 9.97-min-half-life isotope used to "tag" ammonia for PET scans, including quantification of myocardial infarction. Consider an intravenous injection incorporating \(20.0 \mathrm{mCi}\) of N-13. Plot a graph of \(\mathrm{N}-13\) activity versus time, with your vertical axis logarithmic and your horizontal axis linear. Why is the graph a straight line? What's the significance of its slope?

You're assessing the safety of an airport bomb-detection system in which neutron activation of the stable nitrogen isotope \(^{15} 7 \mathrm{N}\) turns it into unstable \(^{16}\) ?N. The N-16 decays by beta emission with 7.13-s half-life. How long after activation will the \(\mathbf{N}-16\) activity have dropped by a factor of 1 million?

Carbon- 11 -labeled acetate shows promise in \(\mathrm{PET}\) scans for determining the extent of metastasized prostate cancer. (a) Given C-11's 20.4-min half-life, how long will it take an initial dose of \(2.0 \mathrm{GBq}\) to decay to \(7.0 \mathrm{kBq}\) (roughly the natural radioactivity of the human body)? (b) What nucleus remains after \(C\) -1 1 decays by positron emission?

At the Oklo site today, you would expect to find measurable amounts of a. strontium- 90 b. cesium-137. c. plutonium- 239 d. none of the above.

See all solutions

Recommended explanations on Physics Textbooks

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.