4.2 Molecular orbitals and atomic orbitals Consider a linear chain of four identical atoms representing a hypothetical molecule.

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1 4. Molecular orbitals and atomic orbitals Consider a linear chain of four identical atoms representing a hypothetical molecule. Suppose that each atomic wavefunction is 1s wavefunction. This system of identical atoms has a center of symmetry C with respect to the center of the molecule (midway between the second and the third atom), and all molecular wavefunctions must be either symmetric or antisymmetric about C. a. Using LCAO principle, sketch the possible molecular orbitals. b. Sketch the probability distribution ψ c. If more nodes in the wavefunction lead to greater energies, order the energies of the molecular orbitals. Note: The electron wavefunctions, and the related probability distributions, in a simple potential energy well that are shown in Figure 3.15 can be used as a rough guide towards finding the appropriate molecular wavefunctions in the four-atom symmetric molecule. For example, if we were to smooth the electron potential energy in the four-atom molecule into a constant potential energy, that is, generate a potential energy well, we should be able to modify or distort, without flipping, the molecular orbitals to somewhat resemble ψ 1 to ψ 4 sketched in Figure Consider also that the number of nodes increases from none for ψ 1 to three for ψ 4 in Figure Author's Note to the Instructor: This has to changed to four atoms to simplify the problem.

2 Solution Center of symmetry for the PE O S ψ a Combinations of 1,,3 and A ψ b S ψ c ( ) A ψ d Left: Molecular orbitals based on LCAO for a linear array of 4 atoms. S is symmetric and A is antisymmetric. Right: Probability distributions. All sketches are rough schematic illustrations. Energy increases with the number of nodes. In the above figures, energy increases downwards from ψ a to ψ d. Lowest for the top MO (molecular orbital) and highest for the bottom MO.

3 S ψ a Wavefunctions in a PE well O A ψ b O S ψ c O A ψ d O Comparison of molecular orbitals based on LCAO for a linear array of 4 atoms and the first four wavefunctions in a potential energy well. Notice the resemblance.

4 Author's Note: Generally, the actual molecular oribitals are not a simple linear combination but may be a linear combination in which each individual atomic wavefunction is scaled or weighted by a certain amount. The similarity between the molecular oribitals and the wavefunctions in a PE well is more apparent in the above figure from waves, Atoms and Solids, D.A. Davies, Longman (England), 1978, Figure 7.5, p14

5 4.10 Temperature dependence of the Fermi energy a. Given that the Fermi energy for Cu is 7.0 ev at absolute zero, calculate the E F at 300 K. What is the percentage change in E F and what is your conclusion? b. Given the Fermi energy for Cu at absolute zero, calculate the average energy and mean speed per conduction electron at absolute zero and 300 K, and comment. Solution a. The Fermi energy in ev at 0 K is given as 7.0 ev. The temperature dependence of E F is given by Equation 4.3. Remember that E FO is given in ev. E F = E FO π 1 1 kt E 3 ( ) π ( J/K)( 300 K) FO ( )( ) E F = 7.0 ev 1 9 = ev ev J/eV eV 7.0 ev % difference = 100% = % 7.0 ev This is a very small change. The Fermi energy appears to be almost unaffected by temperature. b. The average energy per electron at 0 K is: E av (0 K) = 3/5 (E FO ) = 4. ev The average energy at 300 K can be calculated from Equation 4.6: 3 Eav( T ) = E 5 FO 5π ( ) 5π ( J/K)( 300 K) E av (300 K) This is a very small change. kt E FO 7.0 ev = E av (300 K) = ev ( )( ) ev J/eV sume that the mean speed will be close to the effective speed v e. Effective speed at absolute zero is denoted as v eo, and is given by: 1 E = av( 0 K) q meveo

6 9 qeav (0 K) ( J/eV)( 4. ev) = = v eo = m/s 31 m At 300 K, the effective speed is v e : m/s Comparing the values: e - ( kg) 9 ( J/eV)( ev) qeav(300 K) v e = = = m e - ( kg) m/s m/s % difference = 100% = % m/s The mean speed has increased by a negligible amount (0.003%) from 0 K to 300 K. Note: For thermal conduction this tiny increase in the velocity is sufficient to transport energy from hot regions to cold regions. This very small increase in the velocity also allows the electrons to diffuse from hot to cold regions giving rise to the Seebeck effect.

