Briefly explain the difference between self diffusion and inter diffusion.

Short Answer

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Question: What is the main difference between self-diffusion and inter-diffusion, and provide an example for each process. Answer: The main difference between self-diffusion and inter-diffusion is the type of atoms or molecules involved in the movement. Self-diffusion occurs among the same type of atoms or molecules within a homogeneous material, such as a drop of water spreading evenly on a flat surface due to the random movement of water molecules. On the other hand, inter-diffusion involves the movement between different types of atoms or molecules, such as when a drop of dye is added into a glass of water and the dye molecules diffuse into the water molecules, resulting in a uniform color.

Step by step solution

01

Definition of Self-diffusion

Self-diffusion occurs when atoms or molecules of the same species move randomly within a homogeneous material, spreading out over time. This diffusion process is driven by the atoms' or molecules' internal kinetic energy, resulting in the spreading of their positions within the material.
02

Definition of Inter-diffusion

Inter-diffusion occurs when atoms or molecules of different species or types move into one another, resulting in a mixing or blending of the materials. This process often takes place when two or more different materials come into contact, and the concentration gradients in the materials cause the atoms or molecules to spread from high concentration areas to low concentration areas, ultimately leading to a uniform distribution.
03

Comparison of Self-diffusion and Inter-diffusion

The main difference between these two diffusion processes is the type of atoms or molecules involved. In self-diffusion, the movement is observed among the same type of atoms or molecules within a homogeneous material, while in inter-diffusion, the movement is observed between different types of atoms or molecules.
04

Examples of Self-diffusion and Inter-diffusion

An example of self-diffusion process would be when a drop of a liquid (e.g., water) is placed on a flat surface and spreads evenly due to the random movement of its molecules. An example of inter-diffusion process would be when a drop of dye is placed in a glass of water and the dye molecules diffuse into the water molecules, eventually giving the water a uniform color.

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Key Concepts

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

Self-diffusion
Understanding self-diffusion is like observing how a group of people, all wearing red shirts, move around in an otherwise empty room. With time, you'd notice they spread out evenly, with no particular direction or pattern. Similarly, self-diffusion in materials science refers to the process where atoms or molecules of a single species disperse within a homogeneous material. Imagine a small cube of sugar placed in a cup of tea. Initially, the sugar is concentrated at the bottom, but given enough time, the sugar molecules move about randomly until they are uniformly distributed throughout the tea. This random motion is primarily due to the thermal energy within the molecules themselves, which causes them to wiggle and jiggle, allowing them to explore new areas within the material.

Detecting self-diffusion, however, might not be straightforward because all the participating atoms or molecules are identical. Scientists often use isotopic tracing, where a detectable isotope of the diffusing species is introduced and tracked over time. This method provides insights into how compound substances like alloys or polymers internally reorganize their constituent atoms or molecules, even under stable conditions.
Inter-diffusion
If self-diffusion can be thought of as the movement of identical red-shirted individuals within a room, inter-diffusion is like introducing people with blue shirts into the same space and watching how red and blue mix over time. In more technical terms, inter-diffusion, also known as 'impurity diffusion,' involves the movement of atoms or molecules of different species. This process is essential in the formation of solid solutions and various other alloying phenomena in materials.

For example, when a brass coin, composed of copper and zinc, is produced, atoms of copper and zinc intermingle at the microscopic level through inter-diffusion. The temperature and concentration gradients play critical roles in driving this diffusion process. Atoms move from regions of high concentration, where there are many of them, to regions of low concentration, where there are fewer. This movement aims to achieve a homogenous mixture, an outcome observable in the seamless color and structure of the finished brass coin.
Atom migration
At the heart of both self-diffusion and inter-diffusion lies the concept of atom migration. This is the actual physical movement of atoms from one location to another within a material. Think of it as people moving from one house to another within the same neighborhood. Now, each atom typically sits tight in its own 'house' or lattice site within a solid. But with enough energy, such as from heating the material, the atoms can jump to neighboring vacant sites. This jump is also known as an 'atomic hop'.

