<p>19.2) consider a kilomole of 3 He gas atoms under STP conditions</p><p>No. of 3He atoms at STP = N = 1 kilomole = 6.0231026</p><p>Volume at STP =V = 22.4 dm3 </p><p>3 27 Mass of 3He atom = m =  4.9810 6.0231026</p><p> a) What is the Fermi temperature of the gas?</p><p>Fermi Temperature = TF  ?</p><p>We know that</p><p>2 h 2  N  3 TF    2mk 1.504v  2 2 6.6261034   6.0231026  3    27 23   2  4.9810 1.3810 1.504 22.4   0.069K  0.07K</p><p>FermiTemperature  TF  0.07K</p><p>   b) Calculate and exp  kT  kT    ? kT we knowthat</p><p>3    2mkT  2 V   ln2   kT h 2 N     3  2   27 23    2  4.9810 1.3810  273 22.4   ln 2 2 26   6.6261034   6.02310        12.7 kT    exp   e kT  ?  kT </p><p> e kT  3.28105</p><p>3 kT c) Find the average occupancy f  of a single particle state that has energy of  2</p><p>Energy :   3 kT 2 Chemical Potential :   12.7kT AverageOccupancy : f    ? We knowthat 1 f       KT e 1 1  3 12.7 kT  2  e kT 1 1  1.469106 f    6.8107 Average Occupancy  f    6.8107 19.3) For a system of noninteracting electrons, show that the probability f   of finding an electron in a state with energy ∆ above the chemical potential μ is the same as the probability of finding an electron absent from the with energy ∆ below μ at any given temperature T N  1 f        g  e kT 1 where     1 f     e kT 1 for the probability of absent at      1 1 1 f       1  e kT 1 e kT 1  e kT 1 1 1 f        e kT 1 1 1 e kT 1  e kT thereforethey areequal to eachother</p><p>19.4a) Verify that the average energy per fermion is 3  at absolute zero by making 5  F a direct calculation of U 0 N Use equation 19.18 3 U  N forT  0 5 F U 3   T o    N 5 F b) Similarly, prove that the average speed of a fermion gas particle at T=0 is 3  , 4 VF 3 2 where the Fermi velocity V is defined by  F   mvF F 5  v  f v gv dv v   0 N Nv 1 f v    1 2   mv   gv  2  e kT 1 g  d  gv dv</p><p>1 2 3 8 2V  1 2  1 gv dv  m 2  mv   mv  dv h3  2  2 1 2  1 2  1 mv  mv  dv    8 2V 3 2 2 v  m 2   3  1 2    mv   h 0  2  kT   e  1 3 2  3 8 2V  1  3 v v  m dv 3    1 2    mv   h 2 0  2    kT   e  1 need touseint egrationin part!</p><p>19.6) At very low temperature, the electronic specific heat capacity of a metal is </p><p> approximately ce  AT where A can be determined by experiment. For gold, the constant A is found to be A  0.73J kilomole1K 1 . Compare this to the value obtained from Equation (19.20). (The atomic weight of gold is 197 and its density is18.9103 kgm3 ) From eqn (19.20), one identifies  2   Nk 2 TF 2 34 2 3 3 1 1 6.6310   318.910 26  withT        6.0210  F k F 8.62105 eVk 1 29.111031  8 197 </p><p>TF  62028k  2 6.021026 1.3811023 Jk 1 A   2 62028k  0.66J  kilomole1K 2</p><p>19.7 a) Calculate  F for aluminum assuming three electrons per aluminum atom.</p><p>3 Kg N 2.6910 3 atoms  m 6.021026  5.991028 V 27 Kg kilomole kilomole N # Density for electrons  3 V 1.81029</p><p>2 2 6.631034   31.81029  3      5.6 2.1 11.8eV F 31   29.1110  8 </p><p> b) Show that the aluminum at T=100 K, μ differs from  F by less than 0.01%. (The density of aluminum is 2.69103 kgm3 and its atomic weight is 27.) 