Q11P (a) Calculate<1/r1-r2>for ... [FREE SOLUTION]

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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 5: Q11P (page 214)

(a) Calculate $< (1 / |r_{1} - r_{2}|) >$ for the state $蠄_{0}$ (Equation 5.30). Hint: Do $d^{3} r_{2}$ integral
first, using spherical coordinates, and setting the polar axis along $r_{1}$ , so
that
$蠄_{0} (r_{1}, r_{2}) = 蠄 100 (r_{1}) 蠄 100 (r_{2}) = \frac{8}{{蟺补}^{3}} e^{- 2 (r_{1} + r_{2}) / a}$ (5.30).
$|r_{1} - r_{2}| = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} {肠辞蝉胃}_{2}} .$
The $胃_{2}$ integral is easy, but be careful to take the positive root. You鈥檒l have to
break the $r_{2}$ integral into two pieces, one ranging from 0 to $r_{1}$ ,the other from $r_{1} t o 鈭�$
Answer: 5/4a.
(b) Use your result in (a) to estimate the electron interaction energy in the ground state of helium. Express your answer in electron volts, and add it to $E_{0}$ (Equation 5.31) to get a corrected estimate of the ground state energy. Compare the experimental value. (Of course, we鈥檙e still working with an approximate wave function, so don鈥檛 expect perfect agreement.)
$E_{0} = 8 (- 13.6 e V) = - 109 e V$ (5.31).

Short Answer

Expert verified

$(a) = \frac{32}{a^{4}} \{a . {(\frac{a}{4})}^{2} - 2.2 {(\frac{a}{8})}^{3} - a . {(\frac{a}{8})}^{2}\} = \frac{32}{a} (\frac{1}{16} - \frac{1}{128} - \frac{1}{64}) = \frac{5}{4 a} . (b) V_{e e} 鈮� \frac{e^{2}}{4 蟺鈭�'} < \frac{1}{|r_{1} - r_{2}|} > = \frac{5}{4} \frac{e^{2}}{4 蟺鈭�'} \frac{1}{a} = \frac{5 m}{4 h^{2}} {(\frac{e^{2}}{4 蟺鈭�'})}^{2} = \frac{5}{2} (- E_{1}) = \frac{5}{2} (13.6 e V)$

Step by step solution

(a) Calculating <1/r1-r2>for the state ψ0

$< \frac{1}{|r_{1} - r_{2}|} > = {(\frac{8}{{蟺补}^{3}})}^{2} 鈭� \underset{铀�}{\underset{铃�}{[鈭� \frac{e^{- 4 (r_{1} + r_{2}) / a}}{\sqrt{r_{2}^{1} + r_{2}^{2} - 2 r_{1} r_{2} \cos 胃_{2}}} d^{3} r_{2}] d^{3}}} r_{1} . 铀� = 2 蟺 {鈭�}_{0}^{鈭�} e^{- 4 (r_{1} + r_{2}) / a} \underset{\overset{鈭�}{a}}{\underset{铃�}{[{鈭�}_{0}^{蟺} \frac{{蝉颈苍胃}_{2}}{\sqrt{r_{2}^{1} + r_{2}^{2} - 2 r_{1} r_{2} {肠辞蝉胃}_{2}}} {诲胃}_{2}] r_{2}^{2} d}} r_{2} . x$

