Q13P Question: In Problem 4.43聽you c... [FREE SOLUTION]

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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 6: Q13P (page 270)

Question: In Problem 4.43you calculated the expectation value of $r^{s}$ in the state $蠄_{321}$ . Check your answer for the special cases s = 0(trivial), s = -1(Equation 6.55), s = -2(Equation 6.56), and s = -3(Equation 6.64). Comment on the case s = -7.

Short Answer

Expert verified

for: $s = 0 : 1 s = - 1 : \frac{1}{9 a} s = - 2 : \frac{1}{135 a^{2}} s = - 3 : \frac{1}{405 a^{3}}$

s = -7 Or smaller that the result will be undefined due to

negative numbers! = Undefined quantity

Step by step solution

Given information:

Equation 6.55, 6.56, 6.64 are

$<\frac{1}{r}> = \frac{1}{n^{2} a} <\frac{1}{r^{2}}> = \frac{1}{(l + 1 / 2) n^{3} a^{2}} <\frac{1}{r^{3}}> = \frac{1}{l (l + 1 / 2) (l + 1) n^{3} a^{3}}$

Problem 4.43-(a) Construct the spatial wave function $(蠄)$ for hydrogen in the state n = 3 , l=2 . m = 1 Express your answer as a function of $r, 胃, 蠒$ , and a (the Bohr radius) only - no other variables, (p,z etc.) or ,functions, ( y , 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 $r, 胃,$ 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?

Step 2:Check for trivial or non-trivial solution

For n = 3,l = 2,m = 1

Then:

- For s = 0:

$<1> = \frac{6!}{6!} (1)$ 鈥︹赌︹赌�.摆罢谤颈惫颈补濒闭

- For s = -1

$<\frac{1}{r}> = \frac{5!}{6!} {(\frac{3 a}{2})}^{- 1} = \frac{1}{6} (\frac{2}{3 a}) = \frac{1}{3^{3} a} = \frac{1}{n^{2} a}$

- For s = -2:

$<\frac{1}{r^{2}}> = \frac{4!}{6!} {(\frac{3 a}{2})}^{2} = \frac{1}{30} {(\frac{2}{3 a})}^{2} = \frac{2}{135 a^{2}} = \frac{1}{(2 + \frac{1}{2}) 3^{3} a^{2}} = \frac{1}{(1 + \frac{1}{2}) n^{3} a^{2}}$

solve further

-For s = -3

$<\frac{1}{r^{3}}> = \frac{3!}{6!} {(\frac{3 a}{2})}^{- 3} = \frac{1}{120} (\frac{8}{27 a^{3}}) = \frac{1}{2 (2 + \frac{1}{2}) (2 + 1) 3^{3} a^{3}} = \frac{1}{l (l + \frac{1}{2}) (l + 1) n^{3} a^{3}}$

- For s = -7

$<\frac{1}{r^{7}}> = \frac{(- 1)!}{6!} {(\frac{3 a}{2})}^{- 7} (- 1)!$

$undentified 鈮� 鈭�$

Then for s = -7 or smaller that the result will be undefined due to

negative numbers! = Undefined quantity

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91影视

Question: In Problem 4.43you calculated the expectation value of $r^{s}$ in the state $蠄_{321}$ . Check your answer for the special cases s = 0(trivial), s = -1(Equation 6.55), s = -2(Equation 6.56), and s = -3(Equation 6.64). Comment on the case s = -7.

Short Answer

Step by step solution

Given information:

Step 2:Check for trivial or non-trivial solution

solve further

One App. One Place for Learning.

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91影视

Question: In Problem 4.43you calculated the expectation value ofrsin the state蠄321. Check your answer for the special cases s = 0(trivial), s = -1(Equation 6.55), s = -2(Equation 6.56), and s = -3(Equation 6.64). Comment on the case s = -7.

Short Answer

Step by step solution

Given information:

Step 2:Check for trivial or non-trivial solution

solve further

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Force

Physical Quantities And Units

Electromagnetism

Magnetism and Electromagnetic Induction

Work Energy and Power

Study anywhere. Anytime. Across all devices.

Question: In Problem 4.43you calculated the expectation value of $r^{s}$ in the state $蠄_{321}$ . Check your answer for the special cases s = 0(trivial), s = -1(Equation 6.55), s = -2(Equation 6.56), and s = -3(Equation 6.64). Comment on the case s = -7.