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ps13soln - 1 Physics 316 Solution for homework 13 Spring...

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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 . Let’s 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 ( ξ ) = n summationdisplay k = l +1 A ( nl ) k ξ 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 non-zero 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 ( ξ ) = A (10) 1 ξ w 21 ( ξ ) = A (21) 2 ξ 2 w 20 ( ξ ) = A (20) 1 parenleftbigg ξ- 1 2 ξ 2 parenrightbigg w 32 ( ξ ) = A (32) 3 ξ 3 w 31 ( ξ ) = A (31) 2 parenleftbigg ξ 2- 1 6 ξ 3 parenrightbigg w 30 ( ξ ) = A (30) 1 parenleftbigg ξ- 2 3 ξ 2 + 2 27 ξ 3 parenrightbigg The remaining coefficients are determined by the normalization condition: integraldisplay ∞ | R nl | 2 r 2 dr = a integraldisplay ∞ e- 2 n ξ w 2 nl ( ξ ) dξ = a n 2 integraldisplay ∞ e- x w nl parenleftBig nx 2 parenrightBig dx = 1 , (1.4) Where we used ξ = r a and x = 2 ξ n . Since w nl ( ξ ) 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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ps13soln - 1 Physics 316 Solution for homework 13 Spring...

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