Answer:
a) At T = 600 K
D_Cu = (6.48 × 10⁻¹⁹) cm²/sec
D_Al = (8.21 × 10⁻¹³) cm²/sec
b) At their respective melting points
D_Cu at 1083°C = (2.00 × 10⁻⁹) cm²/sec
D_Al at 660°C = (5.70 × 10⁻⁹) cm²/sec
c) Check Explanation.
Explanation:
D = D₀ e^(-ΔH/kT)
k = Boltzmann's constant = (1.38 × 10⁻²³) J/K
And T is in Kelvin
For Cu
D₀ = 0.16 cm²/sec
ΔH = 2.07 eV = (2.07 × 1.602 × 10⁻¹⁹) J
= (3.316 × 10⁻¹⁹) J
For Al
D₀ = 0.0407 cm²/sec
ΔH = 1.28 eV = (1.28 × 1.602 × 10⁻¹⁹) J
= (2.051 × 10⁻¹⁹) J
a) The diffusion coefficient: D_Cu and D_Al at 600K.
D_Cu = D₀ e^(-ΔH/kT)
D₀ = 0.16 cm²/sec
ΔH = (3.312 × 10⁻¹⁹) J
k = (1.38 × 10⁻²³) J/K
T = 600 K
D = 0.16 × e^[(-3.316 × 10⁻¹⁹)/(1.38 × 10⁻²³ × 600)]
D = 0.16 × e^(-40.048)
D = (6.48 × 10⁻¹⁹) cm²/sec
D_Al
D₀ = 0.0407 cm²/sec
ΔH = (2.051 × 10⁻¹⁹) J
k = (1.38 × 10⁻²³) J/K
T = 600 K
D = 0.047 × e^[(-2.051 × 10⁻¹⁹)/(1.38 × 10⁻²³ × 600)]
D = 0.047 × e^(-24.77)
D = (8.21 × 10⁻¹³) cm²/sec
b) D_Cu and D_Al at their melting point. (T_M for Cu is 1083°C, and for Al is 660°C).
D_Cu = D₀ e^(-ΔH/kT)
D₀ = 0.16 cm²/sec
ΔH = (3.312 × 10⁻¹⁹) J
k = (1.38 × 10⁻²³) J/K
T = 1083°C = 1,356.15 K
D = 0.16 × e^[(-3.316 × 10⁻¹⁹)/(1.38 × 10⁻²³ × 1,356.15)]
D = 0.16 × e^(-40.048)
D = (2.00 × 10⁻⁹) cm²/sec
D_Al
D₀ = 0.0407 cm²/sec
ΔH = (2.051 × 10⁻¹⁹) J
k = (1.38 × 10⁻²³) J/K
T = 660°C = 933.15 K
D = 0.047 × e^[(-2.051 × 10⁻¹⁹)/(1.38 × 10⁻²³ × 933.15)]
D = 0.047 × e^(-24.77)
D = (5.70 × 10⁻⁹) cm²/sec
c) The diffusion coefficient for substances increase as we move from solid state to liquid state to gaseous state.
And from the calculations, it is evident that these metals have a higher diffusion coefficient at/close to their melting points.
- So, to obtain high diffusion coefficients for metal contact in electronic circuits, temperatures as high and close to their melting points should be used.
- Of the two metals, Al and Cu, at the two points where the calculations were carried out, Aluminium has a higher diffusion coefficient at both points, hence, it's a better choice for interconnects.
Hope this Helps!!!
