Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

The wave functions for the three states with the dot plots shown in Fig. 39-23, which have n = 2 , l = 1 , and 0, and $m_{l} = 0, + 1, - 1$ , are
$Ψ_{210} (r, θ) = (1 / 4 \sqrt{2 π}) (a^{- 3 / 2}) (r / a) r^{- r / 2 a} \cos θ Ψ_{21 + 1} (r, θ) = (1 / 8 \sqrt{π}) (a^{- 3 / 2}) (r / a) r^{- r / 2 a} (sinθ) e^{+ iϕ} Ψ_{21 - 1} (r, θ) = (1 / 8 \sqrt{π}) (a^{- 3 / 2}) (r / a) r^{- r / 2 a} (sinθ) e^{- iϕ}$

in which the subscripts on $Ψ (r, θ)$ give the values of the quantum numbers n , l , and $m_{l}$ the angles $θ$ and $ϕ$ are defined in Fig. 39-22. Note that the first wave function is real but the others, which involve the imaginary number i, are complex. Find the radial probability density P(r) for (a) $Ψ_{210}$ and (b) $Ψ_{21 + 1}$ (same as for $Ψ_{21 - 1}$ ). (c) Show that each P(r) is consistent with the corresponding dot plot in Fig. 39-23. (d) Add the radial probability densities for $Ψ_{210}$ , $Ψ_{21 + 1}$ , and $Ψ_{21 - 1}$ and then show that the sum is spherically symmetric, depending only on r .

The probability density of the function $Ψ_{210}$ is $P_{210} (r) = (\frac{r^{4}}{8 a^{5}}) r^{- r / a} \cos^{2} θ$ .
The probability density of the function $Ψ_{21 + 1}$ is $P_{21 + 1} (r) = (\frac{r^{4}}{16 a^{5}}) r^{- r / a} (\sin^{2} θ)$ .
The probability P(r) is consistent with the corresponding dot plot.
It is proved that the total probability density depends only on r , and it is spherically symmetric.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Give the expression for probability density function:

The probability density function (PDF) is used to define the probability of a random variable falling into a distinct range of values, as opposed to assuming a single value. The function explains the probability density function of the normal distribution and how the mean and variance exist.

The expression for the probability density of the function $Ψ$ is given by,

$P (r) = {| Ψ |}^{2} (4 {ττr}^{2})$

Step 2: (a) Define the radial probability density P(r) for Ψ210 :

The probability density of the function $Ψ_{210}$ is calculated as follows.

$P_{210} (r) = {|Ψ_{210}|}^{2} (4 {πr}^{2}) = {|(1 / 4 \sqrt{2 π}) (a^{- 3 / 2}) (r / a) r^{- r / 2 a} \cos θ|}^{2} (4 {πr}^{2}) = {(\frac{1}{4 \sqrt{2 π}})}^{2} {(\frac{r}{a^{5 / 2}})}^{2} {(r^{- r / 2 a} \cos θ)}^{2} {(4 π r)}^{2} = (\frac{1}{32 π}) (\frac{r^{2}}{a^{5}}) (r^{- 2 r / 2 a} \cos^{2} θ) (4 {πr}^{2})$

Simplify further.

$P_{210} (r) = (\frac{r^{4}}{8 a^{5}}) r^{- r / a} \cos^{2} θ$

Therefore, the probability density of the function $Ψ_{210}$ is $P_{210} (r) = (\frac{r^{4}}{8 a^{5}}) r^{- r / a} \cos^{2} θ$ .

Step 3: (b) Find the radial probability density P(r) for Ψ21+1 :

Find the square of the function $Ψ_{21 + 1}$ .

${|Ψ_{21 + 1}|}^{2} = {((1 / 8 \sqrt{π}) (a^{- 3 / 2}) (r / a) r^{- r / 2 a} (\sin θ) e^{+ i ϕ})}^{2} = {((1 / 8 \sqrt{π}) (\frac{r}{a^{5 / 2}}) r^{- r / 2 a} (\sin^{2} θ) e^{+ i ϕ})}^{2} = (\frac{1}{64 π}) (\frac{r^{2}}{a^{5}}) r^{- 2 r / 2 a} (\sin^{2} θ) e^{+ 2 i ϕ} = (\frac{r^{2}}{64 {πa}^{5}}) r^{- r / a} (\sin^{2} θ)$

Find the square of the function $Ψ_{21 - 1}$ .

