Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

If orbital angular momentum is measured along, say, a z-axis to obtain a value for $L_{z}$ , show that role="math" localid="1661497092782" ${(L_{x}^{2} + L_{y}^{2})}^{1 / 2} = {[I (I + 1) - m_{I}^{2}]}^{1 / 2} ħ$ is the most that can be said about the other two components of the orbital angular momentum.

It is proved for the other two components of the orbital angular momentum is

${(L_{x}^{2} + L_{y}^{2})}^{1 / 2} = {[I (I + 1) - m_{I}^{2}]}^{1 / 2} ħ$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data:

The orbital angular momentum $\vec{L}$ is measured along a z-axis to get a value $L_{z}$ .

Step 2: Understanding the concept of angular momentum:

In quantum mechanics, angular momentum is a vector operator with well-defined commutation relations between its three components.

Using the concept of orbital angular momentum and the value of the momentum in the z-axis in the resultant value of the momentum, define the required expression of the x- and y-component of the angular momentum.

Formulas:

The magnitude of the orbital angular momentum in terms $ħ$ of is,

$\vec{L} = \sqrt{I (I + 1) ħ}$ ….. (1)

The z-component of the orbital angular momentum is,

$L_{z} = m_{I} ħ$ ….. (2)

The resultant value of the angular momentum is,

${|\vec{L}|}^{2} = L_{x}^{2} + L_{y}^{2} + L_{z}^{2}$ ….. (3)

Step 3: Calculation to get the required equation:

Substituting the values of equations (1) and (2) in equation (3), the required equation to be proved as follows:

${(\sqrt{I (I + 1)} ħ)}^{2} = L_{x}^{2} + L_{y}^{2} + {(m_{I}^{} ħ)}^{2} L_{x}^{2} + L_{y}^{2} = {(\sqrt{I (I + 1)} ħ)}^{2} - {(m_{I}^{} ħ)}^{2} (L_{x}^{2} + L_{y}^{2}) = {[I (I + 1) - m_{I}^{2}]}^{1 / 2} ħ$

For a given value of I, the greatest value of $m_{1}$ is , so the smallest value of $\sqrt{L_{x}^{2} + L_{y}^{2}} is ħ \sqrt{I}$ .

Now, the smallest possible value of $m_{1}$ is zero, thus the largest possible value of is .

$\sqrt{L_{x}^{2} + L_{y}^{2}} is ħ \sqrt{I (I + 1)}$

So, the range is given by:

$ħ \sqrt{I} \leq \sqrt{L_{x}^{2} + L_{y}^{2}} \leq \sqrt{I (I + 1)}$

Hence, the given equation is proved.

Element	λ (pm)	Element	λ (pm)
Ti	275	Co	179
V	250	Ni	166
Cr	229	Cu	154
Mn	210	Zn	143
Fe	193	Ga	134

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Step by Step Solution

Step 1: The given data:

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Step 3: Calculation to get the required equation:

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Fundamentals Of Physics

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Short Answer

If orbital angular momentum is measured along, say, a z-axis to obtain a value for Lz, show that role="math" localid="1661497092782" (Lx2+Ly2)1/2=[I(I+1)-mI2]1/2ħ is the most that can be said about the other two components of the orbital angular momentum.

Step by Step Solution

Step 1: The given data:

Step 2: Understanding the concept of angular momentum:

Step 3: Calculation to get the required equation:

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