Explanation Given information: The probability density function is p ( r ) = 4 a − 3 0 r 2 e − 2 r a 0 . Formula used: Integration by parts: ∫ u d v = u v − ∫ v d u The mean value is μ = ∫ − ∞ ∞ x . p ( x ) d x Proof: p ( r ) = 4 a − 3 0 r 2 e − 2 r a 0 (Given) The required mean value of p ( x ) on [ 0 , ∞ ) is μ = ∫ 0 ∞ r . p ( r ) d r . μ = ∫ 0 ∞ r p ( r ) d r = ∫ 0 ∞ r 4 a 0 − 3 r 2 e − 2 r a 0 d r = 4 a 0 − 3 ∫ 0 ∞ r 3 e − 2 r a 0 d r Substituting u = 2 r a 0 , then d u = 2 a 0 d r ⇒ r = a 0 u 2 . Using these in the above equation, we get μ = 4 a 0 − 3 ∫ 0 ∞ ( a 0 u 2 ) 3 e − u a 0 2 d u = 4 a 0 − 3 a 0 4 16 ∫ 0 ∞ ( u ) 3 e − u d u = a 0 4 ∫ 0 ∞ ( u ) 3 e − u d u By integration by parts, we get μ = a 0 4 ∫ 0 ∞ ( u ) 3

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 9.1, Problem 16E

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To prove: The distance r between the electron and nucleus has mean μ=3a02 .

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Chapter 9.1, Problem 15E

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To prove: The distance r between the electron and nucleus has mean μ = 3 a 0 2 .

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To prove: The distance r between the electron and nucleus has mean μ=3a02 .

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