Midterm 2_S08_sol - MatE 153 Spring 2008 Midterm 2...

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Unformatted text preview: MatE 153 Spring 2008 Midterm 2 SOLUTIONS Class average was 79% 1.) An electron from the 2s shell in B (Z=5) is excited to the 5th n shell. What wavelength of light is given off when the electron returns to the 2s shell? (10 points) EI = mq 4 = 13.6eV 2 8ε 0 h 2 En = − Z2E I n2 1⎞ ⎛1 ΔE = −5 2 ⋅ 13.6eV⎜ 2 − 2 ⎟ = −71.4eV 5⎠ ⎝2 The negative sign represents the light being given off. ΔE = λ= hc λ ΔE = hc 4.136 × 10 −15 eV ⋅ s ⋅ 3 × 10 8 71.4eV m s = 1.7 × 10 −8 m = 17 nm The selection rules are n must change, Δl = ±1, Δm l = 0,±1 . Briefly state why there are selection rules. (5 points) l and ml are related to the angular momentum of the electron. The selection rules are needed because only certain transitions of l and ml are possible to ensure that momentum is conserved. 2.) 3.) Using the selection rules, give an allowed set of 4 quantum numbers for an electron in the 2s of B and an allowed set of 4 quantum numbers for the electron that is excited to the 5th n shell (problem 1). (5 points) For your 2s electron, you could have given either of these: n=2, l=0, ml= 0, ms = + ½ n=2, l=0, ml= 0, ms = - ½ For the 5th shell electron, you could have given any of these: n=5, l=1, ml= 0, ms = + ½ n=5, l=1, ml= 0, ms = - ½ n=5, l=1, ml= 1, ms = + ½ n=5, l=1, ml= 1, ms = - ½ n=5, l=1, ml= -1, ms = + ½ n=5, l=1, ml= -1, ms = - ½ 1 4.) Write the electron configuration for Co (Z=27). (5 points) Co has Z=27 (27 electrons). Using the given chart showing the electron energies, you need to fill in the orbitals with the lowest energies first. This gives: 1s22s22p63s23p64s23d7. 5.) Assuming Co has 2 valence electrons, calculate the Fermi energy at 0K for Co. (10 points) electrons g atoms 2 ⋅ 8.9 3 ⋅ 6.02 × 10 23 # valence ⋅ d ⋅ N A atom mol = 1.818 × 10 23 electrons cm n= = g M at cm 3 58.93 mol 2 ( )( ) ) 2 6.626 × 10 −34 J s ⎡ 3 1.818 × 10 29 m 3 ⎤ 3 h ⎛ 3n ⎞ 3 −18 EF = = ⎜⎟ ⎢ ⎥ = 1.87 × 10 J = 11.7eV −31 π 8m e ⎝ π ⎠ 8 9.1 × 10 kg ⎣ ⎦ Note: To get the units to work out you needed h in Joules and n in m-3 2 2 ( 6.) Does the mean speed of a conduction electron in a metal have a weak or strong dependence on temperature? Briefly explain why. (10 points) This problem was directly testing the concepts you learned in HW 4 Question 1. The mean speed of the electron relates to the average energy of the electron at that temperature. The temperature dependence of the average energy is the temperature dependence of the Fermi energy (equation 4.26). As we saw in our homework, this is a very weak dependence on temperature. That is to say, the Fermi energy, and in return, the average energy and the mean speed of the electron is only weakly dependent on temperature. 7.) Describe what Ψhyb represents in this figure. Explain in words how LCAO resulted in ΨA and ΨB having different energy. (10 points) The valence electrons in the 3s and 3p shells of Si interact to form a new orbital that is an overlap of the sand p orbitals. This new orbital has 4 lobes in the shape of a tetrahedral and is referred to as sp3 hybridization. Ψhyb is the wavefunction of an electron in this new orbital. When the Si atoms come together in a solid, Si covalently bonds with it’s neighbors. This means that the wavefunction from one atom now combines with the wavefunction from the other atom. This combination is known as LCAO (linear combination of atomic orbitals). The Ψhyb wavefunctions of two atoms can combine constructively to form a new wavefunction, ΨB. The Ψhyb wavefunctions of two atoms can combine destructively to form a new wavefunction, ΨA. ΨA is higher in energy because it has nodes (and results in anti-bonding). Extra: Not required for the answer: The figure also shows the valence band and the conduction band. When more than two atoms come together, all the ΨB from all the pairs of atoms covalently sharing electrons shift energy with respect to each other in order not to violate Pauli’s exclusion principle. This results in the broad valence band rather than a discrete state. 2 8.) Calculate the intrinsic carrier concentration in Ge at 150K. (10 points) ⎛ − EG ⎞ n i = N C N V exp⎜ ⎟ ⎝ 2kT ⎠ ⎛ 2πm * kT ⎞ e ⎟ N C = 2⎜ ⎜ h2 ⎟ ⎝ ⎠ 3/ 2 ⎛ 2πm * kT ⎞ h ⎟ N V = 2⎜ ⎜ h2 ⎟ ⎝ ⎠ ⎛ ⎞ J ⎜ 2π ⋅ 0.56 ⋅ 9.1 × 10 −31 kg ⋅ 1.38 × 10 − 23 150K ⎟ K ⎟ = 2⎜ 3 ⎜ ⎟ 2 kg ⋅ m 6.626 × 10 −34 Js ⎜ ⎟ 2 Js ⎝ ⎠ ( 3/ 2 3/ 2 ) ⎛ ⎞ J ⎜ 2π ⋅ 0.4 ⋅ 9.1 × 10 −31 kg ⋅ 1.38 × 10 − 23 150K ⎟ K ⎟ = 2⎜ 3 ⎜ ⎟ 2 kg ⋅ m 6.626 × 10 −34 Js ⎜ ⎟ Js 2 ⎝ ⎠ ( ) = 3.71 × 10 24 m −3 3/ 2 = 2.24 × 10 24 m −3 Note: You needed to use h and k in Joules to get the units to work out. As in the homework problems and example problem 5.1, you need to use the (b) value for effective mass for the density of states calculation. ⎛ ⎞ ⎜ ⎟ − 0.66eV ⎛ − EG ⎞ 24 −3 24 −3 ⎜ ⎟ n i = N C N V exp⎜ ⎟ = 3.71 × 10 m ⋅ 2.24 × 10 m exp ⎜ ⎟ − 5 eV ⎝ 2kT ⎠ 150K ⎟ ⎜ 2 ⋅ 8.62 × 10 K ⎝ ⎠ 13 −3 = 2.38 × 10 m 9.) If Ge is doped with Ga, is it n-type or p-type? (3 points) Ga has 3 valence electrons compared to Ge’s 4 so Ge doped with Ga is p-type. 10.) Using a band diagram, compare the likely relative magnitudes of the band gap energy of Ge with the ionization energy of Ga as a dopant in Ge. (7 points) Conduction Band EG (band gap) ΔEA Ionization energy Valence Band 3 ...
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This note was uploaded on 03/13/2012 for the course MATE 153 taught by Professor Staff during the Fall '08 term at San Jose State University .

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