Solutions to Problem Set 15 (M23)

M2.3 [U = -A r^{-m} + \frac{B}{r^{n}}] (0) long range electrostatic attraction short range repulsion inner electron orbitals, nuclei Cubic unit cell → dF C r₀ → r₀ + dr d²F = A dr² d³F = B dr³ | r=r₀ [du = \frac{dU}{dr} = MAR^{-m} - BnB(n-1)r^{-n-1}] at r = r₀: [\frac{du}{dr} = 0] [⇒ B = \frac{M²A}{n-m}r₀^{n-m}] Substitute into (0) [U = -Ar^{-m} + \frac{M²A}{n-m}r₀^{n-m}r^{-n}] [U(r) = -Ar^{-m} + M²Ar₀^{n-m}r^{-n} = Ar₀^{n-m}(-\frac{1}{r^{m}} + \frac{M²}{r^{n}})] From definition in question [-KT_{m} = Ar₀^{-m} (\frac{m-n}{m})] ∴ [A = -KT_{m}r₀^{m}(\frac{m-n}{m})] [U = \frac{-KT_{m}r₀^{m}(m-n)}{m-n}r^{-m} - \frac{M²KT_{m}r₀^{n-m}}{m-n}r^{-n}] [\frac{dU}{dr} = -mKT_{m}r₀^{m}(\frac{m-n}{m})r^{-m-1} + \frac{nMK_{Tm}r₀^{n-m}}{m-n}r^{-n-1}] [\frac{dF}{dr} = (m+1)mKT_{m}r₀^{m}(\frac{m-n}{m})r^{-m-2} - (n+1)nMK_{Tm}r₀^{n-m}r^{-n-2}] For r₀, [\frac{1}{r₀}\frac{dF}{dr} + (m+1)mKT_{m}r₀^{m}(\frac{m-n}{m})r₀^{-m-2} = (n+1)nMK_{Tm}r₀^{n-m}r₀^{-n-2}] [= (mnKT_{m} + (m+1)mKT_{m})r₀^{-2} = \frac{(mn + m + 1)mKT_{m}}{r₀^{2}}] ∴ [E = \frac{1}{r₀}\frac{dF}{dr} \Big|_{r=r₀} = \frac{mnKT_{m} + (m+1)mKT_{m}}{r₀^{2}}] The purpose of this question is to demonstrate the intrinsic link between moduli and Tm. Diamond, SiC have high E.