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Unformatted text preview: 1 Physics 316 Solution for homework 13 Spring 2005 Cornell University I. EXERCISE 1 A. Compute the radial part R nl ( r ) of the Hydrogen wave function for n < 3 for all possible l . Lets find the radial part of the Hydrogen wave function for n < 3. We know that R nl ( r ) = 1 r e r na w nl parenleftBig r a parenrightBig , (1.1) where w nl ( ) = n summationdisplay k = l +1 A ( nl ) k k (0 l n 1) . (1.2) By substituting this general solution into the Schr odinger, we can derive the following relation for A k [ k ( k + 1) l ( l + 1)] A k +1 = 2 parenleftbigg k n 1 parenrightbigg A k . (1.3) It is easy, now, to find those coefficients in the cases we are interested in: ( n = 1 , l = 0) : A 1 (only nonzero coefficient to be determined by normalization condition) ( n = 2 , l = 1) : A 2 ( n = 2 , l = 0) : A 1 , A 2 = 1 2 A 1 ( n = 3 , l = 2) : A 3 ( n = 3 , l = 1) : A 2 , A 3 = 1 6 A 2 ( n = 3 , l = 0) : A 1 , A 2 = 2 3 A 1 , A 3 = 1 9 A 2 = 2 27 A 1 . Then, using (1.2), we get: w 10 ( ) = A (10) 1 w 21 ( ) = A (21) 2 2 w 20 ( ) = A (20) 1 parenleftbigg  1 2 2 parenrightbigg w 32 ( ) = A (32) 3 3 w 31 ( ) = A (31) 2 parenleftbigg 2 1 6 3 parenrightbigg w 30 ( ) = A (30) 1 parenleftbigg  2 3 2 + 2 27 3 parenrightbigg The remaining coefficients are determined by the normalization condition: integraldisplay  R nl  2 r 2 dr = a integraldisplay e 2 n w 2 nl ( ) d = a n 2 integraldisplay e x w nl parenleftBig nx 2 parenrightBig dx = 1 , (1.4) Where we used = r a and x = 2 n . Since w nl ( ) is a polynomial, this integral is computed using the Gamma function (for an integer argument): n ! = integraldisplay e x x n dx. (1.5) 2 For instance, for R 20 , we find: 1 a =  A  2 integraldisplay e x parenleftbigg x 1 2 x 2 parenrightbigg 2 dx =  A  2 bracketleftbiggintegraldisplay e x x 2 dx integraldisplay e x x 3 dx + 1 4 integraldisplay e x x 4 dx bracketrightbigg =  A  2 (2! 3! + 4! 4 ) A = 1 2 a . (1.6) Finally, we find R 10 ( r ) = 2 a 3 / 2 e r a (1.7a) R 21 ( r ) = 1 2 6 a 3 / 2 r a e r 2 a (1.7b) R 20 ( r ) = 1 2 a 3 / 2 parenleftbigg 1 1 2 r a parenrightbigg e r 2 a (1.7c) R 32 ( r ) = 2 2 81 15 a 3 / 2 parenleftBig r a parenrightBig 2 e r 3 a (1.7d) R 31 ( r ) = 4 2 27 3 a 3 / 2 bracketleftbigg r a 1 6 parenleftBig r a parenrightBig 2 bracketrightbigg e r 3 a (1.7e) R 30 ( r ) = 2 3 3 a 3 / 2 bracketleftbigg 1 2 3 r a + 2 27 parenleftBig r a parenrightBig 2 bracketrightbigg e r 3 a (1.7f) B. Show by direct integration that the Hydrogen state  n = 1 , l = 0 , m = 0 ) is orthogonal to the state  n = 2 , l = 0 , m = 0 ) ....
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This note was uploaded on 09/28/2008 for the course PHYS 316 taught by Professor Hoffstaetter during the Spring '05 term at Cornell University (Engineering School).
 Spring '05
 HOFFSTAETTER
 Physics, Work

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