This document discusses diffusion and Fick's laws of diffusion. It defines diffusion as the net flux of particles down a concentration gradient, with flux depending on concentration gradient and temperature. Diffusion processes like carburization can increase surface hardness by introducing carbon atoms. Diffusing atoms require activation energy to move into new sites. The Arrhenius equation relates diffusion rate to temperature and activation energy. The document also defines Fick's first law of diffusion, relating flux to the diffusion coefficient and concentration gradient. It provides examples calculating diffusion rates, activation energies, and concentration gradients.

1. APPLICATION OF DIFFUSION
ENG. KAREEM. H. MOKHTAR

2. DIFFUSION
• Diffusion refers to the net flux of any species
• The magnitude of this flux depends upon the concentration gradient and
temperature.

3. APPLICATION
• The carburization process can be used to increase surface hardness. In
carburization, a source of carbon, such as a graphite powder or gaseous phase
containing carbon, is diffused into steel components

4. ACTIVATION ENERGY
• A diffusing atom must squeeze past the surrounding atoms to reach its new site.
In order for this to happen, energy must be supplied to allow the atom to move
to its new position

5. ARRHENIUS EQUATION
• Rate = c e-E/RT
• Rate: rate of diffusion (m2/s
or jump per second)
• E: activation energy
• R: gas constant
• T: temperature
• ln(rate) = ln(c) – E/RT
• Straight line equation
Y= m x + c
Ln (rate)
1/T
Slope= -E/R

6. SHEET 4
• Q1) Suppose that atoms are found to move from one site to another at a rate of
109 jumps per second at 400 oC and the C0 is equal to 3.8 x 1016, calculate the
activation energy (Calory) needed.
• Answer
• Rate = c e-E/RT
• 109 = 3.8 x 1016 x e-E/1.98*673
• E= 23257 cal/mol

7. SHEET 4
• Q2) Suppose that atoms are found to move from one site to another at a rate of
5x108 jumps/ second at 500 oC and 8x 1010 jumps/s at 800 oC, calculate the
activation energy needed for the process
• Answer
• Q= 27,800 cal/mol

8. SHEET 4
• Q3) At 300ºC the diffusion and activation energy for Cu in Si are D(300ºC) = 7.8 x
10-11 m2/s E = 41.5 kJ/mol
What is the diffusion at 350ºC?
• Answer
• Rate1 = c e-E/RT  7.8 x 10-11 = C e-41500/8.314* 573
• Rate2 = c e-E/RT  D= C e-41500/8.314* 623
• Get D = 15.7 x 10-11 m2/s

9. DIFFUSION

10. FICK’S FIRST LAW
• The rate at which atoms, ions, particles or other species diffuse in a material can
be measured by the flux J. Here we are mainly concerned with diffusion of ions or
atoms. The flux J is defined as the number of atoms passing through a plane of
unit area per unit time

11. FICK’S FIRST LAW
• J: flux
• D: diffusivity , cm2/s
• dc/dx : concentration gradient atoms / (cm3. cm) OR atomic % / cm

12. SHEET 4
• Q4) Methylene chloride is a common ingredient of paint removers. Besides being
an irritant, it also may be absorbed through skin. When using this paint remover,
protective gloves should be worn.
If butyl rubber gloves (0.04 cm thick) are used, what is the diffusive flux of
methylene chloride through the glove?
• diffusion coefficient in butyl rubber: D = 110 x10-8 cm2/s
• surface concentrations:
• C1 = 0.44 g/cm3
• C2 = 0.02 g/cm3

13. ANSWER

14. SHEET4
• Q5) One step in manufacturing transistors, which function as electronic switches
in integrated circuits, involves diffusing impurity atoms into a semiconductor
material such as silicon (Si). Suppose a silicon wafer 0.1 cm thick, which originally
contains one phosphorus atom for every 10 million Si atom, is treated so that
there are 400 phosphorus atoms for every 10 million Si atoms at the surface.
• calculate the concentration gradient (at % / cm)

15. ANSWER
• Initial concentration in atom %
• Ci=
1𝑃 𝑎𝑡𝑜𝑚
107 𝑎𝑡𝑜𝑚𝑠
x 100 = 0.00001 at %P
• Cs=
400𝑃 𝑎𝑡𝑜𝑚
107 𝑎𝑡𝑜𝑚𝑠
x 100 = 0.004 at%P
•
𝑑𝑒𝑙𝑡𝑎 𝐶
𝐷𝑒𝑙𝑡𝑎 𝑥
=
0.00001−0.004
0.1
= - 0.04 at% P /cm