Mechanical Properties of Si/Ge Nanowire

Transcription

1 Mchancal Pop of /G Nanow 1/10/004 Eunok L, Joohyun L, and Konwook Kang Fnal Pojc fo ME346 Inoducon o Molcula mulaon Inuco: W Ca, anfod Unvy

2 1. Inoducon Rcnly houcu of and G nanow ha bn fabcad nc blvd ha hy how faly good m-conducng popy, whch can b ud a bac buldng block of molcula lconc dvc o bo no. In h aly ag of dvlopng 1D ho-nanow, mchancal popy of h nanow hould alo b udd bcau, a nano cal, w canno aum ha h boh mchancal and lccal bhavo a ndpndn of ach oh. In h m pojc, auho hav pfomd val mulaon fo h non of, G, and -G co-hll nanow o oban h mchancal pop uch a Young' modulu and yld a 0K a a udmnay p fo fuu wok. W alo obv how h dfomaon nad and dvlopd n h nano ucu ung molcula mulaon.. ym Dcpon y = [110] x = [001] z = [110] Fg 1 Nanow, G and -G nanow of dffn dam (5nm, 10nm, 15nm and 5nm) a nally confgud lk fg.1 o ln up along [110] dcon, whch a ypcal gowh dcon. W ch h mulaon box along z dcon, and compu h avag a a funcon of an. nc h mpau a whch mulaon wa don 0K, h ponal ngy ha dmn ym a and conjuga gadn mhod wa ud o follow h local mnmum of ngy and hfo o fnd a laxd a of ym. Podc bounday condon adopd along z dcon o ha nanow condd o hav nfn lngh along ha dcon. Along x and y dcon, up cll condon ud. Two dffn many-body ponal, llng-wb v v and Toff v, a ud o dcb boh and G aom ( appndx). Th mulaon a pfomd ung h MD++ packag v. Mo dald on mulaon gomy a ld n abl 1. W alo cay ou -G co-hll nanow of dam 15nm o wha would happn a h nfac of and G und non (fg ). Dffn colo n

3 fg.1 and fg. how dffn ngy lvl of ach aom. Th d G aom on h ufac n fg.1 hav hgh ngy fom ha of bulk aom vn f hy boh a. 5nm 15nm 5nm 15nm Fg -G co-hll nanow Maal G ucu Damond-cubc Cyal Lac Conan, a (Å) Dam, 5 NP= d (nm) 10 NP= NP= NP= Tmpau, T (K) 0 0 Thckn, (nm) Ponal W, Toff W, Toff Compu Rouc P-IV 3GHz X Tabl 1. ym Dcpon 3. Damond Cubc Cyal ucu Bacally, and G a damond- cubc cyal a hown n fg.3 v and how dffn aomc aangmn whn vwd along dffn dcon. Th xplan why many pop of and G a anoopc. Dffn hnkag ao a (110) plan can alo b xpland by h anoopy. Fo xampl, h and G nanow hnk anoopcally n dam whl und unaxal non, a hown n Fg. 9.

4 [001] dcon [110] dcon [111] dcon Fg 3 Damond Cubc ucu (gnad by Aomy) 4. Daa analy and Dcuon 1) Young Modulu ) lcon by llng-wb Ponal,.5E+04.0E E E an 1.4E+04 1.E E E E+03.0E an 5nm 10nm 15nm 0nm Fg. 4. Th -an cuv fo nanow wh dffn dam a plod. Th cuv n h d dod qua a nlagd n h gh fgu. Young modul a GPa fo 5nm, 1.330GPa fo 10nm, GPa fo 15nm and GPa fo 0nm. W oband h valu by lcng only h gon

