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1 Thermodynamics of spin systems Consider a system of N spin-1/2 particles, each with magnetic dipole moment µ and in a magnetic field of magnitude B. Let n+ the number of particles with spin up. a) Show that the temperature of the system satisfies 1 T = k 2µB ln ! N − E/µB N + E/µB ” and use the result to determine an expression for the energy equation of state E = E(T,N). b) List the range of possible values of E and plot the temperature as a function of E over the entire range. Describe the range of energies for which the temperature is positive and that for which it is negative. c) Using the probabilities with which a single particle is in the spin up state or the spin down state, determine the mean energy E for a single particle. How does this compare to the expression for the energy, E, of the entire system that you obtained earlier? 2 Spin system in equilibrium with an environment Consider a system of N spin-1/2 particles, each with magnetic dipole moment µ and in a magnetic field of magnitude B. Suppose that the system is in equilibrium with an environment at temperature T > 0. a) Explain whether the number of particles in the spin up state is larger than, smaller than or the same as the number in the spin down state. b) Consider various spin systems in contact with environments at various temperatures. As the temperature of the environment increases does the fraction of particles in the spin up state increase, decrease or remain constant? Explain your answer. 3 Spin-1/2 particles in contact with an environment Consider a system of N spin-1/2 particles, each with magnetic dipole moment µ and in a magnetic field of magnitude B. Suppose that the system is in equilibrium with an environment at temperature T > 0. a) Consider a single particle in the ensemble. What temperature would guarantee that the particle is in the spin-up state with certainty? Explain your answer. 1 b) What temperature is such that the particle is equally likely to be in the spin up state versus in the spin down state? Explain your answer. c) Nuclear magnetic resonance (NMR) uses such systems of spin-1/2 particles. In this case the signal provided by any particle with spin-up cancels that provided by any particle with spin down. In order to increase the signal strength would it be better to increase or decrease the temperature of the system? Explain your answer. 4 Einstein solid: high temperature limit For an Einstein solid where q ” N, we found that Ω(N,q) ≈ #eq N $N 1 √ 2πN . Recall that the energy of the solid is E = !ω % q + N 2 & . a) Determine an expression for the temperature of the Einstein solid in terms of the total energy of the solid. b) Use this to determine an expression for the energy equation of state of the solid E = E(N,T). Does the energy have the expected behavior in terms of the number of particles? c) Using q ” N show that the result of the previous part implies that kT ” !ω (this is why it is called the high temperature limit). Determine the approximate energy of one mole of such oscillators at a temperature of 300 K. 5 Einstein solid: low temperature limit For any Einstein solid Ω(N,q) ≈ % 1 + q N − 1 &N−1 % 1 + N − 1 q &q ‘ N + q − 1 2π(N − 1)q ≈ # 1 + q N $N % 1 + N q &q ‘ N + q 2πNq whenever q ” 1 and N ” 1. In the low temperature limit q % N. a) Show that if 1 % q % N then Ω(N,q) ≈ %eN q &q 1 √2πq and S = kq ! ln %N q & + 1″ . 2 b) Use this to determine an expression for the temperature of the solid in terms of energy. Use this to get an expression for the energy in terms of temperature. 6 Gould and Tobochnik, Statistical and Thermal Physics, 4.24, page 210. 7 Energy distributions Two collections of “toy model” molecules have the following energy distributions. Prob A State 01234 Prob B State 01234 Which of these has the larger temperature? Explain your answer. 8 Five state system Consider a particle which could be in one of five possible states; the energies of these states are 0, #, 2#, 3# and 4#. a) Suppose that # = 0.050 eV and T = 100 K. Determine the probabilities with which the particle could be in each state and plot a bar graph of these. b) Suppose that # = 0.050 eV and T = 500 K. Determine the probabilities with which the particle could be in each state and plot a bar graph of these. c) Suppose that # = 0.20 eV and T = 500 K. Determine the probabilities with which the particle could be in each state and plot a bar graph of these. d) Use your results to describe qualitatively the appearance of the entire bar graph as the temperature increases. e) Use your results to describe qualitatively the appearance of the entire bar graph as the energy gap, #, increases. 