Solution Manual for Principles of Electronic Materials and Devices, 4th Edition Safa Kasap.pdf

Solution Manual for Principles of Electronic Materials and Devices, 4th Edition Safa K Answers to "Why?" in the text Page 31: Oxygen has an atomic mass of 16 whereas it is 14 for nitrogen. The O2 molecule is therefore heavier than the N2 molecule. Thus, from 3( ) 2 2 1 2 1 mv  kT , the rms velocity of O2 molecules is smaller than that of N2 molecules. Page 34, footnote 11 For small extensions, the difference between the engineering and instantaneous strains due to a temperature change are the same. Historically, mechanical and civil engineers measured extension by monitoring the change in length, L; and the instantaneous length L was not measured. It is not trivial to measure both the instantaneous length and the extension simultaneously. However, since we know Lo and measure L, the instantons length L = Lo + L. Is the difference important? Consider a sample of length Lo that extends to a final length Ldue to a temperature change fromTotoT. Let  = (LLo) / Lo = L/Lo be the engineering strain. The engineering definition of strain and hence the thermal expansion coefficient is Engineering strain = T L L o    so that thermal expansion from TotoT gives,    T T L L o o o dT L dL   ( ) o o T T L L      ( )   T To  (1) where = L/Lo is the engineering strain as defined above Physics definition of strain and hence the thermal expansion coefficient is Instantaneous train = T L L    so that thermal expansion from TotoT gives,    T T L Lo o dT L dL   ln ( ) o o T T L L             ln1  ( )    T To  (2) We can expand the ln(1 + ) term for small , so that Equation (2)essentially becomes Equation (1) 1.1 VirialtheoremThe Li atom has a nucleus with a +3e positive charge, which is surrounded by a full 1s shell with two electrons, and a single valence electron in the outer 2s subshell. The atomic radius of the Li atom is about 0.17 nm. Using the Virial theorem, and assuming that the valence electron sees the nuclear +3e shielded by the two 1s electrons, that is, a net charge of +e, estimate the ionization energy of Li (the energy required to free the 2s electron).Compare this value with the experimental value of 5.39 eV. Suppose that the actual nuclear charge seen by the valence electron is not +e but a little higher, say +1.25e, due to the imperfect shielding provided by the closed 1s shell. What would be the new ionization energy? What is your conclusion? Solution First we consider the case when the outermost valence electron can see a net charge of +e. From Coulomb’s law we have the potential energy 0 0 0 0 πε r e e πε r Q Q PE 4 ( )( ) 4 1 2     4 (8.85 10 Fm )(0.17 10 m) (1.6 10 C) 12 1 9 19 2           = 1.354  1018 J or 8.46 eV Virial theorem relates the overall energy, the average kinetic energy KE , and average potential energy PE through the relations E  PE  KE and KE PE 2 1   Thus using Virial theorem, the total energy is 0.5 8.46eV 2 1 E  PE    =  4.23 eV The ionization energy is therefore 4.23 eV. Consider now the second case where the electronsees +1.25e due to imperfect shielding. Again the CoulombicPE between +e and +1.25e will be 0 0 0 0 1 2 4π r e e 4π r Q Q PE   (1.25 )( )   4 (85 10 Fm )(0.17 10 m) 1.25 (1.6 10 C) 12 1 9 19 2           π = 1.692  1018 J or10.58 eV The total energy is, 5.29eV 2 1 E  PE   The ionization energy, considering imperfect shielding, is 5.29 eV. This value is in closer agreement with the experimental value. Hence the second assumption seems to be more realistic. 