# 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

#### Questions & Answers

For the question about the scuba instructor's head above the pool, how did you arrive at this answer? What is the process?
as a free falling object increases speed what is happening to the acceleration
of course g is constant
Alwielland
acceleration also inc
Usman
which paper will be subjective and which one objective
jay
photo electrons doesn't emmit when electrons are free to move on surface of metal why?
What would be the minimum work function of a metal have to be for visible light(400-700)nm to ejected photoelectrons?
give any fix value to wave length
Rafi
40 cm into change mm
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
Prema
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
Prema
there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
Prema
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
Prema
this msg is out of mistake. sorry friends​.
Prema
what is physics?
why we have physics
because is the study of mater and natural world
John
because physics is nature. it explains the laws of nature. some laws already discovered. some laws yet to be discovered.
Yoblaze
is this a physics forum
explain l-s coupling
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what's the signeficance of dirac equetion.?
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As refractive index depend on other factors also but if we supply heat on any system or media its refractive index decrease. i.e. it is inversely proportional to the heat.
ganesh
you are correct
Priyojit
law of multiple
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if we heated the ice then the refractive index be change from natural water
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can someone explain normalization condition
Swati
yes
Chemist
1 millimeter is How many metres
1millimeter =0.001metre
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