Chapter 5: Problem 7
(a) Briefly explain the concept of a driving force. (b) What is the driving force for steady-state diffusion?
Chapter 5: Problem 7
(a) Briefly explain the concept of a driving force. (b) What is the driving force for steady-state diffusion?
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Get started for freeThe preexponential and activation energy for the diffusion of chromium in nickel are \(1.1 \times 10^{-4}\) \(\mathrm{m}^{2} / \mathrm{s}\) and \(272,000 \mathrm{~J} / \mathrm{mol}\), respectively. At what temperature will the diffusion coefficient have a value of \(12 \times 10^{-14} \mathrm{~m}^{2} / \mathrm{s} 2\)
When \(\alpha\)-iron is subjected to an atmosphere of ( nitrogen gas, the concentration of nitrogen in the iron, \(C_{\mathrm{N}}\) (in weight percent), is a function of hydrogen pressure, \(p_{\mathrm{N}_{2}}\) (in \(\left.\mathrm{MPa}\right)\), and absolute temperature \((T)\) according to $$ C_{\mathrm{N}}=4.90 \times 10^{-3} \sqrt{p_{\mathrm{N}_{2}}} \exp \left(-\frac{37,600 \mathrm{~J} / \mathrm{mol}}{R T}\right) $$ Furthermore, the values of \(D_{0}\) and \(Q_{d}\) for this diffusion system are \(5.0 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) and \(77,000 \mathrm{~J} / \mathrm{mol}\), respectively. Consider a thin iron membrane 1.5-mm thick at \(300^{\circ} \mathrm{C}\). Compute the diffusion flux through this membrane if the nitrogen pressure on one side of the membrane is \(0.10 \mathrm{MPa}(0.99 \mathrm{~atm})\) and on the
Antimony atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat treatments; the background concentration of \(\mathrm{Sb}\). in this silicon material is known to be \(2 \times 10^{20}\) atoms/m^3. The predeposition treatment is to be conducted at \(900^{\circ} \mathrm{C}\) for \(1 \mathrm{~h}\); the surface concentration of \(\mathrm{Sb}\) is to be maintained at a constant level of \(8.0 \times 10^{25}\) atoms \(/ \mathrm{m}^{3}\). Drive-in diffusion will be carried out at \(1200^{\circ} \mathrm{C}\) for a period of \(1.75 \mathrm{~h}\). For the diffusion of \(\mathrm{Sb}\) in Si, values of \(Q_{d}\) and \(D_{0}\) are \(3.65 \mathrm{eV} /\) atom and \(2.14 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively. (a) Calculate the value of \(Q_{0}\) - (b) Determine the value of \(x_{j}\) for the drive-in diffusion treatment. (c) Also, for the drive-in treatment, compute the position \(x\) at which the concentration of \(S b\) atoms is \(5 \times 10^{23}\) atoms/m \(^{3}\)
Consider a diffusion couple composed of two cobalt-iron alloys; one has a composition of \(75 \mathrm{wt} \%\) Co-25 wt\% Fe; the other alloy composition is \(50 \mathrm{wt} \%\) Co-50 wt \(\% \mathrm{Fe}\). If this couple is heated to a temperature of \(800^{\circ} \mathrm{C}(1073 \mathrm{~K})\) for \(20,000 \mathrm{~s}\) determine how far from the original interface into the \(50 \mathrm{wt} \%\) Co-50 wt \% Fe alloy the composition has increased to \(52 \mathrm{wt} \%\) Co-48 wt Fe. For the diffusion coefficient, assume values of \(6.6 \times 10^{-t}\) \(\mathrm{m}^{2} / \mathrm{s}\) and \(247,000 \mathrm{~J} / \mathrm{mol}\), respectively, for the pre exponential and activation energy.
The diffusion coefficients for nickel in iron are given at two temperatures, as follows: \begin{tabular}{cc} \hline \(\boldsymbol{T}(\boldsymbol{K})\) & \(\boldsymbol{D}\left(\mathrm{m}^{2} / \mathrm{s}\right)\) \\ \hline 1473 & \(2.2 \times 10^{-15}\) \\ \hline 1673 & \(4.8 \times 10^{-14}\) \\ \hline \end{tabular} (a) Determine the values of \(D_{0}\) and the activation energy \(Q_{d^{\prime}}\) (b) What is the magnitude of \(D\) at \(1300^{\circ} \mathrm{C}(1573 \mathrm{~K}) ?\)
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