Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 2

Self-diffusion involves the motion of atoms that are all of the same type; therefore it is not subject to observation by compositional changes, as with inter-diffusion. Suggest one way in which self-diffusion may be monitored.

Short Answer

Expert verified
Answer: Self-diffusion can be monitored in a material containing atoms of the same type by using a radioactive isotope tracing technique. This method involves introducing a small amount of radioactive isotopes into the material and measuring their displacement using sensitive radiation detectors.

Step by step solution

01

Understand the concept of self-diffusion

Self-diffusion occurs when individual atoms of a similar type exchange positions in a given material. Since there are only atoms of the same type present, tracking such changes through composition is not possible.
02

Radioactive isotope tracing technique

A method to monitor self-diffusion is by using radioactive isotopes as tracers. The radioactive isotope of the element should have the same chemical properties as the non-radioactive isotope.
03

Process of using radioactive isotopes for monitoring self-diffusion

In this method, a small amount of radioactive isotope of the element is introduced into the material. The radioactive isotopes will exchange positions with the non-radioactive isotopes through self-diffusion. This displacement of isotopes can be measured using sensitive radiation detectors, enabling monitoring of self-diffusion.
04

Advantages of the radioactive isotope tracing technique

This technique is quite useful because it doesn't change the chemical properties of the element, and it allows for monitoring self-diffusion without altering the material composition significantly. Since radioactive isotopes emit particles or gamma rays, it is easy to detect their movement within the material.
05

In Conclusion

One way to monitor self-diffusion in a material containing atoms of the same type is by using a radioactive isotope tracing technique which involves introducing a small amount of radioactive isotopes into the material and measuring their displacement using sensitive radiation detectors.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

(a) Compare interstitial and vacancy atomic mechanisms for diffusion. (b) Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion.

An FCC iron-carbon alloy initially containing 0.55 wt \(\%\) C is exposed to an oxygen-rich and virtually carbon-free atmosphere at \(1325 \mathrm{K}\) \(\left(1052^{\circ} \mathrm{C}\right) .\) Under these circumstances the carbon diffuses from the alloy and reacts at the surface with the oxygen in the atmosphere that is, the carbon concentration at the surface position is maintained essentially at 0 wt \(\%\) C. (This process of carbon depletion is termed decarburization.) At what position will the carbon concentration be 0.25 wt\% after a 10-h treatment? The value of \(D\) at \(1325 \mathrm{K}\) is \(4.3 \times 10^{-11} \mathrm{m}^{2} / \mathrm{s}\).

(a) Briefly explain the concept of a driving force. (b) What is the driving force for steady-state diffusion?

At approximately what temperature would a specimen of \(\gamma\) -iron have to be carburized for \(4 \mathrm{h}\) to produce the same diffusion result as at \(1000^{\circ} \mathrm{C}\) for \(12 \mathrm{h} ?\)

The diffusion coefficients for nickel in iron are given at two temperatures: $$\begin{array}{cc} \boldsymbol{T}(\boldsymbol{K}) & \boldsymbol{D}\left(\boldsymbol{m}^{2} / \boldsymbol{s}\right) \\ \hline 1473 & 2.2 \times 10^{-15} \\ 1673 & 4.8 \times 10^{-14} \end{array}$$ (a) Determine the values of \(D_{0}\) and the activation energy \(Q_{d}\) (b) What is the magnitude of \(D\) at \(1300^{\circ} \mathrm{C}\) \((1573 \mathrm{K}) ?\)
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Turning Points in Physics

Read Explanation

Electrostatics

Read Explanation

Energy Physics

Read Explanation

Kinematics Physics

Read Explanation

Fundamentals of Physics

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.

Sign-up for free