${|Ψ_{21 + 1}|}^{2} = {((1 / 8 \sqrt{π}) (a^{- 3 / 2}) (r / a) r^{- r / 2 a} (\sin θ) e^{- i ϕ})}^{2} = {((1 / 8 \sqrt{π}) (\frac{r}{a^{5 / 2}}) r^{- r / 2 a} (\sin^{2} θ) e^{- i ϕ})}^{2} = (\frac{1}{64 π}) (\frac{r^{2}}{a^{5}}) r^{- 2 r / 2 a} (\sin^{2} θ) e^{- 2 i ϕ} = (\frac{r^{2}}{64 {πa}^{5}}) r^{- r / a} (\sin^{2} θ)$

Find the probability density of the above two functions.

$P_{21 \pm 1} (r) = {|Ψ_{21 \pm 1}|}^{2} (4 {πr}^{2}) = (\frac{r^{2}}{64 {πa}^{5}}) r^{- r / a} (\sin^{2} θ) (4 {πr}^{2}) = (\frac{r^{4}}{16 a^{5}}) r^{- r / a} (\sin^{2} θ)$

Hence, the probability density of the function $Ψ_{21 + 1}$ is $P_{21 + 1} (r) = (\frac{r^{4}}{16 a^{5}}) r^{- r / a} (\sin^{2} θ)$ .

Step 4: (c) Show that each P(r) is consistent with the corresponding dot plot:

For the probability density in the first case where $m_{l} = 0$ , the probability decreases with radial distance from the nucleus, also the probability is proportional to the factor of $\cos^{2} θ$ which is the maximum along the z-axis of which the angle is $θ = 0$ . This is consistent with the dot plot.

For the probability density in the second and the third cases where $m_{l} = \pm 1$ , the probability decreases with radial distance from the nucleus, also the probability is proportional to the factor of $\sin^{2} θ$ which is the maximum in the xy -plane of which the angle is $θ = 90^{\circ}$ . This is also consistent with the dot plot.

Therefore, the probability P(r) is consistent with the corresponding dot plot.

Step 5: (d) Show that the sum of Ψ210 , Ψ21+1 , and Ψ21-1 is spherically symmetric, and depending only on r :

Add the three probabilities as follows.

$P (r) = P_{210} (r) + P_{21 + 1} (r) + P_{21 - 1} (r) = (\frac{r^{4}}{8 a^{5}}) r^{- r / a} \cos^{2} θ + (\frac{r^{4}}{16 a^{5}}) r^{- r / a} (\sin^{2} θ) + (\frac{r^{4}}{16 a^{5}}) r^{- r / a} (\sin^{2} θ) = (\frac{r^{4}}{8 a^{5}}) r^{- r / a} (\sin^{2} θ + \cos^{2} θ) = (\frac{r^{4}}{8 a^{5}}) r^{- r / a}$

From the above equation it is clear that the total probability density depends only on r , and it is spherically symmetric.

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Short Answer

Step by Step Solution

Step 1: Give the expression for probability density function:

Step 2: (a) Define the radial probability density P(r) for Ψ210 :

Step 3: (b) Find the radial probability density P(r) for Ψ21+1 :

Step 4: (c) Show that each P(r) is consistent with the corresponding dot plot:

Step 5: (d) Show that the sum of Ψ210 , Ψ21+1 , and Ψ21-1 is spherically symmetric, and depending only on r :

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Give the expression for probability density function:

Step 2: (a) Define the radial probability density P(r) for Ψ210 :

Step 3: (b) Find the radial probability density P(r) for Ψ21+1 :

Step 4: (c) Show that each P(r) is consistent with the corresponding dot plot:

Step 5: (d) Show that the sum of Ψ210 , Ψ21+1 , and Ψ21-1 is spherically symmetric, and depending only on r :

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