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solution2 - 90‘0ka Km'22 Roldfw l(“J o& 4N2 hmflm...

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Unformatted text preview: 90‘0ka Km; '22; Roldfw» l (“J o&+ 4N2, hmflm wbédi‘dvx’ cm+an+ 27" N 9° 2. . ‘5‘ - szqffzyolzsl “'“U' WWW/1W ’00 Q0 2, ”£2, 9? NZ‘Sltébzeowdzz—i ’00 GM» £7; :) JlgoCoei' a 0° 22‘ 2:)4N $526” 42 a; a. ’Ooi WV ffgzoii~= 2L) “‘69 L "' :1 :7 4N0¢i4 ! ‘ ,_ I Q. I fig; / ,- 2< I jag/c” 9%: Z? (3%) chu odd® awwg) at» ”“9 0‘“, Mm, M‘gomi -‘ 200W 0" —— oo C) I) J $002 dx = 0.5 H) J 10002 dx = 0.5 0 ~—oo ...) 0.01 0.11 III _ . J «12002 dx = 7.478 x 10 4 N) J 10002 dx = 0.082 — 0.01 0.09 Since the probability density is an even function i and ii must be equal and thus it is not necessary to calculate both. Since there is a node at x=0 the probability of finding the particle near that point is essentially 0 whereas since the wavefunction is maximum at around 0.1 the probability of finding the particle is maximum there. ...
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