7 4. Lattice waves and heat capacity a. Consider an aluminum sample. The nearest separation R ( atomic radius) between the Al-Al atoms in the crystal is 0.86 nm. Taking a to be R, and given the sound velocity in Al as 5100 m s -1, calculate the force constant β in Equation Use the group velocity ν g from the actual dispersion relation, Equation 4.55, to calculate the sound velocity at wavelengths of Λ = 1 mm, 1 µm and 1 nm. What is your conclusion? b. Aluminum has a Debye temperature of 394 K. Calculate its specific heat capacity at 30 C (Darwin, Australia) and at -30 C (January, Resolute Nunavut, Canada). c. Calculate the specific heat capacity of a germanium crystal at 5 C and compare it with the experimental value in table 4.5 Solution a. The group velocity of lattice waves is given by Equation For sufficiently small K, or long wavelengths, such that 1/Ka 0, the expression for the group velocity can be simplified like in Equation 4.6 to ν = a g β M From here we cam calculate the force constant β ν β = M g a The mass of one Al atom is M M = N and finally for the force constant we receive at A 3 ( 7 10 kg mol ) 3 ( mol ) M ν at g 5100 m s = = 9 N A a β m = 14.6 kg s - Now considering the dispersion relation ν g β N M πa Λ A ( Λ) = a cos at 1 K = π and Equation 4.55 we receive Λ Performing the calculations for the given wavelengths, we receive the following results:

8 ( 10 3 m) ( 10 6 m) ( 10 9 m) ν = 5100 m s -1 g ν = m s -1 g ν = m s -1 g It is evident that for the first two wavelengths, 1/Ka 0 and we can use the approximation in Equation For the third wavelength, this is not true and we have to use the exact dispersion relation when calculating the group velocity. b. In summer, the temperature is given to be T = 30 C = 303 K and T/T D is 303/394 = The molar heat capacity of Al at 30 C is C m = 0.9 (3R) =.95 J K -1 mol -1 The corresponding specific heat capacity is (.95 J K mol ) ( 7 g mol ) Cm c s = = = 0.85 J K -1 g -1 M at At -30 C, T = 43 K and T/T D is 43/394 = C m = 1.94 J K -1 mol -1 Cm (.40 J K mol ) and s = = M ( 7 g mol ) c = 0.81 J K -1 g - c. We can find the heat capacity of Ge in the way described in part b. Alternatively, we can find C m performing the integration in Equation 4.64 numerically C m T TD 3 T x 4 e x = 9R dx 3R 3 T = x ( 1) 360 x D 0 e 0 ( e ) = 3. J K -1 mol -1 Thus the specific heat capacity is: ( 3. J K mol ) 3 ( kg mol ) at Cm c s = = = J K -1 kg -1 M at From Table 4.5, the specific heat capacity is 3.38 J K -1 mol -1. x 4 e x dx = 3R ( 0.931)

9 5.13 has a valency of III and has V. When and atoms are brought together to form the crystal, as depicted in Figure 5.54, the three valence electrons in each and the five valence electrons in each are all shared to form four covalent bonds per atom. In the crystal with some 10 3 or so equal numbers of and atoms, we have an average of four valence electrons per atom, whether or, so we would expect the bonding to be similar to that in the Si crystal: four bonds per atom. The crystal structure, however, is not that of diamond but rather that of zinc blende (Chapter 1). a. What is the average number of valence electrons per atom for a pair of and atoms and in the crystal? b. What will happen if Se or Te, from Group VI, are substituted for an atom in the crystal? c. What will happen if Zn or Cd, from Group II, are substituted for a atom in the crystal? d. What will happen if Si, from Group IV, is substituted for an atom in the crystal? e. What will happen if Si, from Group IV, is substituted for a atom in the crystal? What do you think amphoteric dopant means? f. Based on the above discussion, what do you think the crystal structures of the III-V compound semiconductors Al, P, In, InP, and InSb will be?

10 Solution atom (Valency V) atom (Valency III) Valence electron ψ hyb orbitals ψ hyb orbitals Valence electron ion core (+5e) ion core (+3e) Figure 5Q13-1: Bonding Structure Explanation of bonding in : The one s and three p orbitals hybridize to form 4 ψ hyb orbitals. In there are 5 valence electrons. One ψ hyb has two paired electrons and 3 ψ hyb have 1 electron each as shown. In there are 3 electrons so one ψ hyb is empty. This empty ψ hyb of can overlap the full ψ hyb of. The overlapped orbital, the bonding orbital, then has two paired electrons. This is a bond between and even though the electrons come from (this type of bonding is called dative bonding). It is a bond because the electrons in the overlapped orbital are shared by both and. The other 3 ψ hyb of can overlap 3 ψ hyb of neighboring to form "normal bonds". Repeating this in three dimensions generates the crystal where each atom bonds to four neighboring atoms as shown. Because all the bonding orbitals are full, the valence band formed from these orbitals is also full. The crystal structure is reminiscent of that of Si. is a semiconductor. a. The average number of valence electrons is 4 electrons per atom. b. Se or Te replacing will have one additional electron that cannot be involved in any of the four bonds. Hence Se and Te will act as donors. c. Zn or Cd replacing will have one less electron than the substituted atom. This creates a hole in a bond. Zn and Cd will act as acceptors. d. The Si atom has 1 less electron than the atom and when it substitutes for an atom in there is a "hole" in one of the four bonds. This creates a hole, or the Si atom acts as an acceptor. e. The Si atom has 1 more electron than the atom and when it substitutes for a atom in there is an additional electron that cannot enter any of the four bonds and is therefore donated into the CB (given sufficiently large temperature). Si substituting for therefore acts as a donor. f. All these compounds (Al, P, In, InP, InSb) are compounds of III elements and V elements so they will follow the example of.