1.2 Virial theorem and the He atom In Example 1.1 we calculated the radius of the H-atom using the Virial theorem. First consider the He+ atom, which as shown in Figure 1.75a, has one electron in the Ksell orbiting the nucleus. Take the PE and the KE as zero when the electrons and the nucleus are infinitely separated. The nucleus has a charge of +2e and there is one electron orbiting the nucleus at a radius r2. Using the Virial theorem show that the energy of the He+ ion is 2 2 4 2 (He ) (1/ 2) r e E  o    Energy of He+ ion [1.48] Now consider the He-atom shown in Figure 1.75b. There are two electrons. Each electron interacts with the nucleus (at a distance r1) and the other electron (at a distance 2r1). Using the Virial theorem show that the energy of the He atom is         1 2 8 7 (He) (1/ 2) r e E  o Energy of He atom [1.49] The first ionization energy EI1 is defined as the energy required to remove one electron from the He atom. The second ionization energy EI2 is the energy required to remove the second (last) electron from He+ . Both are shown in Figure 1.75 These have been measured and given as EI1 = 2372 kJ mole1 and EI2= 5250 kJ mol1 . Find the radii r1 and r2 for He and He+ . Note that the first ionization energy provides sufficient energy to take He to He+ , that is, He He+ + e  absorbs 2372 kJ mol1 . How does your r1 value compare with the often quoted He radius of 31 pm? Figure 1.75: (a) A classical view of a He+ ion. There is one electron in the K-shell orbiting the nucleus that has a charge +2e.(b) The He atom. There are two electrons in the K-shell. Due to their mutual repulsion, they orbit to void each other. Solution Virial theorem relates the overall energy, the average kinetic energy KE , and average potential energy PE through the relations E PE KE KE PE E PE KE E 2 1 ; 2 1 ; 2 1   ;      (1) Now, consider the PE of the electron in Figure 1.75a. The electron interacts with +2e of positive charge, so that 2 2 2 4 2 4 ( )(2 ) r e r e e PE  o  o     which means that the total energy (average) is 2 2 2 2 4 4 2 (1/ 2) 2 1 (He ) r e r e E PE  o  o       (2) whichis the desired result. Now consider Figure 1.75b. Assume that, at all times, the electrons avoid each other by staying in opposite parts of the orbit they share. They are "diagonally"opposite to each other. The PE of this system of 2 electrons one nucleus with +2e is PE = PE of electron 1 (left) interacting with the nucleus (+2e), at a distance r1 + PE of electron 2 (right) interacting with the nucleus (+2e), at a distance r1 + PE of electron 1 (left) interacting with electron 2 (right) separated by 2r1  4 (2 ) ( )( ) 4 ( )(2 ) 4 ( )(2 ) 1 1 1 r e e r e e r e e PE  o  o  o         2 2 8 7 r e PE  o   From the Virial theorem in Equation (1)         1 2 8 7 (He) (1/ 2) r e E  o (3) We are given, EI1 = Energy required to remove one electron from the He atom = 2372 kJ mole1 = 25.58 eV EI2 = Energy required to remove the second (last) electron from He+ = 5250 kJ mol1 = 54.41 eV The eV values were obtained by using        A eN E E 1 (eV) (J/mole) We can now calculate the radii as follows. Starting with Equation 2 for the ionization of He+ , 2 12 1 19 2 2 2 2 19 2 4 (8.854 10 Fm ) (1.602 10 C) 4 (54.41eV)(1.602 10 J/eV) (He ) r r e E E E o I I                from which, r2 = 2.651011 or 26.5 pm The calculation of r1 involves realizing that Equation (3) is the energy of the whole He atom, with 2 electrons. If we remove 1 electron we are left with He+ whose energy is Equation (2). Thus the (He) (He ) 1  EI  E  E  2 1 2 19 8 7 (24.58eV)(1.602 10 J/eV) (He) (He ) (1/ 2) I o E r e E E                1 12 1 19 2 1 2 19 4 (8.854 10 Fm ) (1.602 10 C) 8 7 (54.41eV 24.58eV)(1.602 10 J/eV) (1/ 2) r r e o                   from r1 = 3.191011 or 31.9 pm very close to the quoted value of 31 pm in various handbooks or internet period tables ____________________________________________________________________________________ 1.3 Atomic mass and molar fractions a.Consider a multicomponent alloy containing Nelements. If w1, w2, ..., wN are the weight fractions of components 1,2,..., N in the alloy and M1, M2, ..., MN, are the respective atomic masses of the elements, show that the atomic fraction of the i-th component is given by N N i i i M w M w M w w M n     ... / 2 2 1 1 Weight to atomic percentage b.Suppose that a substance (compound or an alloy) is composed of N elements, A, B, C,... and that we know their atomic (or molar) fractions nA, nBnC, .... Show that the weight fractions wA, wB, wC,....are given by    ...  A A B B C C A A A n M n M n M n M w    ...  