An important consideration in materials physics is the number of defects that are unavoidable due to statistical considerations.
Defects influence structural, electronic and optical properties. The equilibrium statistical physics of defects in a material can be modeled as a collection of N two-state systems. Every atomic site can be either, (a) occupied with an atom, in which case its energy is zero, or (b) occupied by a defect, in which case its energy is e. The formation energy (€) is the energy needed to remove an atom from the regular lattice position to an interstitial position, from which it diffuses out of the crystal
(a) Determine the partition function for this system, in terms of N, E, and T.
(b) What is the Helmholtz free energy (F) and the internal energy (U) for a sample of N atoms?
(c) What is the average number of defects, in terms of N, E, and T?
(d) Assume the formation energy is 1 eV. If you have a collection of spherical copper nanocrystals with a radius of 50 nm at room temperature, what is the minimum number of nanocrystals that you would need in a sample before you could expect (on average) that only one of them would have a single defect and the others would be defect free?
(a) Determine the partition function for this system, in terms of N, E, and T.
(b) What is the Helmholtz free energy (F) and the internal energy (U) for a sample of N atoms?
(c) What is the average number of defects, in terms of N, E, and T?
(d) Assume the formation energy is 1 eV. If you have a collection of spherical copper nanocrystals with a radius of 50 nm at room temperature, what is the minimum number of nanocrystals that you would need in a sample before you could expect (on average) that only one of them would have a single defect and the others would be defect free?
1 answer:
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Answer:
10.6 mA
Explanation:
t = time interval = 1.00 s
q = magnitude of charge on each ion = 1.6 x 10⁻¹⁹ C
n₁ = number of Na⁺ ions = 2.68 x 10¹⁶
q₁ = charge due to Na⁺ ions = n₁ q = (2.68 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.004288 C
n₂ = number of Cl⁻ ions = 3.92 x 10¹⁶
q₂ = charge due to Cl⁻ ions = n₂ q = (3.92 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.006272 C
i₁ = Current due to Na⁺ ions = =
= 0.004288 A
i₂ = Current due to Cl⁻ ions = =
= 0.006272 A
Current passing between the electrodes is given as
i = i₁ + i₂
i = 0.004288 + 0.006272
i = 0.01056 A
i = 10.6 x 10⁻³ A
i = 10.6 mA