5 n whch h chang lnaly o h an. In oh wod, w had chon h gon ha h an l han 0.1 (.. 10%). Young modulu ha a mall dffnc fo dffn dam of h nanow. Bu h flucuaon can b wll accound fo by h uncany n dmnng h dam of h nanow (du o h dcn of cyal ucu). Th dam do no hav a bg ffc on h Young modulu n agmn wh Lang al. x W fnd ha h avag Young modulu ov h dffn dam 1.531GPa. ) lcon by Toff Ponal 3.0E+04.5E+04.0E E E an 1.4E+04 1.E E E E+03.0E an 5nm 10nm 15nm 0nm Fg. 5. Th -an cuv fo nanow wh dffn dam a plod ung Toff ponal. Young modul a GPa fo 5nm, GPa fo 10nm, GPa fo 15nm, and GPa fo 0nm. Th avag valu GPa ) Gmanum by llng-wb Ponal.50E+04.00E E E E+03 G 0.00E an 1.00E E E E+03.00E+03 G 0.00E an 5nm 10nm 15nm 0nm Fg. 6. Th -an cuv fo G nanow wh dffn dam a plod ung W ponal.

6 Young modul a GPa fo 5nm, GPa fo 10nm, GPa fo 15nm, and GPa fo 0nm. Th avag valu GPa. v) Gmanum by Toff Ponal.5E+04.0E E E+03 G an 1.4E+04 1.E E E E+03.0E+03 G an 5nm 10nm 15nm 0nm Fg. 7. Th -an cuv fo G nanow wh dffn dam a plod ung Toff ponal. Young modul a GPa fo 5nm, GPa fo 10nm, GPa fo 15nm, and fo 0nm. Th avag valu 14.79GPa. v) -G Coaxal by Toff Ponal.5E+04.0E E E+03 Co-hll an 1.4E+04 1.E E E E+03.0E+03 Co-hll an G co co Fg. 8. Th -an cuv fo -G co-hll nanow wh dffn dam a plod. Young modul a GPa fo h -hll/g-co and GPa fo Ghll/-co. In ummay, h Young modulu abou 13GPa by W ponal and 15GPa by Toff ponal fo lcon and 109GPa by W ponal and 15GPa by Toff ponal fo Gmanum. In Gmanum ca, h valu a ll dffn bwn h

7 on by W ponal and on by Toff ponal. Bu Young modulu oband by ju lcng h gon n whch h chang lnaly by h an and fndng h lop mply. Condng how oughly h Young modulu wa oband, uch dffnc n Gmanum accpabl. Th Young modulu m qu aonabl compad wh h xpmn ul x condng ha h xpmn wa don und 300K and h mulaon wa don und 0K. To h qualav compaon, w dvd a mpl fomula of mxng law o g h Young modulu of h coaxal nanow. Th dald dvaon followd n Appndx C E = E ( d / d ) + E (1 ( d / d ) ) W apply h ul fom h mulaon by Toff ponal, bcau h coaxal -G nanow wa compud by Toff ponal. Young Modulu GPa fo and 14.79GPa fo G. And d 1 =5nm and d =15nm. Thn, n ca of -co/g-hll, E = (1/ 3) (1 (1/3) ) = Th valu ag o h mulaon ul fo coaxal -G, 16GPa. In h mulaon, w ud conjuga gadn mhod whch ak no accoun hmal flucuaon. Hnc, w mulad h maal a 0K. Howv, h xpmn a alway don a fn mpau uch a 300K. o w nd o mula h maal ung MD and h would b fuu wok. Th fuu p fo h wok would b o udy h dfomaon ngh of h nanow ung molcula dynamc mulaon. ) Dfomaon Analy 13nm 15.nm 15.nm 14nm (a) (b) Fg. 9. (a) (d=15nm) a 0% an hghlghd wh d abov -3.8V (b) G (d=15nm) a 0 % an hghlghd wh d abov.8v.

8 Whl h nanow xpand along z dcon, hnk along x and y dcon du o fn Poon ao. Bcau of lac anoopy, h hnkag ao along x and y dcon a dffn n fg. 10. I m ha h dmnon along y dcon man conan o nca dung non and mo of conacon occud n x dcon. And hnk mo n x dcon han G a h am an. And h dffnc look lk a ong candda fo h gnaon of vod a h nfac bwn and G a hgh an whn a h co and G a h hll. Fg. 10. (a) hll G co nanow bokn a 30% an hghlghd wh d abov - V (b) G hll co nanow a 35% an, hghlghd abov -.96V In ca of ngl componn nanow, all h aom a wh hgh ngy whn hgh an,.g. 30% an, nally loadd. Dung h CGR mhod, h hgh ngy a lad hough x dcon and fnally h facu nad fom aom a boh nd of majo ax whch ll hav hgh ngy and popagad no h nn gon. Th ndncy g ong a h dam dca and ong n han n G.