A A B B C C B B B n M n M n M n M w Atomic to weight percentage c.Consider the semiconducting II-VI compound cadmium selenide, CdSe. Given the atomic masses of Cd and Se, find the weight fractions of Cd and Se in the compound and grams of Cd and Se needed to make 100 grams of CdSe. d.A Se-Te-P glass alloy has the composition 77 wt.% Se, 20 wt.% Te and 3 wt.% P. Given their atomic masses, what are the atomic fractions of these constituents? Solution a.Suppose that n1, n2, n3,…,ni,…, nN are the atomic fractions of the elements in the alloy, n1 + n2 + n3 +… + nN = 1 Suppose that we have 1 mole of the alloy. Then it has ni moles of an atom with atomic mass Mi (atomic fractions also represent molar fractions in the alloy). Suppose that we have 1 gram of the alloy. Since wi is the weight fraction of the i-th atom, wi is also the mass of i-th element in grams in the alloy. The number of moles in the alloy is then wi/Mi. Thus, Number of moles of elementi = wi/Mi Number of moles in the whole alloy = w1/M1 + w2/M2 +…+ wi/Mi +…+wN/MN Molar fraction or the atomic fraction of the i-th elements is therefore, Totalnumbersof molesin alloy Numebrofmoles ofelementi ni   N N i i i M w M w M w w M n     ... / 2 2 1 1 b.Suppose that we have the atomic fraction ni of an element with atomic mass Mi. The mass of the element in the alloy will be the product of the atomic mass with the atomic fraction, i.e. niMi. Mass of the alloy is therefore nAMA + nBMB+…+nNMN = Malloy By definition, the weight fractionis, wi = mass of the element i/Mass of alloy. Therefore,    ...  A A B B C C A A A n M n M n M n M w c.The atomic mass of Cd and Se are 112.41 g mol1 and 78.96 g mol1 . Since one atom of each element is in the compound CdSe, the atomic fraction, nCd and nSe are 0.5. The weight fraction of Cd in CdSe is therefore = 0.587 or 58.7% Similarly weight fraction of Se is = 0.4126 or 41.3% Consider 100 g of CdSe. Then the mass of Cd we need is Massof Cd = wCdMcompound = 0.587  100 g = 58.7 g(Cd) and Mass of Se = wSeMcompound = 0.413 100 g = 41.3 g(Se) d.The atomic fractions of the constituents can be calculated using the relations proved above. The atomic masses of the components are MSe = 78.6 g mol1 , MTe = 127.6 g mol1 ,and MP = 30.974 g mol1 . Applying the weight to atomic fraction conversion equation derived in part (a) we find,  nSe = 0.794 or 79.4%  nTe = 0.127 or 12.7 %  nP= 0.0785 or 7.9% 1.4 Mean atomic separation, surface concentration and density There are many instances where we only wish to use reasonable estimates for the mean separationbetween the host atoms in a crystal and the mean separationbetween impurities in the crystal. These can be related in a simple way to the atomic concentrationof the host atoms and atomic concentrationof the impurity atoms respectively. The final result does not depend on the sample geometry or volume. Sometimes we need to know the number of atoms per unit area ns on the surface of a solid given the number of atoms per unit volume in the bulk,nb. Consider a crystal of the material of interest which is a cube of side L as shown in Figure 1.76. To each atom, we can attribute a portion of the whole volume, which is a cube of side a. Thus, each atom is considered to occupy a volume of a 3 . Suppose that there are N atoms in the volume L 3 . Thus, L 3 = Na3 . a. If nb is the bulk concentration of atoms, show that the mean separation a between the atoms is given by 3 1/ a  nb . b. Show that the surface concentration ns of atoms is given by 2/ 3 ns  nb . c. Show that the density of the solid is given by nbM NA /   at where Mat is the atomic mass. Calculate the atomic concentration in Si from its density (2.33 g cm3 ) d. A silicon crystal has been doped with phosphorus. The P concentration in the crystal is 1016 cm3 . P atoms substitute for Si atoms and are randomly distributed in the crystal. What is the mean separation between the P atoms? Figure 1.76 Consider a crystal that has volume L 3 . This volume is proportioned to each atom, which is a cube of side a 3 .