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An electron is in state with value I = 3.

Orbital angular momentum quantum number is 'I' . The orbital angular momentum quantum number determines the shape of the orbital and hence the angular distribution. The value of I is dependent on the principal quantum number n .

Using the given value of the orbital angular quantum number, get the magnitude of the orbital angular momentum and dipole moment. As the range of m varies from -I to +I, the maximum value is given by the most positive value of I . Similarly using the concept of the orbital momentum and dipole moment projection along the z-axis, define the required values using the given value in the formula. The semi-classical angle between two vectors can be given using the cosine angle between them.

Formulas:

The magnitude of the orbital angular momentum in terms of $\mathrm{Ä\S }$is,

$\stackrel{→}{\mathrm{L}}=\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}\mathrm{Ä\S }$ ….. (1)

The magnitude of the orbital dipole moment in terms of Bohr’s magnetron ${\mathrm{μ}}_{\mathrm{B}}$,

$\stackrel{→}{{\mathrm{μ}}_{\mathrm{orb}}}=\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}{\mathrm{μ}}_{\mathrm{B}}$ ….. (2)

The z-component of the orbital angular momentum is,

$L_z=m\mathcal{l}\mathrm{Ä\S }$ ….. (3)

The z-component of the orbital magnetic dipole moment is,

${\mathrm{μ}}_{\mathrm{orb},\mathrm{z}}=-{\mathrm{m}}_{\mathrm{I}}{\mathrm{μ}}_{\mathrm{B}}$ ….. (4)

The semi-classical angle between a vector and its z-component is,

$\mathrm{θ}={\cos }^{-1}\left(\frac{{\mathrm{a}}_{\mathrm{z}}}{\mathrm{a}}\right)$ ….. (5)

Define the value of the magnitude of the orbital angular momentum by substituting 3 for I in equation (1) as below.

$$\begin{gathered} \stackrel{→}{\mathrm{L}}=\sqrt{3\left(3+1\right)}\mathrm{Ä\S } \\ =\sqrt{12}\mathrm{Ä\S } \\ =3.46\mathrm{Ä\S } \end{gathered}$$

Hence, the $\mathrm{Ä\S }$multiple value of the orbital angular momentum is 3.46.

Calculate the value of the orbital magnetic dipole moment ny substituting 3 for I into equation (2) as below.

$$\begin{gathered} \stackrel{→}{{\mathrm{μ}}_{\mathrm{orb}}}=\sqrt{3\left(3+1\right)}{\mathrm{μ}}_{\mathrm{B}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=\sqrt{12}{\mathrm{μ}}_{\mathrm{B}} \\ =3.46{\mathrm{μ}}_{\mathrm{B}} \end{gathered}$$

Hence, the ${\mathrm{μ}}_{\mathrm{B}}$multiple value of the dipole moment is 3.46.

Using the concept of range of ${\mathrm{L}}_{\mathrm{z}}={\mathrm{m}}_{\mathrm{I}}\mathrm{Ä\S }$for I values, get the largest value of the ${\mathrm{m}}_{\mathrm{I}}$as follows:

$$\begin{gathered} {\mathrm{m}}_{\mathrm{I}}=+\mathrm{I} \\ =3 \end{gathered}$$

Hence, the largest value of ${\mathrm{m}}_{\mathrm{I}}$is 3 .

Using the largest value of ${\mathrm{m}}_{\mathrm{I}}=3$in equation (3), obtain the $\mathrm{Ä\S }$multiple value of z-component of the orbital angular momentum as follows:

${\mathrm{L}}_{\mathrm{z}}=3\mathrm{Ä\S }$

Hence, the $\mathrm{Ä\S }$multiple value of the z-component is 3 .

Using the largest value of ${\mathrm{m}}_{\mathrm{I}}=3$in equation (3), determine the ${\mathrm{μ}}_{\mathrm{B}}$multiple value of z-component of the orbital angular momentum as follows:

${\mathrm{μ}}_{\mathrm{orb}.\mathrm{z}}=-3{\mathrm{μ}}_{\mathrm{B}}$

Hence, the ${\mathrm{μ}}_{\mathrm{B}}$multiple value of the z-component is -3.

Using the value of $\stackrel{→}{\mathrm{L}}$and from part (a) and (d) calculations, get the semi-classical angle between the two directions ${\mathrm{L}}_{\mathrm{z}}$and $\stackrel{→}{\mathrm{L}}$using equation (5) as follows:

$$\begin{gathered} \mathrm{θ}={\cos }^{-1}\left(\frac{3\mathrm{Ä\S }}{\sqrt{12}\mathrm{Ä\S }}\right) \\ ={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right) \\ =30° \end{gathered}$$

Hence, the value of the angle is $30°$.

Using the concept, we can say that the second largest value of ${\mathrm{m}}_1\mathrm{is}+2$.

Thus, using this value in equation (3), determine the component value as below.

${\mathrm{L}}_{\mathrm{z}}=2\mathrm{Ä\S }$

Now, the angle between the orbital angular momentum and the second largest value of is given using the above value and value of angular momentum from part (a) calculations in equation (5) as follows:

$$\begin{gathered} \mathrm{θ}={\cos }^{-1}\left(\frac{2\mathrm{Ä\S }}{\sqrt{12}\mathrm{Ä\S }}\right) \\ ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ =54.7° \end{gathered}$$

Hence, the value of the angle is $54.7°$.

Using the concept, we can say that the smallest value of ${\mathrm{m}}_{\mathrm{I}}\mathrm{is}-3$ .

Thus, using this value in equation (3), get the component value as:

${\mathrm{L}}_{\mathrm{z}}=-3\mathrm{Ä\S }$

Now, the angle between the orbital angular momentum and the second largest value of is given using the above value and value of angular momentum from part (a) calculations in equation (5) as follows:

$$\begin{gathered} \mathrm{θ}={\cos }^{-1}\left(\frac{-3\mathrm{Ä\S }}{\sqrt{12}\mathrm{Ä\S }}\right) \\ ={\cos }^{-1}\left(\frac{-\sqrt{3}}{2}\right) \\ =150° \end{gathered}$$

Hence, the value of the angle is $150°$.