Explanation Write the expression for wave function. ψ 200 = 1 32 π r 0 3 ( 2 − r r 0 ) e − r 2 r 0 Here, ψ 200 is the wave function, r is the radius of the sphere and r 0 is Bohr radius. Square both sides of the equation. | ψ 200 | 2 = 1 32 π r 0 3 ( 2 − r r 0 ) 2 e − r r 0 Write the expression for volume. d V = 4 π r 2 ( d r ) Here, d V is the volume, 4 π r 2 is the cross sectional area and d r is change in thickness. For ψ 200 to be normalized, integrate | ψ 200 | 2 d V over all the space. ∫ all space | ψ 200 | 2 d V = ∫ 0 ∞ 1 32 π r 0 3 ( 2 − r r 0 ) 2 e − r r 0 d V Here, ψ 200 is a wave function, r is the radius and r 0 is the Bohr radius. Substitute 4 π r 2 ( d r ) for d V in the above expression. ∫ all space | ψ 200 | 2 d V = ∫ 0 ∞ 1 32 π r 0 3 ( 2 − r r 0 ) 2 e − r r 0 ( 4 π r 2 d r ) (I) Let r r 0 be equal to x . r r 0 = x r = x r 0 The differential of r is, d r = r 0 d x Substitute x for r r 0 , x r 0 for r and r 0 d x for d r in equation (I). ∫ all space | ψ 200 | 2 d V = ∫ 0 ∞ 4 π 32 π r 0 3 ( 2 − x ) 2 e − x r 0 2 x 2 r 0 d x = 1 8 ∫ 0 ∞ ( 2 − x ) 2 e − x x 2 d x = 1 8 ∫ 0 ∞ ( 4 x 2 − 4 x 3 + x 4 ) e − x d x = 1 2 ∫ 0 ∞ x 2 e − x d x − 1 2 ∫ 0 ∞ x 3 e − x d x + 1 8 ∫ 0 ∞ x 4 e − x d x (II) Integrate the terms separately. Integration of first term is shown below. ∫ x 2 e − x d x = − x 2 e − x + ∫ 2 x e − x d x = − x 2 e − x + 2 ( − e − x ) ( x + 1 ) = − e − x ( x 2 + 2 x + 2 ) Integration of second term is shown below

Modified Mastering Physics with Pearson eText -- Standalone Access Card -- for Physics for Scientists & Engineers with Modern Physics

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ISBN: 9780133882261

Author: Douglas C. Giancoli

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Chapter 39, Problem 20P

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The proof that ψ200 as given by equation 39-8 is normalized.

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Chapter 39, Problem 19P

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