Learning Objectives PDF

AI&DS A 1 Dr Prita Nair Learning Objectives PH 1001 T Lec # 26_27 5_6.11.24 At the end of this session you should be able to 10 Fermi Dirac Statistics Problems...

AI&DS A 1 Dr Prita Nair Learning Objectives PH 1001 T Lec # 26_27 5_6.11.24 At the end of this session you should be able to 10 Fermi Dirac Statistics Problems  Derive expression for density of states  Average energy of electron at 0K (application).  AI&DS A Fermi Dirac Distribution 2 Dr Prita Nair PH 1001 T Lec # 26_27 5_6.11.24 The Fermi-Dirac Distribution Function (1926) Applicable to fermions, i.e identical particles that obey Pauli’s exclusion principle Represents probability that a particular quantum state of energy E is occupied by an electron and has been derived by Fermi and Dirac to be F(E) = 1 1+ e(E-EF/kT) Where EF is a constant reference energy level called the Fermi level What is EF? Let us take the case of E>EF and T=0K, then F(E) = 1 = 1 = 0 1+ e(E-EF/0) 1+ e i.e for all E>EF at T=0K, F(E)=0 or probability of occupation by an electron for all energy levels higher that Fermi level is zero.  AI&DS A Fermi Dirac Distribution contd,,, 3 Dr Prita Nair PH 1001 T Lec # 26_27 5_6.11.24 F(E) = 1 = 1 F(E) = 1 T=0K 1+ e(E-EF/0) 1+ e- kT1 T1>0K 1 T2>T1 i.e for all E0K, and E=EF, F(E) = 1 = 1 = 1/2 1+ e(0/kT) 1+ 1 EF i.e for all T>0K, Fermi level represents the energy level at which the probability of occupation by an electron is 0.5 or 50%. T=0K i.e for all T>0K, electrons from levels kT below the Fermi level can acquire energy kT and move to higher levels thereby giving T>0K non-zero probability of occupation above EF upto EF+kT and kT lowers the probability of occupation of levels upto EF-kT kT EF  AI&DS A Fermi Levels & Carrier densityPH 1001 T Lec # 26_27 5_6.11.24 4 Dr Prita Nair Therefore a Fermi level can be defined as The highest energy level occupied by an electron in a solid at 0K and it also represents , for all temperatures above 0K, the energy level at which the probability of occupation by an electron is 0.5 Number density of electrons in a solid in an interval of energy dE about E is therefore = density of allowed states with energy in the interval dE about E x probability that it is occupied by an electron = g(E) F(E) dE General expression for number density of electrons in a solid in an interval of energy E1 to E2 is density of allowed states with energy in the interval dE about E [g(E)] x probability that it is occupied by an electron [F(E)] E2 =  g ( E ) F ( E )dE E1  AI&DS A Sample problems on Fermi Distribution 5 Dr Prita Nair PH 1001 T Lec # 26_27 5_6.11.24 1. Use the Fermi function to obtain the value of F(E) for (E-EF) =0.01eV at 200K. Ans:0.36 Given (E-EF) =0.01eV= 0.01x1.6x10-19 J = 0.016x10-19 J , T=200K, k=1.38x10-23 J/K F(E) = 1 1+ e(E-EF/kT) E T=200K EF 1 1 1 F( E) = = = = 0.36 0.016 𝑥 10−19 1+ 𝑒 0.579 2.785 1+ 𝑒 1.38 𝑥 10−23 𝑥 200  AI&DS A Sample problems on Fermi Distribution 6 Dr Prita Nair PH 1001 T Lec # 26_27 5_6.11.24 2. Using the Fermi function, evaluate the temperature at which there is 1% probability that an electron in a solid will have an energy 0.5eV above EF of 5eV. Ans: 1262K Given (E-EF) =0.5eV=0.5x1.6x10-19=0.8x10-19, F(E)=.01 k=1.38x10-23 J/K, T=? E 5.5 eV E = 0.5eV 5 eV F(E) = 1 EF 1+ e(E-EF/kT) 1 0.01 = 1+𝑒 𝐸−𝐸𝐹 /𝑘𝑇 (𝐸−𝐸𝐹 /𝑘𝑇) ) 0.01 (1+ 𝑒 = 1 0.01 𝑒 (𝐸−𝐸𝐹 /𝑘𝑇) ) = (1- 0.01)= 0.99 𝑒 (𝐸−𝐸𝐹 /𝑘𝑇) ) = 99 (𝐸 − 𝐸𝐹 )/kT = ln (99)=4.595 𝐸−𝐸𝐹 0.8 𝑥 10−19 T= = = 1261.57 K ≈ 1262 K 𝑘𝑙𝑛(99) 1.38 𝑥 10−23 𝑥 4.595  AI&DS A Summary 7 Dr Prita Nair PH 1001 T Lec # 26_27 5_6.11.24 Fermi Dirac distribution, F(E) represents probability that a particular quantum state