1.2 c. = or 58.7% Similarly weight fraction of Se is. = or 41.3%

Transcription

1 1. c Consider the semiconducting II-VI compound cdmium selenide,. Given the tomic msses of nd, find the weight frctions of nd in the compound nd grms of nd needed to mke 100 grms of. The tomic mss of nd re 11.1 g mol -1 nd g mol -1. Since one tom of ech element is in the compound, the tomic frction, n nd n re 0.5. The weight frction of in is therefore nm g mol w or 58.7% n M + n M g mol Similrly weight frction of is nm 0.5 w 0.16 or 1.% n M + n M g mol Consider 100 g of. Then the mss of we need is Mss of w M compound g 58.7 g () nd Mss of w M compound g 1. g () 1. The covlent bond Consider the H molecule in simple wy s two touching H toms s depicted in Figure 1.7. Does this rrngement hve lower energy thn two seprted H toms? Suppose tht electrons totlly correlte their motions so tht they move to void ech other s in the snpshot in Figure 1.7. The rdius of the hydrogen tom is nm. The electrosttic potentil energy PE of two chrges Q 1 nd Q seprted by distnce r is given by Q 1 Q /( r). Using the Viril Theorem s in Exmple 1.1, consider the following:. Clculte the totl electrosttic potentil energy (PE) of ll the chrges when they re rrnged s shown in Figure 1.7. In evluting the PE of the whole collection of chrges you must consider ll pirs of chrges nd, t the sme time, void double counting of interctions between the sme pif chrges. The totl PE is the sum of the following: electron 1 intercting with the proton t distnce on the left, proton t on the right, nd electron t distnc + electron intercting with proton t nd nother proton t + two protons, seprted by, intercting with ech other. Is this configurtion energeticlly fvorble? b. Given tht in the isolted H-tom the PE is (-1.6 ev), clculte the chnge in PE in going from two isolted H-toms to the H molecule. Using the Viril theorem, find the chnge in the totl energy nd hence the covlent bond energy. How does this compre with the experimentl vlue of.51 ev?

2 Solution. Consider the PE of the whole rrngement of chrges shown in the figure. In evluting the PE of ll the chrges, we must void double counting of interctions between the sme pif chrges. The totl PE is the sum of the following: Electron 1 intercting with the proton t distnce on the left, with the proton t on the right nd with electron t distnc + Electron on the fr left intercting with proton t nd nother proton t + Two protons, seprted by, intercting with ech other PE + ( ) + Substituting nd clculting, we find PE J or -6.5 ev The negtive PE for this prticulr rrngement indictes tht this rrngement of chrges is indeed energeticlly fvorble compred with ll the chrges infinitely seprted (PE is then zero). b. The potentil energy of n isolted H-tom is ev or -7. ev. The difference between the PE of the H molecule nd two isolted H-toms is, PE - (6.5) ev - (-7.) ev9.1ev We cn write the lst expression bove s the chnge in the totl energy. E 1 1 PE ( 9.1eV).56eV This chnge in the totl energy is negtive. The H molecule hs lower energy thn two H-toms by.56 ev which is the bonding energy. This is very close to the experimentl vlue of.51 ev. (Note: We used vlue from quntum mechnics - so the clcultion ws not totlly clssicl) (b) Gold hs the FCC crystl structure, density of 19. g cm - nd n tomic mss of g mol -1. Wht is the tomic concentrtion, lttice prmeter, nd tomic rdius of gold? b. Gold hs the FCC crystl structure, hence, there re toms in the unit cell (s shown in Tble 1.). The lttice prmeter is M ρn The tomic concentrtion is 1/ ( 10 kg mol ) 1 ( kg m )( mol ) t A 1/ m nm

3 n t cm m - 0 (.077 For n FCC cell, the lttice prmeter nd the rdius of the tom R re in the following reltion (shown in Tble 1.): ( R m 0.1 nm 1.5 Dimond nd zinc blende Si hs the dimond nd GAs hs the zinc blende crystl structure. Given the lttice prmeters of Si nd GAs, 0.5 nm nd nm, respectively, nd the tomic msses of Si, G, nd As s 8.08, 69.7 g/mol, nd 7.9, respectively, clculte the density of Si nd GAs. Wht is the tomic concentrtion (toms per unit volume) in ech crystl? Solution Referring to the dimond crystl structure in Figure 1Q5-1, we cn identify the following types of toms 8 corner toms lbeled C, 6 fce center toms (lbeled FC) nd inside toms lbeled 1,,,. The effective numbef toms within the unit cell is: (8 Corners) ( 1 / 8 C-tom) + (6 Fces) ( 1 / FC-tom) + toms within the cell (1,,, ) 8 Figure 1Q5-1: The dimond crystl structure. The lttice prmeter (lengths of the sides of the unit cell) of the unit cell is. Thus the tomic concentrtion in the Si crystl (n Si ) is 8 8 n Si toms per m - 9 (0.5 If M t is the tomic mss in the Periodic Tble then the mss of the tom (m t ) in kg is m t (10 - kg/g)m t /N A (1) where N A is Avogdro s number. For Si, M t M Si 8.09 g/mol, so then the density of Si is ρ (numbef toms per unit volume) (mss per tom) n Si m t or 8 ρ (10 kg/g) M N A Si

4 i.e. -1 ( 8.09 g mol ) ( ) 8 (10 kg/g) ρ. 10 kg m - or. g cm m mol In the cse of GAs, it is pprent tht there re G nd As toms in the unit cell. The concentrtion of G (or As) toms per unit volume (n G ) is n G m - 9 (0.565 Totl tomic concentrtion (counting both G nd As toms) is twice n G. n Totl n G m - There r G-As pirs per m. We cn clculte the mss of the G nd As toms from their reltive tomic msses in the Periodic Tble using Eqution (1) with M t M G 69.7 g/mol for G nd M t M As 7.9 g/mol for As. Thus, or ρ ρ (10 kg/g)( M G + M N A As ( (10 m) ) kg/g)(69.7 g/mol g/mol) mol 9 i.e. ρ kg m - or 5. g cm -

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