Chapter 3 EE145 Spring 2002 Homework 3 Solution Prof Ali Shakouri Second Edition ( 2001 McGraw Hill) Chapter 3 3 10 Line spectra of hydrogenic atoms Spectra of hydrogen like atoms are classified in t[.]
Homework Solution EE145 Spring 2002 Prof Ali Shakouri Second Edition ( 2001 McGraw-Hill) Chapter 3.10 Line spectra of hydrogenic atoms Spectra of hydrogen-like atoms are classified in terms of electron transitions to a common lower energy level a All transitions from energy levels n = 2, 3, to n = (the K shell) are labeled K lines and constitute the Lyman series The spectral line corresponding to the smallest energy difference (n = to n = 1) is labeled the Kα line, next is labeled Kβ , and so on The transition from n = ∞ to n = has the largest energy difference and defines the greatest photon energy (shortest wavelength) in the K series; hence it is called the absorption edge Kαe What is the range of wavelengths for the K lines? What is Kαe? Where are these lines with respect to the visible spectrum? b All transitions from energy levels n = 3, 4, to n = (L shell) are labeled L lines and constitute the Balmer series What is the range of wavelengths for the L lines (i.e., Lα and Lαe)? Are these in the visible range? c All transitions from energy levels n = 4, 5, … to n = (M shell) are labeled M lines and constitute the Paschen series What is the range of wavelengths for the M lines? Are these in the visible range? d How would you expect the spectral lines to depend on the atomic number Z? Solution Consider the expression for the energy change for a transition from n = n2 to n = n1, 1 ∆E = − Z E I − n2 n1 (1) where EI = 13.6 eV is the ionization energy from the ground state (n = 1) and Z is the atomic number; for hydrogen Z = The absorption wavelength is given by λ = hc / ∆E a and For the Lyman series (K-shell), n1 = For Kα, n2 = 2, which gives ∆E = 10.2 eV from Eqn (1) λ (Kα ) = −34 −1 hc (6.626 × 10 J s)(3.0 × 10 m s ) = 122 nm = ∆E (10.2 eV)(1.602 × 10−19 J/eV) For Kαe, n2 = ∞ and ∆E = 13.6 eV and −34 −1 hc (6.626 × 10 J s)(3.0 × 10 m s ) = 91.2 nm λ (Kαe ) = = ∆E (13.6 eV)(1.602 × 10 −19 J/eV ) The wavelengths range from 91.2 nm to 122 nm; much shorter than the visible spectrum b For the Balmer series (L-shell), n1 = For La, n2 = 3, which gives ∆E = 1.889 eV and −34 −1 hc (6.626 × 10 J s )(3.0 × 10 m s ) = 657 nm λ (Lα ) = = ∆E (1.889 eV )(1.602 × 10 −19 J/eV ) 3.1 Homework Solution EE145 Spring 2002 Prof Ali Shakouri For Lαe, n2 = ∞ and ∆E = 3.4 eV and −34 −1 hc (6.626 × 10 J s )(3.0 × 10 m s ) = 365 nm λ (Lαe ) = = ∆E (3.4 eV)(1.602 × 10−19 J/eV) The wavelengths range from 365 nm to 657 nm Part of this is in the visible spectrum (blue and green) c For the Paschen series (M-shell), n1 = For Ma, n2 = 4, which gives ∆E = 0.6611 eV and λ (Mα ) = −34 −1 hc (6.626 × 10 J s )(3.0 × 10 m s ) = 1877 nm = ∆E (0.6611 eV)(1.602 × 10 −19 J/eV ) For Mαe, n2 = ∞ and ∆E = 1.511 eV and −34 −1 hc (6.626 × 10 J s )(3.0 × 10 m s ) = 821 nm λ (Mαe ) = = ∆E (1.511 eV)(1.602 × 10−19 J/eV) The wavelengths range from 821 nm to 1877 nm which is in the infrared region d The transition energy depends on Z2, and therefore the emitted photon wavelength of the spectral lines depends inversely on Z2 3.13 Hund's rule For each of the following atoms and ions, sketch the electronic structure, using a box for an orbital wavefunction and an arrow (up or down) for an electron: a Manganese, [Ar]3d54s2 b Cobalt, [Ar]3d74s2 c Iron ion, Fe+2, given that Fe is [Ar]3d64s2 d Neodymium ion, Nd+3, given that Nd is [Xe]4f46s2 Solution Mn [ Ar] 3d5 4s2 Cobalt [ Ar] 3d7 4s2 Fe + [ Ar] 3d4 4s2 4s 3d 4s 3d 4s 3d 4f 6s Nd+3 [ Xe] 4f 6s2 Figure 3Q13-1 Electronic structures of various atoms 3.2 Homework Solution EE145 Spring 2002 Prof Ali Shakouri Chapter 4.1 Phase of an atomic orbital a What is the functional form of a 1s wavefunction, ψ(r)? Sketch schematically the atomic wavefunction ψ1s(r) as a function of distance from the nucleus b What is the total wavefunction Ψ1s(r,t)? c What is meant by two wavefunctions Ψ1s(A) and Ψ1s(B) that are out of phase? d Sketch schematically the two wavefunctions Ψ1s(A) and Ψ1s(B) at one instant Solution ψ1s(r) decays exponentially as exp(−r/ao), where ao is the Bohr radius (Table 3.2 in the a textbook) ψ1s(r) r Figure 4Q1-1 Atomic wavefunction as a function of distance from the nucleus b Ψ1s(r,t) = ψ1s(r) × exp(−jEt/h) = ψ1s(r) × exp(−jωt) where ω = E/h is an angular frequency (Section 3.2.2 of the textbook) The total wavefunction is harmonic in time Recall that exp(jθ) = cosθ + jsinθ so that exp(jωt) = cosωt + jsinωt c Two sine waves of the same frequency will have a certain phase difference which represents the time delay between the time oscillations of the two waves Two waves will be in phase if their maxima coincide and out of phase if the maximum of one coincides with the minimum of the other Two 1s wavefunctions, Ψ1s(A) and Ψ1s(B) will have the same frequency Like two sine waves of the same frequency, the waves can be in phase or out of phase When they are out of phase, if Ψ1s(A) = then