# 30.8 Quantum numbers and rules

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• Define quantum number.
• Calculate angle of angular momentum vector with an axis.
• Define spin quantum number.

Physical characteristics that are quantized—such as energy, charge, and angular momentum—are of such importance that names and symbols are given to them. The values of quantized entities are expressed in terms of quantum numbers    , and the rules governing them are of the utmost importance in determining what nature is and does. This section covers some of the more important quantum numbers and rules—all of which apply in chemistry, material science, and far beyond the realm of atomic physics, where they were first discovered. Once again, we see how physics makes discoveries which enable other fields to grow.

The energy states of bound systems are quantized , because the particle wavelength can fit into the bounds of the system in only certain ways. This was elaborated for the hydrogen atom, for which the allowed energies are expressed as ${E}_{n}\propto 1/{n}^{2}$ , where $n=1, 2, 3, ...$ . We define $n$ to be the principal quantum number that labels the basic states of a system. The lowest-energy state has $n=1$ , the first excited state has $n=2$ , and so on. Thus the allowed values for the principal quantum number are

$n=1, 2, 3, ....$

This is more than just a numbering scheme, since the energy of the system, such as the hydrogen atom, can be expressed as some function of $n$ , as can other characteristics (such as the orbital radii of the hydrogen atom).

The fact that the magnitude of angular momentum is quantized was first recognized by Bohr in relation to the hydrogen atom; it is now known to be true in general. With the development of quantum mechanics, it was found that the magnitude of angular momentum $L$ can have only the values

$L=\sqrt{l\left(l+1\right)}\frac{h}{2\pi }\phantom{\rule{1.00em}{0ex}}\left(l=0, 1, 2, ...,\phantom{\rule{0.25em}{0ex}}n-1\right)\text{,}$

where $l$ is defined to be the angular momentum quantum number    . The rule for $l$ in atoms is given in the parentheses. Given $n$ , the value of $l$ can be any integer from zero up to $n-1$ . For example, if $n=4$ , then $l$ can be 0, 1, 2, or 3.

Note that for $n=1$ , $l$ can only be zero. This means that the ground-state angular momentum for hydrogen is actually zero, not $h/2\pi$ as Bohr proposed. The picture of circular orbits is not valid, because there would be angular momentum for any circular orbit. A more valid picture is the cloud of probability shown for the ground state of hydrogen in [link] . The electron actually spends time in and near the nucleus. The reason the electron does not remain in the nucleus is related to Heisenberg’s uncertainty principle—the electron’s energy would have to be much too large to be confined to the small space of the nucleus. Now the first excited state of hydrogen has $n=2$ , so that $l$ can be either 0 or 1, according to the rule in $L=\sqrt{l\left(l+1\right)}\frac{h}{2\pi }$ . Similarly, for $n=3$ , $l$ can be 0, 1, or 2. It is often most convenient to state the value of $l$ , a simple integer, rather than calculating the value of $L$ from $L=\sqrt{l\left(l+1\right)}\frac{h}{2\pi }$ . For example, for $l=2$ , we see that

$L=\sqrt{2\left(2+1\right)}\frac{h}{2\pi }=\sqrt{6}\frac{h}{2\pi }=0\text{.}\text{390}h=2\text{.}\text{58}×{\text{10}}^{-\text{34}}\phantom{\rule{0.25em}{0ex}}\text{J}\cdot s.$

It is much simpler to state $l=2$ .

As recognized in the Zeeman effect, the direction of angular momentum is quantized . We now know this is true in all circumstances. It is found that the component of angular momentum along one direction in space, usually called the $z$ -axis, can have only certain values of ${L}_{z}$ . The direction in space must be related to something physical, such as the direction of the magnetic field at that location. This is an aspect of relativity. Direction has no meaning if there is nothing that varies with direction, as does magnetic force. The allowed values of ${L}_{z}$ are

#### Questions & Answers

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it good to use mercury because mercury does not wet glass and it does not evaporate easily
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0
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SFAR Sifar SIFAT -<SIFST
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albert
24÷6.022×10²³
albert
@Albert is wrong
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Aki
1.204×10^-22
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