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UCSC PHYS 112 - PHYS 112 Homework 8

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PHYSICS 112Homework 8Due in class, Tuesday March 6There will be one more homework.Note th at the Final Exam will be on Tuesday, March 20, 12:00–3: 00 pm.1. Ionization of the hydrogen atom(a) Consider the formati on of atomic hydrogen from a proton and an electron, i.e. according tothe reactione−+ H+←→ HShow that the equilibrium concentration of the reactants isne−nH+nH= nQ,eexp−∆EkBT, (1)wherenQ,e=mkBT2π¯h23/2i.e. nQ,e≡ 1/VQ,ein which VQ,eis the quantum volume of the electrons, and ∆E is theionization energy.Note: Just consider the ground state of the hydrogen atom and the ionized stat e ; ignor ehigher excited (bound) states of the atom. Neglect the spin of the p art i c l es , which does notaffect the fi nal answer. Equation (1) is called the Saha equation.(b) Note that i f all the electrons and protons arise from t he ionization of hydrogen atoms, wehave ne−= nH+and sone−= n1/2Hn1/2Q,eexp−∆E2kBT. (2)Note th at:i. The exponent in Eq. (2) involves ∆E/2, not ∆E, so there’s more to it than just theBoltzmann factor.ii. The electron concentration in Eq. (2) is proportional to the square root of the concen-tration of neutral hydrogen atoms.iii. If we add extra electrons, so they are no longer all due to the ionization of hydrogenatoms, then we have to go back to Eq. (1) which shows that the concentration of protonsactually goes down.Hence show that the temperature Tiat which half the hydrogen atoms are ionized is givenbyexp−∆EkBTi=nHnQ,e. (3)Note: You might have thought that the c ondi t i on would be e−βi∆E≃ 1, or equivalentlykBTi≃ ∆E, but Eq. (3) also involves the density. Furthermore, if the density is low, i.e.nH/nQ,e≪ 1, then Tiis much less than ∆ E/kB. The re ason is that, at low densities,the ionized state has much more entropy than the bound atom, so the probability that theatom ionizes is much greater than the Bolt zm ann factor ex p( −β∆E), which only includes1the effects of the energy. This result is important, for example, in the early universe wherethe temperature at which electrons and protons combined to form n eut r al hydrogen (andallow the thermal radi ati on, now observable as cosmic background radiation, to decouplefrom matter) is about 3000 K, far less than the ionization ener gy (13.6 eV) divided by kBwhich is about 150,000 K.2. Particle-antiparticle equilibriumConsider the reaction of a particle with its anti-particleA++ A−= 0,where A+and A−may, for example be (i) electrons and positrons, or (ii) electrons and holes ina semiconduct or . Denote the energy to create a particle-antiparticle pair by ∆. For the electron-position case, ∆ = 2mc2, where m is the electron mass. For the case of case of electrons and holesin a semiconductor, ∆ would be a property of the materi al . (For simplicity ne gl e ct the spin of theparticles.)(a) Calculate the equilibrium density n = n+= n−(assuming that the densities of parti c l e s andantiparticles are equal).Note: it is convenient t o distribute the minimum energy of the particle-antiparticle pair (∆E)equally between the particle and hole (i.e. half each).(b) Estimate the density of electrons and positrons produced thermally at the c enter of the sun,assuming that T = 1.5 × 107K.Note: You are are given that the mass of the electron is 0.51 Mev, and 1 ev ≡ 11, 600 K.3. Thermal expansion near absolute zero.(a) Prove the fol l owing three Maxwell relations∂V∂TP= −∂S∂PT, (4a)∂V∂NP= +∂µ∂PN, (4b)∂µ∂TN= −∂S∂NT. (4c)Note that quantities which are kept constant on both sides of the equation are not explicitlyindicated. For example, Eq. (4a), should really be written∂V∂TP,N= −∂S∂PT,N.(b) Show from Eq. (4a) and the third law of thermodynamics that the volume coefficient ofthermal expansionα =1V∂V∂TPapproaches zero as T → 0.24. Calculation of dT/dP for waterFrom the vapor pressure equat i on1PdPdT=LkBT2,which comes from the Clausius-Clapeyron equation with certai n assumptions, calc ul at e the valueof dT/dP at P = 1 atm for the liq u i d-vapor equilibrium of water. The heat of vaporization at100◦C is 2260 J g−1. Express your result in kelvin/atm.5. Phase boundary between liquid and solid3He.The figure below is a sketch of the coexistence line between the solid (S) and liquid (L) phases of3He at low temperature.ST (K)L0230P (atm)It has the fol l owing properties(a) For T → 0, dP/dT = 0.(b) For T small but non-zero, dP/dT < 0.Explain what these resul ts tell you about the difference in entropy between the sol i d and l i q ui dphases. Is the r es ul t for part (b) surpr i s i

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