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1184_Physics ProblemsTechnical Physics

1184_Physics ProblemsTechnical Physics - Chapter 42 Section...

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Chapter 42 525 Section 42.5 The Wave Functions of Hydrogen P42.18 ψ π 1 0 3 1 0 s ra r a e a f = (Eq. 42.22) Pr r a e s 1 2 0 3 2 4 0 af = (Eq. 42.25) FIG. P42.18 P42.19 (a) ψπ 22 2 0 0 3 0 44 1 0 dV r dr a re d r zz z == F H G I K J Using integral tables, 2 0 2 0 0 2 0 0 2 0 2 2 2 2 2 1 0 dV a er a r a a a z =− + + F H G I K J L N M M O Q P P F H G I K J F H G I K J = so the wave function as given is normalized. (b) d r a d r aa a a a a 00 0 0 0 0 0 23 2 2 2 32 0 3 2 1 2 F H G I K J πψ Again, using integral tables, P a a r a a e a e a a a 0 0 0 2 2 2 17 4 5 4 0497 0 2 0 0 2 2 0 2 3 0 2 1 0 2 23 2 −− + + F H G I K J L N M M O Q P P F H G I K J F H G I K J L N M M O Q P P = .. P42.20 = 1 3 1 2 0 0 2 0 a r a e bg so r r a e r 24 2 2 0 5 0 . Set dP dr a r a e =+ F H G I K J L N M M O Q P P = 4 24 4 1 0 0 5 34 0 . Solving for r , this is a maximum at = 4 0 . P42.21 = 1 0 3 0 a e 2 0 5 0 0 r d dr e ra = d dr a e a 2 2 0 7 0 2 11 0 F H G I K J = = 2 0 2 0 2 0 2 12 4 m a ra e r E e ψψ But a me e 0 2 0 2 4 = = so = e a E 2 8 or E ke a e 2 0 2 . This is true, so the Schrödinger equation is satisfied.
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