Ψ1s(B) = −1 and vice versa d Ψ1s(B) Ψ1s(A) r B r A Figure 4Q1-2 Two 1s wavefunctions which are out of phase (sketches at the same instant) 3.3 Homework Solution EE145 Spring 2002 Prof Ali Shakouri 4.2 Molecular orbitals and atomic orbitals Consider a linear chain of five 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 molecule (third atom from the end) 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 Solution POSSIBLE MOLECULAR ORBITALS IN INCREACING ENERGY ORDER PROBABILITY DISTRIBUTIONS A B C D E A B C D E A B C D E A B C D E A B C D E A B C D E A B C D E A B C D E A B C D E A B C D E 3.4 EE145 Spring 2002 Homework Solution Prof Ali Shakouri 4.3 Diamond and Tin Germanium, silicon, and diamond have the same crystal structure, that of diamond Bonding in each case involves sp3 hybridization The bonding energy decreases as we go from C to Si to Ge, as noted in Table 4Q3-1 Table 4Q3-1 Property Diamond Silicon Melting temperature, 0C 3800 Covalent radius, nm 0.077 Bond energy, eV 3.60 First ionization energy, eV 11.26 Bandgap, eV ? Germanium 1417 0.117 1.84 8.15 1.12 Tin 937 0.122 1.7 7.88 0.67 232 0.146 1.2 7.33 ? a What would you expect for the band gap of diamond? How does it compare with the experimental value of 5.5 eV? b Tin has a tetragonal crystal structure, which makes it different than its group members, diamond, silicon, and germanium Is it a metal or a semiconductor? What experiments you think would expose its semiconductor properties? Solution Given the properties in Table 4Q3-1, we have the following plots: 3.5 Homework Solution EE145 Spring 2002 Energy Gap, eV Energy Gap, eV Diamond 5 Diamond 3 Prof Ali Shakouri Ge Si Tin 0 200 1200 2200 -2 (a) Energy gap versus melting temperature From the plot, it seems that diamond has an Eg = 3.4 eV and Sn has Eg ≈ or is a metal Energy Gap, eV Diamond Si Ge -1 Tin -2 0.05 (b) Energy gap versus bond energy From the plot, it seems that diamond has an Eg = 6.8 eV and Sn has Eg ≈ −0.9 or is a metal Diamond Bond Energy, eV Energy Gap, eV Tin -1 3200 Melting Temperature °C Si Ge 0.1 Covalent Radius, nm Ge 0.15 Si Tin -1 (c) Energy gap versus covalent radius From the plot, it seems that diamond has an Eg = 4.7 eV and Sn has Eg ≈ −1.5 eV or is a metal 10 11 12 First Ionization Energy, eV (d) Energy gap versus first ionization energy From the plot, it seems that diamond has an Eg = 6.3 eV and Sn has Eg ≈ −0.3 eV or is a metal Figure 4Q3-1 Each is a plot of the band gap Eg (or energy gap) vs some property The straight line in each is drawn to pass through Si and Ge Diamond and tin points are then located on this straight line at the intersections with the vertical lines representing the corresponding properties of diamond and tin 3.6 Homework Solution EE145 Spring 2002 Prof Ali Shakouri a Diamond has an Eg greater than Si and Ge Averaging the four Eg for diamond we find 5.3 eV which is close to the experimental value of 5.5 eV b (1) All the four properties indicate that tin has Eg ≤ or is a metal (2) Tin’s semiconductivity can be tested by examining its electrical conductivity and optical absorption (see Chapter in the textbook) For example, for metals the conductivity should NOT be thermally activated over a wide temperature range, whereas for semiconductors there will be an Arrhenius temperature dependence over at least some temperature range Further, semiconductors have an absorption edge that corresponds to hυ > Eg (Chapter in the textbook) 4.4 Compound III-V semiconductors Indium as an element is a metal It has a valency of III Sb as an element is a metal and has a valency of V InSb is a semiconductor, with each atom bonding to four neighbors, just like in silicon Explain how this is possible and why InSb is a semiconductor and not a metal alloy (Consider the electronic structure and sp3 hybridization for each atom.) Solution The one s and three p orbitals hybridize to form ψhyb orbitals In Sb there are valence electrons One ψhyb has two paired electrons and ψhyb have electron each as shown in Figure 4Q4-1 In In there are electrons so one ψhyb is empty This empty ψhyb of In can overlap the full ψhyb of Sb The overlapped orbital, the bonding orbital, then has two paired electrons This is a bond between In and Sb even though the electrons come from Sb (this type of bonding is called dative bonding) It is a bond because the electrons in the overlapped orbital are shared by both Sb and In The other ψhyb of Sb can overlap ψhyb of neighboring In to form "normal bonds" Repeating this in three dimensions generates the InSb crystal where each atom bonds to four neighboring atoms as shown As 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, as all the valence electrons are in bonds Since it is similar to Si, InSb is a semiconductor In Sb Sb atom (Valency V) ψhyb orbitals Valence electron Sb ion core (+5e) In atom (Valency III) ψhyb orbitals Valence electron In ion core (+3e) Figure 4Q4-1 Bonding structure of InSb 3.7 In Sb In Sb Sb In Sb In In Sb In Sb Sb In Sb In EE145 Spring 2002 Homework Solution 3.8 Prof Ali Shakouri