$\overset{鈭�}{a = \frac{1}{r_{1} r_{2}}} \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} {肠辞蝉胃}_{2}} |_{0}^{蟺} = \frac{1}{r_{1} r_{2}} [\sqrt{r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2}} - \sqrt{r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2}}] . = \frac{1}{r_{1} r_{2}} [(r_{1} + r_{2}) - |r_{1} - r_{2}|] = \{\begin{cases} 2 / r_{1} (r_{2} < r_{1}) \\ 2 / r_{2} (r_{2} > r_{1}) \end{cases} . 铀� = 4 {蟺别}^{- 4 r_{1} / a} [\frac{1}{r_{1}} {鈭�}_{0}^{r_{1}} r_{2}^{2} e^{- 4 r_{2} / a} {dr}_{2} + {鈭�}_{r_{1}}^{鈭�} r_{2} e^{- 4 r_{2} / a} {dr}_{2}] . \frac{1}{r_{1}} {鈭�}_{0}^{r_{1}} r_{2}^{2} e^{- 4 r_{1} / a} {dr}_{2} = \frac{1}{r_{1}} [- \frac{a}{4} r_{2}^{2} e^{- 4 r_{2} / a} + \frac{a}{2} {(\frac{a}{4})}^{2} e^{- 4 r_{2} / a} (- \frac{4 r_{2}}{a} - 1)] |_{0}^{r_{1}} = - \frac{a}{4 r_{1}} [r_{1}^{2} e^{- 4 r_{1} / a} + \frac{{ar}_{1}}{2} e^{- 4 r_{1} / a} + \frac{a^{2}}{8} e^{- 4 r_{1} / a} - \frac{a^{2}}{8}] . {鈭�}_{r_{1}}^{鈭�} r_{2} e^{- 4 r_{2} / a} {dr}_{2} = {(\frac{a}{4})}^{2} e^{- 4 r_{2} / a} (- \frac{4 r_{2}}{a} - 1) |_{r_{1}}^{鈭�} = \frac{{ar}_{1}}{4} e^{- 4 r_{1} / a} + \frac{a^{2}}{16} e^{- 4 r_{1} / a} = 4 蟺 \{\frac{a^{3}}{32 r_{1}} e^{- 4 r_{1} / a} + [- \frac{ar}{4} - \frac{a^{2}}{8} - \frac{a^{3}}{32 r_{1}} + \frac{{ar}_{1}}{4} + \frac{a^{2}}{16}] e^{- 8 r_{1} / a}\}$

$= \frac{{蟺补}^{2}}{8} \{\frac{a}{r_{1}} e^{- 4 r_{1} / a} - (2 + \frac{a}{r_{1}}) e^{- 8 r_{1} / a}\} .$

$< \frac{1}{|r_{1} - r_{2}|} > = \frac{8}{{蟺补}^{4}} . 4 蟺 {鈭�}_{0}^{鈭�} [\frac{a}{r_{1}} e^{- 4 r_{1} / a} - (2 + \frac{a}{r_{1}}) e^{- 8 r_{1} / a}] r_{1}^{2} {dr}_{1 .} = \frac{32}{a^{4}} \{a . {鈭�}_{0}^{鈭�} r_{1} e^{- 4 r_{1} / a} {dr}_{1} - 2 {鈭�}_{0}^{鈭�} r_{1}^{2} e^{- 8 r_{1} / a} {dr}_{1} - a {鈭�}_{0}^{鈭�} r_{1} e^{- 8 r_{1} / a} {dr}_{1}\} . = \frac{32}{a^{4}} \{a . {(\frac{a}{4})}^{2} - 2.2 {(\frac{a}{8})}^{3} - a . {(\frac{a}{8})}^{2}\} = \frac{32}{a} (\frac{1}{16} - \frac{1}{128} - \frac{1}{64}) = \frac{5}{4 a} .$

(b) Estimating the electron interaction energy

$V^{e e} 鈮� \frac{e^{2}}{4 蟺鈭�'} < \frac{1}{|r_{1} - r_{2}|} > = \frac{5}{4} \frac{e^{2}}{4 蟺鈭�'} \frac{1}{a} = \frac{5}{4} \frac{m}{h} {(\frac{e^{2}}{4 蟺鈭�'})}^{2} = \frac{5}{2} (- E_{1}) = \frac{5}{2} (13.6 e V) . E_{0} + V_{e e} = (- 109 + 34) e V = - 75 e V, w h i c h i s p r e t t y c l o s e t o t h e e x p e r i m e n t a l v a l u e - 79 e V) .$

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Short Answer

Step by step solution

(a) Calculating <1/r1-r2>for the state ψ0

(b) Estimating the electron interaction energy

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