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# Lect21 - Lecture 21 p 1 Miscellaneous Notes • The end is...

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Unformatted text preview: Lecture 21, p 1 Miscellaneous Notes • The end is near – don’t get behind. • All Excuses must be taken to 233 Loomis before noon, Friday, Dec. 3. • The PHYS 213 final exam times are * 7-9 PM, Monday, Dec. 13 and * 8-10 AM, Tuesday, Dec. 14 . The deadline for changing your final exam time is 10pm, Tuesday, Nov. 30. • Homework 6 is due Saturday , Dec. 4 at 8 am. • Course Survey = 2 bonus points (accessible at the top of HW6) • Review Lecture: Thursday, Dec. 9, 3-5 PM in 228 NHB. Lecture 21, p 2 Lecture 21 Review & Examples Gas p 1 T p Phase Diagram: p 3 p 2 Solid Liquid f p (T 2 ) 0.5 1 p T 1 T 3 T 2 T 3 > T 2 > T 1 Lecture 21, p 3 Example: p + e ↔ H at 3000K We pick this reaction because each component (proton, electron, hydrogen atom) has no internal modes (except spin, not important here) so we can write down the component μ ’s easily. What is the relation between n p and n H at T = 3000 K? Lecture 21, p 4 Solution We pick this reaction because each component (proton, electron, hydrogen atom) has no internal modes (except spin, not important here) so we can write down the component μ ’s easily. What is the relation between n p and n H at T = 3000 K? ln ln ln Also: because they are produced in pairs. p e H pQ eQ HQ p e n n n kT kT kT n n n n n + =- ∆ ⇐ = ( 29 2 because p e H kT pQ eQ HQ H kT p pQ eQ eQ H HQ pQ HQ n n n e n n n n n n n n n e n n n ∆- ∆- = = ≈ ≈ 2 4 3 10 at 3000 K p H n m n- ≈ Start with the exact equilibrium rule: μ p + μ e = μ H Now use that these are almost ideal monatomic components, (as long as the density is low). So: Now use: ∆ =13.6eV and n eQ = 8x10 26 m-3 at 3000K On the Sun: ρ ~ ρ H2O . So, n H ~ 6 × 10 29 /m 3 . ⇒ n p ~ 6 × 10 29 /m 3 , not very large. Lecture 21, p 5 Example: Adsorption of Atoms At T = 300 K and p = 1 atm , it is observed that 50% of the binding sites on a particular metal surface are occupied. When the temperature is raised to 320 K at constant pressure, only 25% of the sites are occupied. (You may assume that p Q is constant over this small temperature range.) What is the binding energy of a site on the surface? k = 8.617 × 10-5 eV/K Lecture 21, p 6 k = 8.617 × 10-5 eV/K Solution ( 29 ( 29 ( 29 ( 29 / (300 ) / (320 ) / 1/320 1/300 1 1 Atm (300 ) (300 ) 1 Atm 2 1 Atm (300 ) What is p (320K)? (320 ) 3 Atm : 3 ln 3 / 1/300 1/320 0.454 eV k K Q k K Q k f K p K p e p K p K p e So e k-∆-∆- ∆- = = = = + = = =- = → ∆ = At T = 300 K and p = 1 atm , it is observed that 50% of the binding sites on a particular metal surface are occupied. When the temperature is raised...
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Lect21 - Lecture 21 p 1 Miscellaneous Notes • The end is...

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