9) For the 2p, orbital, what is the most probable point (r, 8, d) where an e- will be found. R₁, Y₁ = fi ce -$12 3 2,1 1,0 √24 a. pe 3/2 "pe²³ ) ² = 0 d Z dx √24 a. pe 4T COSO 2 (COSO )² = 0
9) For the 2p, orbital, what is the most probable point (r, 8, d) where an e- will be found. R₁, Y₁ = fi ce -$12 3 2,1 1,0 √24 a. pe 3/2 "pe²³ ) ² = 0 d Z dx √24 a. pe 4T COSO 2 (COSO )² = 0
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Question 9: For the 2p\(_z\) orbital, what is the most probable point \((r, \theta, \phi)\) where an electron will be found?**
The expression for the radial and angular parts is:
\[
R_{2,1}Y_{1,0} = \left(\frac{1}{\sqrt{24}}\right)\left(\frac{z}{a_0}\right)^{3/2} pe^{-r/2}\sqrt{\frac{3}{4\pi}} \cos\theta
\]
To find the most probable point, the first derivative of the probability density with respect to the position \(x\) is set to zero:
\[
\frac{d}{dx} \left(\sqrt{\frac{3}{4\pi}} \cos\theta \right)^2 = 0
\]
And:
\[
\frac{d}{dx} \left(\frac{1}{\sqrt{24}}\left(\frac{z}{a_0}\right)^{3/2} pe^{-r/2} \right)^2 = 0
\]
This involves determining where the derivative of the probability density function is zero, indicating points of maximum probability.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1fed7a52-65d8-44f4-988f-6c762cda0c09%2F7c937233-99ad-402e-aa8f-d8c463d55a3b%2Fd0rzhfp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 9: For the 2p\(_z\) orbital, what is the most probable point \((r, \theta, \phi)\) where an electron will be found?**
The expression for the radial and angular parts is:
\[
R_{2,1}Y_{1,0} = \left(\frac{1}{\sqrt{24}}\right)\left(\frac{z}{a_0}\right)^{3/2} pe^{-r/2}\sqrt{\frac{3}{4\pi}} \cos\theta
\]
To find the most probable point, the first derivative of the probability density with respect to the position \(x\) is set to zero:
\[
\frac{d}{dx} \left(\sqrt{\frac{3}{4\pi}} \cos\theta \right)^2 = 0
\]
And:
\[
\frac{d}{dx} \left(\frac{1}{\sqrt{24}}\left(\frac{z}{a_0}\right)^{3/2} pe^{-r/2} \right)^2 = 0
\]
This involves determining where the derivative of the probability density function is zero, indicating points of maximum probability.
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