11 5.4 Schottky junction a. Consider a Schottky junction diode between Au and n-si, doped with donors cm - 3. The cross- sectional area is 1 mm. Given the work function of Au as 5.1 ev, what is the theoretical barrier height, Φ B, from the metal to the semiconductor? b. Given that the experimental barrier height Φ B is about 0.8 ev, what is the reverse saturation current and the current when there is a forward bias of 0.3 V across the diode? (Use Equation 4.37.) Solution a. The number of donors (N d = 10 m -3 ) is related to the energy difference from the Fermi level ( E = E c E F ) by: E Nd = Nc exp kt N E = kt ln N d c From Table 5.1, N c = m -3, and we are given and N d = 10 m -3. suming temperature T = 300 K: E = m 1 ( J/K)( 300 K) ln m J/eV The work function of n-si (Φ nsi ) must be less than that of Au (Φ Au = 5.1 ev) in order to have a Schottky junction. Φ nsi is given as Φ nsi = E + χ, where χ = 4.01 ev (see Example 5.19) is the electron affinity of Si. Φ nsi = ev ev = 4.15 ev This is indeed less than Φ Au and therefore we have a Schottky junction. The effective barrier height Φ B is then: Φ B = Φ Au χ = 5.1 ev 4.01 ev = 1.09 ev b. The experimental potential barrier is given as Φ B = 0.8 ev. sume temperature T = 300 K. The reverse saturation current (I o ) is given by (see Example 5.19) (where B e = A K - m - is the effective Richardson-Dushman constant) I o = AB T e Φ B exp kt = ev

12 I o 6 6 ( m )( A K m )( 300 K) exp 9 ( 0.8 ev)( J/eV) = 10 3 I o = A ( J/K)( 300 K) The forward bias voltage V f is given as 0.3 V. The forward current I f is then: I f 9 9 ( ) ( C)( 0.3 V) A exp ev f I exp 1 = 3 kt = o I f = A ( J/K)( 300 K) 1

13 *6.15 Ultimate limits to device performance a. Consider the speed of operation of an n-channel FET-type device. The time required for an electron to transit from the source to the drain is τ t = L/v d, where L is the channel length and v d is the drift velocity. This transit time can be shortened by shortening L and increasing v d. the field increases, the drift velocity eventually saturates at about v dsat = 10 5 m s -1 when the field in the channel is equal to E c 10 6 V m -1. A short τ t requires a field that is at least E c. 1. What is the change in the PE of an electron when it traverses the channel length L from source to drain if the voltage difference is V DS?. This energy must be greater than the energy due to thermal fluctuations, which is of the order of kt. Otherwise, electrons would be brought in and out of the drain due to thermal fluctuations. Given the minimum field and V DS, what is the minimum channel length and hence the minimum transit time? b. Heisenberg's uncertainty principle relates the energy and the time duration in which that energy is possessed through a relationship of the form (Chapter 3) E t > ħ. Given that during the transit of the electron from the source to the drain its energy changes by ev DS, what is the shortest transit time, τ, satisfying Heisenberg's uncertainty principle? How does it compare with your calculation in part (a)? c. How does electron tunneling limit the thickness of the gate oxide and the channel length in a MOSFET? What would be typical distances for tunneling to be effective? (Consider Example 3.10). Solution sume temperature T = 300 K. The saturation velocity is given as v dsat = 10 5 m/s and the saturation field is given as E c = 10 6 V/m. a. (1) The change in the PE is PE. This is the charge times the voltage, i.e. PE = ev DS. () The lower limit to PE due to thermal fluctuations is kt. Therefore, substituting into the equation above: ev DS = kt V DS = kt/e = ( J/K)(300 K)/( C) = V This is the lower limit to V DS. The minimum channel length L can now be found from the minimum electric field, given by E c = V DS /L: L = V DS /E c = ( V)/(10 6 V/m) = m The minimum transit time τ t is then, τ t = L/v dsat = ( m)/(10 5 m/s) = s

14 The above limit is the thermal fluctuation limit (thermal noise limit). b. Consider the Heisenberg uncertainty principle. Let τ = t be the transit time. During this time the energy changes by E = ev DS. We are given E t > ħ, therefore, substituting for the shortest transit time: ev DS τ = ħ 34 ħ J s τ = = = s ev DS 9 ( C)( V) The uncertainty limit allows a shorter transit time down to ps. Thus thermal fluctuation limit will operate at room temperature. c. If the oxide becomes too thin then the electron tunneling will allow gate charge to tunnel into the channel. This will lead to a gate current. The field effect will fail. Similarly, if the source and drain are very close there will then be a tunneling current, a drain current, even when the transistor is off. From Example 3.10, one can guess that the thickness should be less than 1 nm or 10 Å depending on various material properties. The same order of magnitude also applies to the minimum source-drain separation.