3 a) First rlale enugy to nt= number spin up 2020 HW16 Q1 E= -MBN+ + MBn- and n-=N-A =E=MB(N-20+) E 2 uB 2S Ther = JE and S= kkin 2CE)] willgive temperatre N! Hoe R- ()= ) N? S= kh kh N!! – In – laN.vCi[ (W-N) it یہ = k /NIAN -(N- A+) 1n (N-n2) -Aaln[ Now se se tue DE ant aE and ant YFU-tUUI (tn-N(nI+ as = = -kein no – InlN-n)} nt N-nt T TU T N-A+ 2MB k In = 2MB N-V+ Nint [3 2019 HW 18 QI Thus withh M (N) N-A+ = (N we get k N-EMB T ZµB N+E/MB Alternative expressions ave h T 2µB N-A 1k = T 24B In ( To invert Hus: 2MB n KT N-E/MB N+E/u E/uB 2MBET N-E/uB e = =D Ne +E e* = N-E/uB N+E/MB MB =0 E MB /1-e 2MB/CT = E = MBN Ta b) The range of nt is O< N+ SN Thus when A+=0 E MBN when nt= N E = -M8N =D – MBN S EE MBN Now consider I as a furction o Е. If E 20 then N-E/MB <1 =D o = T>0 N+E/uB Thus If EZO then T<о If Es0 thern I>O Ther as E-DMBN 2402N n ]0-0 TO0 frong regahive as E-D-MBN -4 T-D0 fram tue. as E 0 =-4O =D T -0 ±00 T T DE c P+= 2MB/T – It e-us/KT e?MB/kT +1 e-248/Eт | P-= 1+e-24B/KT e?MB/kT +1 E = E+p+ + E-P- = -MB P+ +MBP- = MB C 2MB/K7 C?MB/KT емв MB 1-e 48/et 1+e 2MB/T Same as energy belore with N=1 The probabilities are 2020 нW16 Q2 spin up P+= 1+e-2MB/KT Spin dowh P-= e-2MB/kT 1 te-zмB/кT 1+ e2MB/кT. a) When T>O E-MB/кт < = 1 + е – гивает 2 = p+> 2 More in spin up b) As T increaxs -2MB increases =D 1+e-2MB CT KT increases = P+ clecreases fraction in spin up decreases 2020 HW 16 Q3 a) P+= 1+e-2,40/KT we reed P+=1 =D e-2MB/cT= =0 1n (-2,4B/C7) = In (0) -8 =D 2MB/CT = 8 =D T=O We need T=0 b) we need p+= 1/½ e-2/MB/KT = ( =D 1n(e-2mо/ст) =1 =0 -2MB/LT=O = T= 00 We neel T=00 c) Based on a,b) and reed for p+ to increase we reed to decrease temperatve a) ( ond we need S= kin2 Then N = S=k In =n()] S = k N[ne+ lng-hN] – h (2TN) зе ве зе se se be = But E 9= N be T =0 hw JE tiw NK So = =D Now E =hw(q+N/) T= E Nk Thwq Nk Phys 362 2020 HW 16 Q4 and q>N =D E twg Tuus E-hWN exact: T= Nk. E=NKT [exact: E= NET+ thON] Yes it is proportonal to N c) E= NkT = hwg +N/2) =0 q kT N ħw اد Thus k T >»ћiw N = NA= 6.02×1023 E= 602x(023x 1,38×10-233/cx 300K E= 2500J 2019 HW 16 94 a) N = (i+ N )H )b if 1gN< =0 9» h(2ta) So S = k a [In(+1] )b 2019 HW 18 Q5 EN E=hw (q+ N/2) =8 9= hw 2 se T ЭЕ Tbe ве де se be Then = DE thw se be = k + ka q N N Then =k k= )(nI k in (第) 0 = ( tiw hw 一/n() hw -/in()/10(一) E-=e-hw/CT Nhw E = Nhw[e-hw/kт +4] G+T Cu4 Prob 4.24 The abundance can be detemined by the probabinties. The fiaction in state A e-EAB is fa= PA= 근 where EA is the erergy of state A. Let A = trons Then C B fA T=300K 러 40-2 873-2/ B= cis. = e – (EB-Eл)B = -AEB =e= p. – 4180K/7 tB 4180 PA e 306 = 8,9×10-7 4180 T=1000K fo e 1000 0.015 fA Tuen useng NA= # in trans stalte N3= # 1 cis state NA= FAN etc,.. =D NB = 8.9×10-7 at 30ок NA No 0.015 at 10ooK NA -ESB Ps = e Z 2020 HW16 Q3 Consicler successive probabilihes: Psti = 0-(ES1-Es)B Ps e The drop in probability is larger when (Est1-Es)B is larger So the dep is larger wher B is lerger But B=KT implies. larger drep =D smaller temp 2020 HW 16 94a T =100K e=0.05eV k= 8.62E-05 6 epsilon= 0.05 T= 1.00E+02 state energy Boltzmann factor prob 0 이 1 0.99698 1.2 1 0.05 0.003020045 0.003011 1 2 0.1 9.12067E-06 9.09E-06 0.8 3 0.15 2.75448E-08 2.75E-08 20.6 4 0.2 8.31867E-11 8.29E-11 0.2 0.4 Ο 이 0.05 Z= 1.003029194 0.1 Energy Series1 0.15 0.2 2019 HW 19 246 k= 8.62E-05 8. epsilon= 0.05 T= 5.00E+02 state energy Boltzmann faactor prob 0 1 0.68875 1 0.0505 0.313330513 0.215806 2 0.1 0.098176011 0.067619 3 0.15 0.03076154 0.021187 4 0.2 0.009638529 0.006639 0.4 0.5 0.6 0.7 07 0.8 Z= 1.451906593 0.1 0.2 0.3 T =500K e=0.05eV 0.05 0.1 Energy Series1 0.15 0.2 k= 8.62E-05 epsilon= 0.2 T= 5.00E+02 state energy Boltzmann factor prob 0 이 1 0.990361 1.2 1 0.2 0.009638529 0.009546 1 0:4 9.29012E-05 9.2E-05 25 805131F-07 27-07 Z= 1.009732334 0.2 0.4 T =500K e=0.20eV 이 0.2 0.4 Energy Series1 0.6 0.8 d) The higher ereigy states become mave probable barp eregy higher T lower T. e) The lower erergy states become mave probabde proh N smaller 6 larger E ewgy 2019 HW 19 G4