9 5. Concluon Th Young modulu w oband fom h of molcula mulaon do no how appcabl dffnc accodng o h ponal w ud and h z of /G nanow. And ho valu a whn h ang of Young modulu fo h bulk /G maal a 300K. And n h coaxal -G ca, h ul fom h mulaon vy mla o h xpcd valu ung a mpl mxng valu. To h mpau dpndnc of Young modulu, ou fuu wok would b o pfom MD mulaon wh a fn mpau. Du o h dffn valu of h Poon ao along h dffn dcon, h vod occu a h nfac bwn coaxal -G. Th facu, nad a h nfac, can anm o nwad o ouwad n ca of -hll/g-co o G-hll/-co pcvly. Mo goou analy fo h mulaon ul wll b pfomd n h fuu.

10 Appndx A. Toff Ponal Toff Ponal a many-body ponal, alhough look lk a modfd body ponal.(rf. v) 1 E = E = V j wh V = f ( )[ f ( ) +b f ( )] C R A : bondng ngy and : h danc bwn h aom and j f R ( ) = Axp ( λ) : pulv pa ponal = Bxp µ : aacv pa ponal f A f C ( ) ( ) 1, 0, < R 1 1 ( ) = + co[ π( R )/( R )], b = ζ = n n 1/ n ( 1+ β ζ ) k,j f C χ > ( ) ω g( θ ) k k : dpndnc k R < < : mooh cuoff funcon : ffcv coodnaon numb of aom, akn no accoun h lav danc of wo nghbo k and h bond-angl θ. c c g( θk ) = 1+ d d + h coθ ( θ) ( ) k g ha a mnmum whn h = co θ, d dmn how hap h dpndnc on angl, and c how h ngh of h angula ffc. C G A (V) B (V) λ (Å -1 ) µ (Å -1 ) β 1.574E E E-007

11 C G n c E E E+005 d h R (Å) (Å) χ C = χ G =

12 B. llng-wb (W) Ponal (Rf. v) Φ 1 ( 1,,L,N ) = V( ) + V3 (,k,jk ), j, j,k V c 1 wh -body m ( ) = f ( ) [ A φ( ) A φ( ) ] f and c ( ) = xp µ < R Rc 0 R φ c c λ ( ) = 3-body m (,, ) = Zψ( ) ψ( ) g( θ ) V 3, k jk k : cu-off funcon = 1, ψ( ) = [ f ( )] α c and g( θ ) = ( coθ coθ ) 0 n md++ G G n md++ A bb=a1/aa= bb=a1/aa= A aa= aa= λ 1 4 ho=4 4 ho=4 λ 0 0 R C = aσ (Å) acu= acu=3.958 Z = ελ plam= plam=59.83 α 1. pgam= pgam=.617 µ (Å).0951 p= p=.181 θ 0 (dg)

13 C. Effcv Young modulu of coaxal maal und non d d 1 Cond wo maal 1 and n a co-hll ucu wh Young modul of E 1 and E, a chon abov. Th oal nl foc f and h wo maal conbu f 1 and f pcvly. f = σ A = Eε A, f1 = σ1a1 = E1ε 1A1, f = σ A = Eε A To oban Young modulu, w focu on h gon n whch h chang lnaly whn h an. o w can aum h an o mall ha h aa chang nglgbl. Thfo, h aa fo ach lmn a followd. = π d, A π d d1 A 1 1 / 4 = ( ) / 4, A = π d / 4 In a non, boh maal a dfomd by h am an,.. Thn, ε = ε1 = ε 1 σ1 1 σ 1ε1π 1 ε π 1 / / 4 ( 1 ) / 4 1( 1 / ) (1 ( 1 / ) ) f = f + f = A + A = E d / 4 + E ( d d ) / 4 Eεπ d = E επ d + E επ d d E = E d d + E d d

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