of energy E is occupied by an electron F(E) = 1 1+ e(E-EF/kT) Fermi level, EF can be defined as The highest energy level occupied by an electron in a solid at 0K and it also represents , for all temperatures above 0K, the energy level at which the probability of occupation by an electron is 0.5 General expression for number density of electrons in a solid in an interval of energy E1 to E2 is density of allowed states with energy in the interval dE about E [Z(E)] x probability that it is occupied by an electron[F(E)] E2 =  g ( E ) F ( E )dE E1  AI&DS A 8 Dr Prita Nair Estimation of Carrier density PH 1001 T Lec # 26_27 5_6.11.24 Number density of electrons in a solid in an interval of energy dE about E is therefore = density of allowed states with energy in the interval dE about E x probability that it is occupied by an electron = g(E) F(E) dE General expression for number density of electrons in a solid in an interval of energy E1 to E2 is density of allowed states with energy in the interval dE about E [g(E)] x probability that it is occupied by an electron [F(E)] E2 E2 5.5 eV =  g ( E ) F ( E )dE E1 EF E1 5 eV  AI&DS A Density of states 9 Dr Prita Nair PH 1001 T Lec # 26_27 5_6.11.24 Expression for density of states between E and E+dE  Energy of any allowed level n for electron confined in cubical box is given by the formula (recall particle in a box) 2 2 2 p2 1  h  1  h2  1   2  K.E = =   =   =   2k 2 n2h2 n 2m 2m  dB  2m  2dB  2m  dB  En = = as k= 2m 8mL2 L = 1 2 2  k where k = 2 , p = k 2m dB Therefore no. of levels between E and E+dE = no. of allowed values between k and k+dk i.e. = allowed values between n and n+dn; En 8mL2 n = nx + n y + nz = 2 2 2 2 where n can take values 1,2,3……. h2 Let us represent each n value in a 3D space whose axes are defined by nx, ny and nz. (Remember multiplying n by a constant /L gives k and this space is therefore called the momentum space.) In this space, all the points that have the same n value, will lie on the surface of a spherical shell. But same n value indicates same energy E. Thus a spherical shell of radius n in momentum space represents a equi-potential surface of energy E.  AI&DS A Density of states (contd ….) PH 1001 T Lec # 26_27 5_6.11.24 10 Dr Prita Nair THIS SLIDE IS FOR YOUR UNDERSTANDING ONLY NOT TO BE WRITTEN FOR EXAMS! ny E0 (1,1,1) (1,2,1) E1 What did you want to (2,1,1) find? No. of levels n+dn (1,1,2) between n & n+dn n = vol enclosed E2 (2,2,1) between n & n+dn (2,1,2) (1,2,2) -nx nx E3 (2,2,2) Unit volume in the positive quadrant of this All states of energy E1 will be on the momentum space contains surface of a spherical shell. 1 allowed energy level -nz side vol no. of allowed states All states of energy E2 will be on the -ny surface of a spherical shell. 1 1 1, E0 2 8 8, E0,E1,E2,E3 Spherical shell of radius n- Momentum space representation of E  AI&DS A Density of states (contd ….) PH 1001 T Lec # 26_27 5_6.11.24 11 Dr Prita Nair E+dE can be represented by a sphere of radius n+dn in this space ny Unit volume in the positive quadrant of this space contains 1 allowed energy level n No. of levels between E & E+dE = no. of levels between n and n+dn =volume enclosed between the spheres of radius n and n+dn -nx nx = 4/3 [(n+dn)3-n3] = 4 n2dn ignoring higher powers of dn nz -ny or alternatively volume enclosed between the spheres of radius n and n+dn =volume of shell of thickness dn =surface area x thickness =4n2dn Only (+ve quadrants) of the sphere contributes since only +ve integer values of nx,ny and nz are allowed, therefore, the no. of allowed states in the interval n & n+dn = Vol of shell of thickness in the +ve quadrant =1/8th the volume of the spherical shell = 4 n2dn /8 = n2dn /2  AI&DS A Density of states (contd ….) 