91影视

It is a correspondence problem with quantum numberI.

Magnetic resonance, absorption or radiation by electrons or atomic nuclei in response to the use of other magnetic fields.

Using the concept of orbital angular momentum, get the relation and the value of the orbital angular quantum number. Now, using the concept of allowed orientation considering the orbital quantum number, define the allowed or possible electrons states for the magnetic quantum number. At last for the value of minimum half-angle, use the value of the component of angular component and the orbital angular momentum.

Formulas:

The magnitude of the orbital angular momentum,

$\left|\stackrel{鈫�}{\mathrm{L}}\right|=\sqrt{I\left(I+1\right)}魔$ 鈥�.. (1)

The magnitude of the z-component of the angular momentum,

$$\begin{gathered} \left|{\mathrm{L}}_{\mathrm{z}}\right|=m_I魔 \\ \mathrm{Here},{\mathrm{m}}_{\mathrm{I}}=-\left|\mathrm{to}+\right| \end{gathered}$$ 鈥�.. (2)

The number of allowed orientations for the given number I ,

${\mathrm{N}}_{\mathrm{I}}=2\mathrm{I}+1$ 鈥�.. (3)

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

$\mathrm{胃}={\cos }^{-1}\left(\frac{{\stackrel{鈫�}{\mathrm{L}}}_{\mathrm{z}}}{\stackrel{鈫�}{\mathrm{L}}}\right)$ 鈥�.. (4)

The value of $\mathcal{l}$satisfies the equation (1) that is equation of the orbital angular momentum.

Thus, the value of the quantum number I for the orbital motion of Earth around the Sun is given using equation (1) as follows:

$$\begin{gathered} \left|\stackrel{鈫�}{\mathrm{L}}\right|=\sqrt{I\left(I+1\right)}魔 \\ 鈮�\sqrt{{\mathrm{I}}^2}魔 \\ =\mathrm{I}魔 \\ \mathrm{I}=\frac{\mathrm{L}}{\mathrm{魔}} \\ =3脳{10}^{74} \end{gathered}$$

Hence, the value of the quantum number is $3脳{10}^{74}$.

Using the concept and equation (3), the number of allowed orientations of the plane of Earth鈥檚 orbit as follows:

$$\begin{gathered} {\mathrm{N}}_{\mathrm{I}}=2\mathrm{I}+1 \\ 鈮�2\mathrm{I} \\ =2\left(3脳{10}^{74}\right) \\ =6脳{10}^{74} \end{gathered}$$

Hence, the value of the number of orientations is $6脳{10}^{74}$.

The equation (4) represents the semi-classical angle between the component of the angular momemtum and the angular momentum, which can further be simplified using equations (1) and (2) as follows:

$$\begin{gathered} \mathrm{肠辞蝉胃}=\left(\frac{{\mathrm{m}}_{\mathrm{Ima}}{}_{}\mathrm{魔}}{\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}{}_{}\mathrm{魔}}\right) \\ =\frac{1}{\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}} \\ 鈮�1-\frac{1}{2\mathrm{I}} \\ =1-\frac{1}{2\left(3脳{10}^{74}\right)} \end{gathered}$$

Now, the value of the half angle considering the half-angle approximations can be calculated as follows:

$$\begin{gathered} \cos {\mathrm{胃}}_{\min }=1-\frac{{{\mathrm{胃}}^2}_{\min }}{2} \\ 鈮�1-\frac{{10}^{-74}}{6} \\ {\mathrm{胃}}_{\min }=\sqrt{\frac{{10}^{-74}}{3}} \\ =6脳{10}^{-38}\mathrm{rad} \end{gathered}$$

The correspondence principle requires that all the quantum effects vanish as.

In this case, is extremely small so the quantization effects are barely existent, with

$$\begin{gathered} {\mathrm{胃}}_{\min }饾啅{10}^{-38}\mathrm{rad} \\ 鈮�0 \end{gathered}$$

Hence, the value of the half-angle of the smallest cone is $6脳{10}^{-38}\mathrm{rad}$.