Atom migration can be significantly influenced by external factors like temperature, applied pressure, or the presence of an electric field. Higher temperatures, for instance, give atoms more energy to overcome the barriers that keep them in place, increasing the rate at which they move. Various defects in a material's lattice structure, like vacancies or dislocations, also provide pathways that can facilitate or hinder atom migration. Understanding this process is fundamental, as it's crucial for the design of materials in applications ranging from electronics to structural engineering, where the reliability and durability of the material are paramount.
Concentration gradients
Concentration gradients are the driving force behind diffusion. They are to atom migration what slopes are to skiers – the greater the slope, the faster the skier goes down the hill. In materials science, this slope refers to how much the concentration of atoms or molecules changes over a given distance within a material. Molecules naturally move from areas where they are more crowded to areas where they have more space – that is, from high to low concentration.

A familiar experience of concentration gradients in action is when you add a spoonful of honey to tea. Initially, the honey is highly concentrated at the bottom. On its own, without stirring, it will gradually diffuse, creating a concentration gradient as the honey molecules spread upwards. Ultimately, the molecules move down the gradient until the honey concentration is the same throughout the tea, eliminating the gradient. The same principle explains why the scent of a perfume diffuses through an entire room, or how oxygen moves from the air in our lungs into the bloodstream. Understanding these gradients allows scientists to predict how materials will behave in different environments, influencing how they develop new materials for industrial, medical, and technological applications.

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Most popular questions from this chapter

A sheet of \(\mathrm{BCC}\) iron \(2 \mathrm{mm}\) thick was exposed to a carburizing gas atmosphere on one side and a decarburizing atmosphere on the other side at \(675^{\circ} \mathrm{C}\). After having reached steady state, the iron was quickly cooled to room temperature. The carbon concentrations at the two surfaces of the sheet were determined to be 0.015 and 0.0068 wt \(\% .\) Compute the diffusion coefficient if the diffusion flux is \(7.36 \times 10^{-9}\) \(\mathrm{kg} / \mathrm{m}^{2}\) -s. Hint: Use Equation 4.9 to convert the concentrations from weight percent to kilograms of carbon per cubic meter of iron.

For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration to 0.35 wt \(\%\) at a point \(2.0 \mathrm{mm}\) from the surface. Estimate the time necessary to achieve the same concentration at a 6.0 -mm position for an identical steel and at the same carburizing temperature.

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, \(p_{\mathrm{N}_{2}}(\text { in } \mathrm{MPa}),\) and absolute temperature \((T)\) according to $$C_{\mathrm{N}}=4.90 \times 10^{-3} \sqrt{p_{\mathrm{N}_{2}}} \exp \left(-\frac{37.6 \mathrm{kJ} / \mathrm{mol}}{R T}\right)$$.Furthermore, the values of \(D_{0}\) and \(Q_{d}\) for this diffusion system are \(3.0 \times 10^{-7} \mathrm{m}^{2} / \mathrm{s}\) and \(76,150 \mathrm{J} / \mathrm{mol}\), respectively. Consider a thin iron membrane \(1.5 \mathrm{mm}\) thick that is 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 other side \(5.0 \mathrm{MPa}(49.3 \mathrm{atm})\).

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 \(4.3 \times 10^{-11} \mathrm{m}^{2} / \mathrm{s}\).

The steady-state diffusion flux through a metal plate is \(7.8 \times 10^{-8} \mathrm{kg} / \mathrm{m}^{2}-\mathrm{s}\) at a tem- perature of \(1200^{\circ} \mathrm{C}(1473 \mathrm{K})\) and when the concentration gradient is \(-500 \mathrm{kg} / \mathrm{m}^{4}\). Calculate the diffusion flux at \(1000^{\circ} \mathrm{C}(1273 \mathrm{K})\) for the same concentration gradient and assuming an activation energy for diffusion of \(145,000 \mathrm{J} / \mathrm{mol}\).

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