12 Dr Prita Nair PH 1001 T Lec # 26_27 5_6.11.24 THIS SLIDE IS FOR YOUR UNDERSTANDING ONLY NOT TO BE WRITTEN FOR EXAMS! Blue/Red continuous arrows indicate +ve axes of sphere Red dotted lines indicate –ve axes ½ a spherical shell Therefore no. density of allowed states in the interval n & n+dn = Vol of shell in the +ve quadrant =1/8th the volume of the spherical shell. No. of levels between E & E+dE = 4n2dn/ 8 = n2dn/ 2 1.1 =1 2 4 8 of the spherical shell  AI&DS A Density of states (contd ….) 13 Dr Prita Nair PH 1001 T Lec # 26_27 5_6.11.24 No. of levels between E & E+dE = = n2dn/ 2 Expressing n in terms of known quantities of the material. Volume of material we were considering = L3 h2 2 8mL 2 8mL2 E= 2 n Therefore n2 = E and n= 2 E 8mL h 2 h 8mL2 4mL2 Differentiating n2 2ndn = 2 dE ndn = 2 dE h h 2 2 8mL 8mL Therefore n 2 dn = n.ndn = 2 E. 2 dE h 2h 3/ 2 1  8mL2  =  2  E dE 2 h  No. of levels between E & E+dE = n2dn/ 2 3/ 2 3/ 2  1  8mL  2  1  2.2mL  2 2  23 = . 2  E dE = . 2  E dE = 3 (2m )3 / 2 L3 E dE 2 2 h  2 2 h  4h  AI&DS A Density of states (contd ….) 14 Dr Prita Nair PH 1001 T Lec # 26_27 5_6.11.24 No. of levels between E & E+dE in unit volume of material =  (2m )3 / 2 8L3 E dE = 2 (2 m )3/ 2 E dE 3 3 4h L h3 No. density of states between E & E+dE taking into account 2 possible spin states  2  4 = 2. 3 (2m ) E dE  = ( ) 3/ 2 3/ 2 2 m E dE h  h3 Density of states represents all the states per unit volume of material that can be occupied by charge carriers and is given by 4 m* represents effective mass of electron. It captures g ( E ) dE = 3 (2m ) * 3/ 2 E dE the change in velocity of e- in the vicinity of the ion cores h However no. of charge carriers per unit volume will depend on the probability that the charge acquires sufficient energy to occupy these states and is given by Fermi Dirac distribution function F(E).  AI&DS A 15 Dr Prita Nair PH 1001 T Lec # 26_27 5_6.11.24 Number density of electrons in a solid in an interval of energy dE about E is therefore = (density of allowed states with energy in the interval dE about E) x (probability that it is occupied by an electron) F(E) = 1 = g(E) F(E) dE where 1+ e(E-EF/kT) 4 g ( E ) dE = h3 (2m ) * 3/ 2 E dE General expression for number density of electrons in a solid in an interval of energy E1 to E2 is therefore E2 =  g ( E ) F ( E ) dE E1  AI&DS A Summary 16 Dr Prita Nair PH 1001 T Lec # 26_27 5_6.11.24 No. density of states with energy E=g(E)/dE is  E1/2 Graphical representation No. density of electrons with energy E= g(E)/dE x F(E) Prob. of occupation No. density of of level E by an T=0K states with electron energy E 1 T1>0K g(E)/dE F(E) x EF E EF E No. density of electrons with energy E T=0K F(E).g(E)/dE = T1>0K EF E  AI&DS A Dr Prita Nair Problem PH 1001 T Lec # 26_27 5_6.11.24 17 Find electron density for metals at 0K. For metals at 0K , all the levels up to energy level EF is occupied. Therefore the number density of electrons available at 0K are those present in energy levels ranging from 0 to EF. Hence expression for number density of electrons in a metal at 0K, nc is EF EF nc=  g(E) F(E) dE = (4/h3)(2m*e)3/2 0  E1/2.1 dE since F(E) =1 upto EF 0 EF 2 3/ 2  = (4/h3)(2m*e)3/2 [E3/2 / (3/2)] =   EF 0 K  where  = (4/h3)(2me)3/2 3  = (8/3) (2m*e/h2)3/2 EF3/2 OR Expression for Fermi Energy at 0K for metal is EF(0K) = (h2/2m*e) (3nc/ 8) 2/3 For temperatures above 0K the full expression for F(E) has to be put in during integration. EF(T) that will be obtained in such as case is given in summary for completion sake. As kBT is