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 ˇ ˇ640J 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 ˇ ? = 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? ˇ/vYb ˇ c ˙ -5P? ˇ: 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 ˇ 5P Yb ˇ?I B (1)c IIP ˇ (2)??5P ’currency1ˇ 1 c IP : = 0 + (2h=m0c) sin2 (’=2) E : m = 0 + 2h=m0c d c IIP ˇ ?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 ’ ˇ 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 (ˇ: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@: _currency1ˇ: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(ˇ h:n K8p c: z- I ) 5P currency1 I Yb ˇ B? W . 1 5P currency1 I Yb ˇ d{ (?” f 5P currency1ˇp1d ? p p2c2 +m2ec4 mec2 = mec2 p =p3mec @ 5P ? W : = h=p = h=p3mec = 0=p3 5 ? - -Pcurrency1 ?:6:12 1012eV ? -P ? W -P Yb ˇ: mnc2 = 1:675 10 27 9 1016 9:42 108eV -P currency1ˇp1d? 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 P?1- P p: (x) = 1pa cos 3 x2a ( a x a) 1 B P(x = a2 ? ?? 2 ( a5 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 ˇ/ P$ * ˇK ( P PX( ee *IP ˇc}I — N K P N ?` d ee IP * ee IP ˇ ?M ?? /currency1? I 7 IP 2 + ? P? @ ? I y? T ? ? I/ ?r @ IPw eIP h? y’ s ?M /currency1? ?? G ? 3 H a \ Pp?l Pp?l a / H T Pp?l Ih \i( - Pp ˙ N Pp Pp?l a Pp?l – P \i( 3 4 CO2 Ih ? I :10:6 m.(1)B? $* (2) w 7 sa CO2 S )? :300K. B PpN1 PpN2K < 1 $ E = hc = 6:626 10 34 3 108 10:6 10 6 1:875 10 20J 0:117eV 2 1 ?) kT 0:0253eV @ N1 N2K ’ 2 ˙/a !? I } ? ’ 3 ˙If /T \( I Ur’} 6 I \i( P @: Nh:? Ei Pp N1 >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