21-Mat-B10 Properties and Processing of Micro- and Nanomaterials · May 2013
Question 1 of 6: Porosity of a Silver-Infiltrated Tungsten Contact
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — 10-Met-B10, Advanced Electronic Materials — May 2013, 3 hours. Six questions; Question 6 (40 marks) is mandatory and any 4 of the remaining 5 questions (15 marks each) complete the paper. All six are answered below.
Reference texts: S.O. Kasap, Principles of Electronic Materials and Devices; W.D. Callister, Materials Science and Engineering: An Introduction.
Two points matter: Question 4 prints the GaAs carrier density per m³ (not per cm³), which sets the conductivity scale; and the formula sheet prints Planck's constant as h = 4.375×10-15 eV·s, which is not the physical value (4.136×10-15 eV·s) — the standard value is used in Question 5.1 and the result with the printed value is also given.
Question 1: Porosity of a Silver-Infiltrated Tungsten Contact (15 marks)
Given. Mass of tungsten (W), mW = 125 g; mass of infiltrated silver (Ag), mAg = 105 g; final composite density ρcomp = 13.8 g/cm³.
Given data (bulk densities from standard tables)
Quantity
Symbol
Value
Tungsten density
ρW
19.3 g/cm³
Silver density
ρAg
10.49 g/cm³
Mass of W
mW
125 g
Mass of infiltrated Ag
mAg
105 g
Final composite density
ρcomp
13.8 g/cm³
Find. The volume fraction of the original compact that is (a) interconnected (open) porosity and (b) closed porosity.
Approach. Get the total composite volume from its final mass and density, subtract the solid-tungsten volume and the infiltrated-silver volume (which equals the interconnected pore volume the liquid could reach), and whatever volume remains is closed porosity.
Total composite volume from the final density. The composite mass is the sum of the tungsten and infiltrated silver (closed pores stay empty and contribute no mass):
$$V_{total}=\dfrac{m_W+m_{Ag}}{\rho_{comp}}=\dfrac{125+105}{13.8}=16.667\ \text{cm}^3$$
Interconnected (open) pore volume. This is exactly the volume the liquid silver could reach and fill:
$$V_{Ag}=\dfrac{m_{Ag}}{\rho_{Ag}}=\dfrac{105}{10.49}=10.010\ \text{cm}^3$$
Substituting into the volume-fraction definition,
$$f_{interconnected}=\dfrac{V_{Ag}}{V_{total}}=\dfrac{10.010}{16.667}=\boxed{0.601\ (60.1\%)}$$
Closed porosity by difference. Whatever volume is neither solid W nor infiltrated Ag must be an isolated pore the silver never reached:
$$V_{closed}=V_{total}-V_W-V_{Ag}=16.667-6.477-10.010=0.180\ \text{cm}^3$$
$$f_{closed}=\dfrac{V_{closed}}{V_{total}}=\dfrac{0.180}{16.667}=\boxed{0.0108\ (1.08\%)}$$
As a check, the three volume fractions (solid W 38.9%, interconnected pores 60.1%, closed pores 1.1%) sum to 100%, confirming the partition is complete.
