大連理工大物作業(yè)答案.pdf
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1 I. 40TH 1 S I T 86g ( S 5 iS 2 h:n UrIg I5H Zh:( UrI F/ I: LI5H 40-1 -*cn : H U 1 1 O65 dcn :A /C (I: N U M e 0 IPp q 5A:_1 ETH:A. 3 sa zT b :4cm2 T 640J BzT ) ( = 5:67 10 8W=m _K;b = 2:898 10 3m _K) 1fly y ; : M = T4 E P = A T4 d T = 4 r P A = 4 s 640J=60s 4 10 4m2 5:67 10 8W=m K 828K 4 -8 *5P 4:2eV = 200nm + g hb B (1) P (2) O65 (3)I5P currency1 e = 1:6 10 19C;h = 6:626 10 34J s 1 P 0 = c 0 = hcE = 6:626 10 34 3 108 4:2 1:6 10 19 m 2:96 10 7m = 296nm 2 O65 Uc = h e U0 = hc U0 = 6:622 10 34 3 108 200 10 9 1:6 10 19 4:2 2V 3 I5P currency1 1 2mv 2 m = eUc = 1:6 10 19C 2V = 3:2 10 19J = 2eV 5 ( S - v Ur , 10:6 m 00:4 m v; , 1 iM mT = b T2 T1 = m1 m2 1fly y M = T4 M2 M1 = ( T2 T1 ) 4 = ( m1 m2 ) 4 = (0:6 m 0:4 m) 4 = 5:0625 6 H/n H n H c I ; y T n H SX (5P c (c I- d e I e I e I y 1 c I c 2 n : c i( P p 2 7 (n c - 5P :0:6c (c -5P /vYb c -5P : Ek = E E0 = m0c2= p 1 v2=c2 m0c2 Ek E0 = 1p 1 v2=c2 1 = 1p 1 (0:6c)2=c2 1 = 0:25 8 ( c - g IIP 5P Yb I B (1)c IIP (2)5P currency1 1 c IP : = 0 + (2h=m0c) sin2 (=2) E : m = 0 + 2h=m0c d c IIP hc= 0 = m0c2 0:511MeV Emin = hc m = hc 0 + 2hm0c = hc 0 =(1 + 2(hc= )m 0c ) = m0c 2 3 0:17MeV 2 5P Emax = E0 + hc 0 hc m = m0c2 +m0c2 m0c 2 3 = 5 3m0c 2 currency1 pm = 1c q E2max m20c4 = 43m0c 3:6 10 22kg m=s 3 II. 41TH 1 P (:m currency1 :Ek Q”H Bv W 1 p = h c = h = hp = hp2mE k 2 2 5PT-5P W :0:1nm B 5 1 5P /* d (” f Ek = p 2 2m = h2 2m 2 = eU U = h 2 2m 2e = (6:626 10 34)2 2 9:11 10 31 10 20 1:6 10 19 151V 3 41-1: _currency1:p 5P :a ( :R n gIO BO M 7- . a d. 1 W s = h=p U M a a : asin k = k 1 R d, sin 1 d=2R. / d = 2Rsin 1 = 2R =a = 2Rh=pa 4 0 = hmec :5P n (me:5P Yb( h:n K8p c: z- I ) 5P currency1 I Yb B W . 1 5P currency1 I Yb d (” f 5P currency1p1d p p2c2 +m2ec4 mec2 = mec2 p =p3mec 5P W : = h=p = h=p3mec = 0=p3 5 - -Pcurrency1 :6:12 1012eV -P W -P Yb : mnc2 = 1:675 10 27 9 1016 9:42 108eV -P currency1p1d Ek = p p2c2 +m2nc4 mnc2 4 d p = (1=c) p Ek(Ek + 2mnc2) 1 W s -P W : = hp = hcpE k(Ek + 2mnc2) 6:626 10 34 3 108 p6:12 1012 (6:12 1012 + 2 9:42 108) 1:6 10 19 2:03 10 19m 1 Ek m0c2 d p. 5 III. 42TH 1 :2a P1- P p: (x) = 1pa cos 3 x2a ( a x a) 1 B P(x = a2 2 ( a5 x a5 P 1 p b p R 1 P(x = a2 p(x = a=2) =j (x = a=2)j2 = 1a cos(3 4 ) 2 = 12a 2 ( a5 x a5 P p( a5 x a5) = Z a=5 a=5 j (x)j2dx = 1a Z a=5 a=5 cos2 3 x2a dx = 1a Z a=5 0 1 + cos 3 xa dx = 1a ha 5 + a 3 sin(3 =5) i = 15+ 13 sin(3 5 ) 0:3 2 P( P1-currency1 42-1 : P p (x) (1) P p (2)(pf B P Mn 1 P p (x) = (q 2 a sin( 3 x a ); (0 xa) 0; (xa) 2 P Mn1dj (x)j2dx = 0 s sin(3 xa ) cos(3 xa ) = 0 sin(3 xa ) = 0 / 0 cos(3 xa ) = 0 / 1cos(3 xa ) = 0 3 xa = 2; 3 2 ; 5 2 ,sx = a6;a2; 5a6 . 