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Q11P

Expert-verifiedFound in: Page 1247

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**If orbital angular momentum is measured along, say, a z-axis to obtain a value for ${{\mathbf{L}}}_{{\mathbf{z}}}$, show that role="math" localid="1661497092782" ${\mathbf{(}{\mathbf{L}}_{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{L}}_{\mathbf{y}}^{\mathbf{2}}\mathbf{)}}^{\mathbf{1}\mathbf{/}\mathbf{2}}{\mathbf{=}}{\mathbf{[}\mathbf{I}\left(I+1\right)\mathbf{-}{\mathbf{m}}_{\mathbf{I}}^{\mathbf{2}}\mathbf{]}}^{\mathbf{1}\mathbf{/}\mathbf{2}}{\mathit{\u0127}}$ 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

${({\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2})}^{1/2}={[\mathrm{I}(\mathrm{I}+1)-{\mathrm{m}}_{\mathrm{I}}^{2}]}^{1/2}\u0127$.

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

**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 $\u0127$ of is,

$\overrightarrow{\mathrm{L}}=\sqrt{I\left(I+1\right)\u0127}$ ….. (1)

The z-component of the orbital angular momentum is,

${\mathrm{L}}_{\mathrm{z}}={\mathrm{m}}_{\mathrm{I}}\mathrm{\u0127}$ ….. (2)

The resultant value of the angular momentum is,

${\left|\overrightarrow{\mathrm{L}}\right|}^{2}={\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}+{\mathrm{L}}_{\mathrm{z}}^{2}$ ….. (3)

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

${\left(\sqrt{\mathrm{I}(\mathrm{I}+1)}\u0127\right)}^{2}={\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}+{\left({\mathrm{m}}_{\mathrm{I}}^{}\mathrm{\u0127}\right)}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}={\left(\sqrt{\mathrm{I}(\mathrm{I}+1)}\mathrm{\u0127}\right)}^{2}-{\left({\mathrm{m}}_{\mathrm{I}}^{}\mathrm{\u0127}\right)}^{2}\phantom{\rule{0ex}{0ex}}\left({\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}\right)={\left[\mathrm{I}(\mathrm{I}+1)-{\mathrm{m}}_{\mathrm{I}}^{2}\right]}^{1/2}\mathrm{\u0127}$

For a given value of I, the greatest value of ${\mathrm{m}}_{1}$ is , so the smallest value of $\sqrt{{\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}}\mathrm{is}\mathrm{\u0127}\sqrt{\mathrm{I}}$.

Now, the smallest possible value of ${\mathrm{m}}_{1}$ is zero, thus the largest possible value of is .

$\sqrt{{\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}}\mathrm{is}\mathrm{\u0127}\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}$

So, the range is given by:

$\mathrm{\u0127}\sqrt{\mathrm{I}}\le \sqrt{{\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}}\le \sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}$

Hence, the given equation is proved.

**Here are the ${{\mathbf{K}}}_{{\mathbf{\alpha}}}$ wavelengths of a few elements:**

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 |

**Make a Moseley plot (like that in Fig. 40-16) from these data and verify that its slope agrees with the value given for C in Module 40-6.**

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