# 8.1 The hydrogen atom  (Page 4/11)

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## What are the allowed directions?

Calculate the angles that the angular momentum vector $\stackrel{\to }{L}$ can make with the z -axis for $l=1$ , as shown in [link] .

## Strategy

The vectors $\stackrel{\to }{L}$ and ${\stackrel{\to }{L}}_{z}$ (in the z -direction) form a right triangle, where $\stackrel{\to }{L}$ is the hypotenuse and ${\stackrel{\to }{L}}_{z}$ is the adjacent side. The ratio of ${L}_{z}$ to | $\stackrel{\to }{L}$ | is the cosine of the angle of interest. The magnitudes $L=|\stackrel{\to }{L}|$ and ${L}_{z}$ are given by

$L=\sqrt{l\left(l+1\right)}\hslash \phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{L}_{z}=m\hslash .$

## Solution

We are given $l=1$ , so ml can be $+1,0,\text{or}\phantom{\rule{0.2em}{0ex}}-1.$ Thus, L has the value given by

$L=\sqrt{l\left(l+1\right)}\hslash =\sqrt{2}\hslash .$

The quantity ${L}_{z}$ can have three values, given by ${L}_{z}={m}_{l}\hslash$ .

${L}_{z}={m}_{l}\hslash =\left\{\begin{array}{l}\phantom{\rule{0.5em}{0ex}}\hslash ,\text{}\phantom{\rule{0.2em}{0ex}}{m}_{l}=+1\hfill \\ \phantom{\rule{0.5em}{0ex}}0,\phantom{\rule{0.5em}{0ex}}{m}_{l}=0\hfill \\ -\hslash ,{m}_{l}=-1\hfill \end{array}$

As you can see in [link] , $\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ={L}_{z}\text{/}L,$ so for $m=+1$ , we have

$\text{cos}\phantom{\rule{0.2em}{0ex}}{\theta }_{1}=\frac{{L}_{Z}}{L}=\frac{\hslash }{\sqrt{2}\hslash }=\frac{1}{\sqrt{2}}=0.707.$

Thus,

${\theta }_{1}={\phantom{\rule{0.2em}{0ex}}\text{cos}}^{-1}0.707=45.0\text{°}.$

Similarly, for $m=0$ , we find $\text{cos}\phantom{\rule{0.2em}{0ex}}{\theta }_{2}=0;$ this gives

${\theta }_{2}={\text{cos}}^{-1}0=90.0\text{°}.$

Then for ${m}_{l}=-1$ :

$\text{cos}\phantom{\rule{0.2em}{0ex}}{\theta }_{3}=\frac{{L}_{Z}}{L}=\frac{\text{−}\hslash }{\sqrt{2}\hslash }=-\frac{1}{\sqrt{2}}=-0.707,$

so that

${\theta }_{3}={\phantom{\rule{0.2em}{0ex}}\text{cos}}^{-1}\left(-0.707\right)=135.0\text{°}.$

## Significance

The angles are consistent with the figure. Only the angle relative to the z -axis is quantized. L can point in any direction as long as it makes the proper angle with the z -axis. Thus, the angular momentum vectors lie on cones, as illustrated. To see how the correspondence principle holds here, consider that the smallest angle ( ${\theta }_{1}$ in the example) is for the maximum value of ${m}_{l},$ namely ${m}_{l}=l.$ For that smallest angle,

$\text{cos}\phantom{\rule{0.2em}{0ex}}\theta =\frac{{L}_{z}}{L}=\frac{l}{\sqrt{l\left(l+1\right)}},$

which approaches 1 as l becomes very large. If $\text{cos}\phantom{\rule{0.2em}{0ex}}\theta =1$ , then $\theta =0º$ . Furthermore, for large l , there are many values of ${m}_{l}$ , so that all angles become possible as l gets very large.

Check Your Understanding Can the magnitude of ${L}_{z}$ ever be equal to L ?

No. The quantum number $m=\text{−}l,\text{−}l+1\text{,…},0\text{,…},l-1,l.$ Thus, the magnitude of ${L}_{z}$ is always less than L because $<\sqrt{l\left(l+1\right)}$

## Using the wave function to make predictions

As we saw earlier, we can use quantum mechanics to make predictions about physical events by the use of probability statements. It is therefore proper to state, “An electron is located within this volume with this probability at this time,” but not, “An electron is located at the position ( x , y , z ) at this time.” To determine the probability of finding an electron in a hydrogen atom in a particular region of space, it is necessary to integrate the probability density $|{\psi }_{nlm}{|}^{2}$ over that region:

$\text{Probability}={\int }_{\text{volume}}|{\psi }_{nlm}{|}^{2}dV,$

where dV is an infinitesimal volume element. If this integral is computed for all space, the result is 1, because the probability of the particle to be located somewhere is 100% (the normalization condition). In a more advanced course on modern physics, you will find that $|{\psi }_{nlm}{|}^{2}={\psi }_{nlm}^{*}{\psi }_{nlm},$ where ${\psi }_{nlm}^{*}$ is the complex conjugate. This eliminates the occurrences of $i=\sqrt{-1}$ in the above calculation.

Consider an electron in a state of zero angular momentum ( $l=0$ ). In this case, the electron’s wave function depends only on the radial coordinate r . (Refer to the states ${\psi }_{100}$ and ${\psi }_{200}$ in [link] .) The infinitesimal volume element corresponds to a spherical shell of radius r and infinitesimal thickness dr , written as

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