Q43P (a) Construct the spatial wave f... [FREE SOLUTION]

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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 4: Q43P (page 192)

(a) Construct the spatial wave function $(蠄)$ for hydrogen in the state $n = 3, I = 2, m = 1 .$ Express your answer as a function of $r, 胃, 蠒, and a$ (the Bohr radius) only鈥攏o other variables ( $p, z,$ etc.) or functions ( $p, v,$ etc.), or constants ( $A, c_{0},$ etc.), or derivatives, allowed (蟺 is okay, and e, and 2, etc.).
(b) Check that this wave function is properly normalized, by carrying out the appropriate integrals over, $胃, and 蠒 .$
(c) Find the expectation value of $r^{s}$ in this state. For what range of s (positive and negative) is the result finite?

Short Answer

Expert verified

$(a) 蠄_{321} = R_{32} Y_{2}^{1} = - \frac{1}{\sqrt{蟺}} \frac{1}{81 a^{7 / 2}} r^{2} e^{- r / 3 a} {蝉颈苍胃肠辞蝉胃别}^{颈蠒} .$

(b)The wave function is properly normalized by 1

$(C) <r^{s}> = - 7 .$

Step by step solution

(a) Constructing the spatial wave function for hydrogen.

$蠄_{321} = R_{32} Y_{2}^{1} = \frac{4}{81 \sqrt{30}} \frac{1}{a^{3 / 2}} {(\frac{r}{a})}^{2} e^{- r / 3 a} [- \sqrt{\frac{15}{8 蟺}} {蝉颈苍胃肠辞蝉胃别}^{颈蠒}] = - \frac{1}{\sqrt{蟺}} \frac{1}{81 a^{7 / 2}} r^{2} e^{- r / 3 a} {蝉颈苍胃肠辞蝉胃别}^{颈蠒} .$

Step2: (b) Checking the wave function is properly normalized.

$鈭� {|蠄|}^{2} d^{3} r = \frac{1}{蟺} \frac{1}{蟺 {(81)}^{2} a^{7}} 鈭� (r^{4} e^{- 2 r / 3 a} \sin^{2} 胃 \cos^{2} 胃) r^{2} \sin 胃 d r d 胃 d 蠒 . = \frac{1}{蟺 {(81)}^{2} a^{7}} 2 蟺 {鈭�}_{0}^{鈭�} r^{6} e^{- 2 r / 3 a} d r {鈭�}_{0}^{蟺} (1 - \cos^{2} 胃) \cos^{2} 胃 \cos^{2} 胃 d 胃 . = \frac{2}{{(81)}^{2} a^{7}} [6! {(\frac{3 a}{2})}^{7}] {[- \frac{\cos^{3} 胃}{3} + \frac{\cos^{5} 胃}{5}]}_{0}^{蟺} . = \frac{2}{3^{8} a^{7}} 6.5.4.3.2 \frac{3^{7} a^{7}}{2^{7}} [\frac{2}{3} - \frac{2}{5}] = \frac{3.5}{4} . \frac{4}{15} = 1 .$

Step3: (c)Finding the exceptional value of rs

$<r^{s}> = {鈭�}_{0}^{鈭�} r^{s} {|R_{32}|}^{2} r^{2} dr = {(\frac{4}{81})}^{2} \frac{1}{30} \frac{1}{a^{7}} {鈭�}_{0}^{鈭�} r^{s + 6} e^{- 2 r / 3 a} dr . = \frac{8}{15 {(81)}^{2} a^{7}} (s + 6)! {(\frac{3 a}{2})}^{s + 7} = (s + 6)! {(\frac{3 a}{2})}^{5} \frac{1}{720} = \frac{(s + 6)!}{6!} {(\frac{3 a}{2})}^{3} . Finite for S > - 7 .$

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Short Answer

Step by step solution

(a) Constructing the spatial wave function for hydrogen.

Step2: (b) Checking the wave function is properly normalized.

Step3: (c)Finding the exceptional value of rs

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