# hw6 - 60 C HAPTER . M ATTER AvEs 6 w 6.2 r [ new] ) , : h...

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60 CHAPTER 6. MATTER wAvEs 6.2 r [new] ), : h/mu : (6.63 x 10-34 J.s)/l(60 x 10-3 ks) x (30 m/s)l : Impossible to detect, 3.68 x 10-34 m 6.! o [old 7.11 Since K: p2/2m,p = /2mR and ): n/p: n/JzrnX : hc/Jffik : (tzao ev.nm)/@ : to^obs ;f 6.4 r fold 7.2] Since p= hlA, the KE is K = p2l2rn: h2 /(2m)2) = (hc)2 /(2rne2\2): (1240 eV' nm)2 /[2 x (0.511 x 106 eV) x (450 nm)z] = lZ.a x rO-6 eV. I 6,5. As in Problem 6.4, K = (hc)z /(2mc2)2), and with mcz : 940 MeV, this gives 6.6 o lold 7.41 As in Problem 6.3, ) : hc/r/z*P X. For an electron with mc2 : 0.5 MeV and K = 3 eY, this gives f" = lO.n ,,rn;l for a neutron with rnc2 = 940 MeV but the same KE, it gives ), = | o.ot?l-*l Th","tiol. ffi = 6J"" :@ 6.7 o [old 7.5] With K: 15 keV, the electron is still reasonably nonrelativistic, and as in Problem 6.3, .\" = hc/,[email protected] = l'b^om;] Since ) o< 1.lJm,^t,: A./J207 = F.o * ro-n n*ll cl""rly A.f ^p : \/207 : 114.4. I 6.8 o [old 7.6] For a nonrelativistic massive particle K = (hc)2 /(2m.'?]'?).,- =l-in Problem 6.4. For a neutron (rn c2 = 940 MeV) with ,\ = 0.05 nm, this gives K' : IO.-53 "Vt I for an electron (rnc2 = 0.511 MeV), K" : 600 eV. (These answers show that both particles are nonrelativistic.) The photon (with rn: 0) is always relativistic, and we must use 1( : E : hcl\:124,800 eV, I 6.e oo [old 7.7) pc: Joz-:6Ay : ^/F=Og MeV : 1.e4 MeV. Thus ),: hclpc: (1240 eV' nm)l(.ea x 106 eV) : l640 d.l 6.10 oo [old 7.S] With K =2MeY, E: K *mc2:2.511 MeV. Thus pc = Mev. Therefore, ),: [email protected]): (1240 MeV.fm)/(2.46 MeV) : F04 ft*l 6.lf oo fold 7.9] In all cases we use .l

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## This note was uploaded on 02/10/2010 for the course PHSY 70 taught by Professor -1 during the Fall '10 term at Stanford.

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hw6 - 60 C HAPTER . M ATTER AvEs 6 w 6.2 r [ new] ) , : h...

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