3 currency1 P p: (x) = Ae ax (x 0); 0 (x 0): v-a: 0 8p n R p A p R a 1 = Z 1 1 j (x)j2dx =jAj2 Z 1 0 e 2axdx = 12ajAj2 K A =p2a 4 (:a P1-currency1 P p: (x) = ( 0 (xa) q 2 a sin n x a (0 x a) B 1 P (a3 x 2a3 2 ;Ppn = 2 P Mn p R 1 P (a3 x 2a3 p = Z 2a 3 a 3 j 1(x)j2dx = 13 + p3 2 6 2 n = 2 P Mn 2 x a = 2; 3 2 x = a4; 3a4 5 P ;Ppn = 4 (1) vhS currency1 (2) l = 3 v currency1( : q 1 ;Ppn = 4 P l = 0;1;2;3;L = p l(l + 1) h = 0;p2 h;p6 h;2p3 h 2 l = 3 m = 3; 2; 1;0;1;2;3:Lz = m h = 3 h; 2 h; h;0; h;2 h;3 h: 6 5P 3d (1) hS currency1 (2) ;Pp/ 1 hS currency1 L = p l(l + 1) h = p 2(2 + 1) h =p6 h 2 ;Ppn = 3. 7 fi P Ppl = 2 1 B currency1L 2 Ppm Lz 3 ( 42-2-;vY L v mLz 1 currency1 L = p l(l + 1) h = p 2(2 + 1) h =p6 h 2 Pp m = 2; 1;0;1;2;Lz = m h = 2 h; h;0; h;2 h: 3 8 P-5P p v4*Pp s p c * pA c *p 1 3; 1; 1;1 2 2 1;1;0;1 2 3 3;1;1;0 4 1;0;1 2; 1 2 1 1 l 0, : 3;1; 1;1 2 2 1 ln n = 1/ 1m 20n = 1 m = 20n = 1 max = c 21 = 1R(1=12 1=22) = 43R = 43 1:096776 107m 1 = 1:2157 10 7m = 121:57nm m =10n = 1 11 = c 11 = 1R(1=12 12) = 1R = 0:9118 10 7m = 91:18nm 2 ,3 P N I1 ; ,3 P N I1: n = 4!n = 3; 43 = c 43 = 1R(1=32 1=42) = 1447R 1875:63nm n = 4!n = 2; 42 = c 42 = 1R(1=22 1=42) = 163R 486:27nm n = 4!n = 1; 41 = c 41 = 1R(1=12 1=42) = 1615R 97:25nm n = 3!n = 2; 32 = c 32 = 1R(1=22 1=32) = 365R 656:47nm n = 3!n = 1; 31 = c 31 = 1R(1=12 1=32) = 98R 102:57nm n = 2!n = 1; 21 = c 21 = 1R(1=12 1=22) = 43R 121:57nm 1 I :390nm n P l = 0;1;2;3 l 1ql* 5P n v 9 P-n = 2 -5P h p n;l;ml;ms / v n = 2 l = 0;1 Sl = 1 ml = 0; 1 Sl = 0 ml = 0 ms = 12 n = 2 -5P h p: 2;1;1;1 2 ; 2;1;1; 1 2 ; 2;1;0;1 2 ; 2;1;0; 1 2 2;1; 1;1 2 ; 2;1; 1; 1 2 ; 2;0;0;1 2 ; 2;0;0; 1 2 E /8 v 9 V. 44TH 1 H/ H/ T L: P 0 N *IP IP / P$ * K ( P PX( ee *IP cI N K P N d ee IP * ee IP M /currency1 I 7 IP 2 + P I y T I/ r IPw eIP h y s M /currency1 G 3 H a Ppl Ppl a / H T Ppl Ih i( - Pp N Pp Ppl a Ppl P i( 3 4 CO2 Ih I :10:6 m.(1)B $* (2) w 7 sa CO2 S ) :300K. B PpN1 PpN2K N2 N4 N3 T 1 $* K Ppl 2 I 3 gI I1 * 1 3 4 K Ppl 2 I = E4 E3h 3 gI I1 6* 4!3;4!2;4!1;3!2;3!1;2!1: 7 H/, J S , J S 5:6/ H T , J S fl J SUvP UB( :w P u“% h , J S 5:6 , 0z AP 8 + S S J S : T S KClI PvS Cl2;CO2I J S p J S n J S p J S ; 5P 0zE2K 86 IP +: = jE2 E1 jh ; = c = hcjE 2 E1 j EE, J S 86 I max;o = hcE g = 6:626 10 34 3 108 1:2 1:6 10 19 m 10:35 10 7m 1035nm n J S 86 I max;n = hc E = 6:626 10 34 3 108 0:045 1:6 10 19 m 276:08 10 7m = 27608nm- 1.請(qǐng)仔細(xì)閱讀文檔,確保文檔完整性,對(duì)于不預(yù)覽、不比對(duì)內(nèi)容而直接下載帶來(lái)的問(wèn)題本站不予受理。
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