Question

Q3. Calculate the theoretical density of the following metals in g/cmº given their atomic radius and crystal structure. Compare those results with the accepted density. (8 points) Metal Crystal Structure FCC Aluminum Copper Nickel Silver Atomic Radius (m x 10) 0.1413 0.1278 0.1246 0.1445 FCC FCC Q4. Explain how slipping of atoms in metallic bonds lead to ductile behavior. (2 points) Q5. Explain why a polymer (plastic) described as thermoset is more difficult to recycle when compared to one described as a thermoplastic. (3 points) Q6. A tensile load of 190 kN is applied to a round metal specimen that has an initial diameter of 16-mm and a gauge length of 50 mm. Under this load, the bar elastically deforms so that the gauge length increases to 50.1349-mm and the diameter decreases to 15.993 mm. Using the information given determine the modulus of elasticity and Poisson's ratio for this metal. (8 points) w h atween initial tangent modulus, secant modulus, and actual modulus.

Answer 1

3.(a). We know that

density (d) = Effective number of atoms (Z) x Atomic mass of metal (A) / Volume of cube (a³) x Avogadro's number (N_A)

Effective number of atoms (Z) in FCC = 4

Atomic mass of metal (A) of Al = 27 g/mol

Given radius (r) = 0.1413 x 10^-9 m = 1.413 x 10^-8 cm

Edge length (a) for FCC = 4r / (2)^1/2 = 4 x  1.413 x 10^-8 cm /1.414 = 3.997 x 10^-8 cm

So,

d = 4 x 27 g/mol / (3.997 x 10^-8 cm)³.(6.022 x 10²³)

d = 2.81 g/cm³

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(b). We know that

density (d) = Effective number of atoms (Z) x Atomic mass of metal (A) / Volume of cube (a³) x Avogadro's number (N_A)

Effective number of atoms (Z) in FCC = 4

Atomic mass of metal (A) of Cu = 63.5 g/mol

Given radius (r) = 0.1278 x 10^-9 m = 1.278 x 10^-8 cm

Edge length (a) for FCC = 4r / (2)^1/2 = 4 x  1.278 x 10^-8 cm /1.414 = 3.61 x 10^-8 cm

So,

d = 4 x 63.5 g/mol / (3.61 x 10^-8 cm)³.(6.022 x 10²³)

d = 8.97 g/cm³

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(c). We know that

density (d) = Effective number of atoms (Z) x Atomic mass of metal (A) / Volume of cube (a³) x Avogadro's number (N_A)

Effective number of atoms (Z) in FCC = 4

Atomic mass of metal (A) of Ni = 58.69 g/mol

Given radius (r) = 0.1246 x 10^-9 m = 1.246 x 10^-8 cm

Edge length (a) for FCC = 4r / (2)^1/2 = 4 x  1.246 x 10^-8 cm /1.414 = 3.52 x 10^-8 cm

So,

d = 4 x 58.69 g/mol / (3.52 x 10^-8 cm)³.(6.022 x 10²³)

d = 8.94 g/cm³

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(d). We know that

density (d) = Effective number of atoms (Z) x Atomic mass of metal (A) / Volume of cube (a³) x Avogadro's number (N_A)

Effective number of atoms (Z) in FCC = 4

Atomic mass of metal (A) of Ag = 108 g/mol

Given radius (r) = 0.1445 x 10^-9 m = 1.445 x 10^-8 cm

Edge length (a) for FCC = 4r / (2)^1/2 = 4 x  1.445 x 10^-8 cm /1.414 = 4.09 x 10^-8 cm

So,

d = 4 x 108 g/mol / (4.09 x 10^-8 cm)³.(6.022 x 10²³)

d = 10.5 g/cm³