2   2  T      1    0 12  T    F   2      2       1000k     0 1  12  11.8eV     5 1     8.6210 eVk   5  0 1 4.3810  lessthan0.01%</p><p> c) Calculate the electronic contribution to the specific heat capacity of aluminum at room temperature and compare it to 3R</p><p>  T   2  kT      Ce  Nk   Nk  2  TF  2   F    8.617 105 eVk 1  2.43103    3 n  R   2    297J  kilomole  k 1</p><p>19.8) In sodium there are approximately 2.61026 conduction electrons per cubic meter, which behave as a free electron gas. Give an approximate value for the electronic specific heat of sodium at room temperature. (The atomic weight of sodium is 23.) In the series </p><p> expansion force , show that at this temperature the cubic term is negligible compared with the linear term. N  2.61028 V 2 6.631034 2  3 2.61028  3     F 29.111031  8   5.6eV  0.579  3.243eV   k  298  2 C  Nk   R  0.008 e 2  3.243  2  325J kilomole1k 1</p><p>19.10) calculate the isothermal compressibility of the fermion gas consisting of the free electrons in silver. Compare your answer with the experimental value for silver of 0.991011 Pa 1</p><p>1  2v       V  2 p T 2 NkT fromeq19.24 P  F 5 V 2 2 h 2  N  3  1  3 T      eq19.8 F 2mk 1.504  V </p><p>2  5 2 h2  N  3  1  3 P  Nk      5 2mk 1.504  V  2 Nh 2  N  3 1 P     5 5m 1.504  V 3 2 2 3 5 Nh  N  1 V 3     5m 1.504  p 3 2 5  Nh2  N  3  1 V       3 5m 1.504   p 5 3 2 5 dv  Nh 2  N  3   3 8         P 5 dp 5m 1.504    5 3 2 5 1  Nh2  N  3   3 8            P 5 v 5m 1.504    5 3 this equationleadsto  P with P  2.11010 Pa fromtextbook 5 3 3   P  2.11010 Pa 1.261011 Pa 1 5 5 6 19.12) For the white dwarf star Sirius B, do the computation leading to Rmin  7 10 m Use Equation (19.28) and (19.30) with the appropriate parameter value.</p><p>6 Rmin  7 10 m We knowthat 2a R  19.32 min b calculating b 3 b  GM 2 19.30 5 G  Gravitational energy  6.6731011 M  Mass of the sun  2.09 1030 kg</p><p>3 2 b  6.6731011 2.091030  5 b  1.751050 calculating a 2 2 3h  9  3 5 a    N 3 10me  32 2  calculating N 22 20 5 1.82 10 7.2310 56 N   6.1010 2 5.331014 so valueof ais</p><p>2 2 34 3 5 36.62610   9  56 a    6.1010  3 10 9.11031  32 2  a  1.451037  0.0933 4.3881094 a  5.921056 put values of a and bin19.32 25.921056  R  min 1.751050 6 6 Rmin  6.77 10  7 10 19.13) consider the collapse of the sun into a white dwarf. For the sun, M= 21030 kg R= 7108 m , V=1.41027 m3 a) Calculate the Fermi energy of the sun’s electrons</p><p>21030 No.of electrons   1.2051057 1.66 10 27 1 #of electrons  of nucleous 2 N 1.205 21057  V 1.41027 2 h 2  3N  3  F    2me  8V  2 1.205 7.231027  3   0.33m v     F e  27   1.26 1.410  5  F  0.33mev  6.24810</p><p> F  20.6eV</p><p> b) What is the Fermi Temperature?  20.6eV T  F   2.39105 k F k 8.62105 eVk 1 c) What is the average speed of the electrons in the fermion gas (see problem 19-4). Compare your answer with the speed of light. N 1.205 21057 Density  9.111031   9.111031  V 1.41027 kg  0.39 3 m</p>