91影视

Electron in hydrogen atom.

In order t to find the minimum angle that the total angular momentum vector J makes with the Z -axis for a 3d state of hydrogen, the expressions for the magnitude of the total angular momentum vector J, and the z projection of $J,J_z$, are needed.

The magnitude J of the total angular momentum vector J is, $J=\sqrt{j(j+1)}h$.

Here $\mathrm{鈩�}$ is reduced Planck constant, and J is the total angular momentum quantum number, which is restricted to the values,

$j=|l-s|,|l-s|+1,鈥�.,l+s-1,l+s$.

With lbeing he orbital angular momentum quantum number, and s is the spin quantum number is, $J_z=m_j\mathrm{鈩�}$.

Here $\mathit{鈩�}$ is reduced Planck constant, and $m_j$ is he quantum number for the Z -projection of J , which has the values, $m_j:-j,-j+1,鈥�鈥�,j-1,j$.

With J being the total angular momentum quantum number.

First, it's required to know what l and sis, so as to be able to figure out what j is.

Since they say that they're looking at the d shell, that corresponds to an l value of 2.

And given that we're looking at an electron, that means that the spin swill be$1/2$ .

Next, determine whether the L and S will be aligned or anti-aligned, so as to figure out how to combine the l and s.

Since J should be as close as possible to the Z -axis, it will help to have as many J vectors as possible.

This can be achieved with L and S being aligned, such that l and s will add, and thus give a great range for j.

So, j will be:

$\begin{array}{rcl}j& =& l+s\\ & =& \left(2\right)+\left(\frac{1}{2}\right)\\ & =& \frac{5}{2}\\ & & \end{array}$

The magnitude of J :

$\begin{array}{rcl}J& =& \sqrt{j(j+1)}\mathrm{鈩�}\\ & =& \sqrt{\left(\frac{5}{2}\right)\left(\left[\frac{5}{2}\right]+1\right)}\mathrm{鈩�}\\ & =& \frac{\sqrt{35}}{2}\mathrm{鈩�}\\ & & \end{array}$

For simplicity, use the J closer to the positive Z-axis, which means that $m_j$ is just equal to$j$ , or just $5/2$.

Find the projection of J onto the Z-axis.

$\begin{array}{rcl}{J}_z& =& m_j\mathrm{鈩�}\\ & =& \left(\frac{5}{2}\right)\mathrm{鈩�}\\ & =& \frac{5}{2}\mathrm{鈩�}\\ & & \end{array}$

The magnitudes of and the angle can be summed up with the picture in figure 1.

Figure 1

Solve for the angle as follows

$\begin{array}{rcl}\cos 胃& =& \frac{J_z}{J}胃\\ & =& {\cos }^{-1}\left(\frac{J_z}{J}\right)\\ & & \end{array}$

Use the values$\frac{5}{2}\mathrm{鈩�}$for$J_z$, and$\frac{\sqrt{35}}{2}\mathrm{鈩�}$for$J$ .

$\begin{array}{rcl}胃& =& {\cos }^{-1}\left(\frac{J_x}{J}\right)\\ & =& {\cos }^{-1}\left[\frac{\left(\frac{5}{2}\mathrm{鈩�}\right)}{\left(\frac{\sqrt{33}\mathrm{鈩�}}{2}\right)}\right]\\ & =& {\cos }^{-1}\left[\frac{5}{\sqrt{35}}\right]\\ & =& 32.3^{鈭�}\\ & & \end{array}$

Therefore, minimum angle that J makes with the Z -axis in the 3d state of hydrogen is $胃=32.3^{鈭�}$.