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Chapter Review

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31-5   The Quantum Mechanical Hydrogen Atom

Aside from relativistic effects, the results from solving Schrodinger's equation for the hydrogen atom produces the best model of this atom that we have. From this analysis we now describe the hydrogen atom using four parameters called quantum numbers.

The principle quantum number, n

The principle quantum number takes on all integer values, n = 1, 2, 3, ..., and determines the total energy of a given state. This total energy is given by, approximately, the same result produced by Bohr's model:

.

The orbital angular momentum quantum number,

For an electron in a state of principle quantum number n, the orbital angular momentum of the electron can only have certain values as determined by the orbital angular momentum quantum number , which takes on the values

= 0, 1, 2, ...,(n - 1).

The magnitude of the angular momentum of an electron with a given value of is

.

Notice that an orbiting electron is allowed to have an angular momentum of zero.

The magnetic quantum number, m

When a hydrogen atom is in an external magnetic field, the allowed energy values of the electron, with specific values of n and , depend on an additional quantum number m. The range of values for this quantum number is from - to + in increments of 1:

m = -. - + 1, - +2, ... -1, 0, 1, ..., - 2, - 1, .

This magnetic quantum number specifies the component of the orbital angular momentum along a single direction; this direction is usually chosen to be the z-axis. This result is

.

One of the new consequences of the complete quantum theory is that only one component of the angular momentum can be known precisely, at a given time.

The electron spin quantum number, m_s

This last quantum number results from the fact that electrons contain intrinsic angular momentum that is a property of the electron itself (like its mass and charge). This type of angular momentum is called the spin angular momentum of the electron. The quantum number m_s takes on two possible values:

.

When the values of all four quantum numbers are specified we call that a particular state of the hydrogen atom. Notice that several different states can have the same energy. The solution of the Schrodinger equation suggests that, when the atom is in a particular state, we should think of the matter wave of the electron in terms of a three-dimensional probability cloud instead of moving in a specific path as suggested by Bohr's model. The probability cloud means that, at any given time, the electron has some finite probability of being anywhere around the nucleus.

Physlet Illustration: Hydrogen Atom Wavefunctions
n= l= m=
In this simulation, several views of the quantum mechanical hydrogen atom are shown.  The plot in the upper left is a polar plot showing how the wavefunction depends upon angle.  The plot in the upper right shows the radial (distance) dependence.  And the plot in the lower left illustrates the electron density.  The text boxes in the lower right allow the user to change the quantum numbers for this hydrogen atom.  How does the probability of finding the electron in a particular location depend on the quantum numbers?
Hints: What rules for picking quantum numbers must be followed? Which combinations of the quantum numbers results in spherically symmetric patterns?  Which produce lobes of various kinds? For quantum numbers which produce spherically symmetric distributions, what affect does the principle quantum number have on the radial distribution? For a fixed principle quantum number what affect does changing the angular momentum quantum and magnetic quantum numbers have?  You will want to use a sufficiently large principle quantum number to allow some experimentation.

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Example 31.3   The Ground State of Hydrogen:   For an electron in the ground state of hydrogen find (a) the total energy, and (b) the magnitude of the orbital angular momentum. (c) Since it is possible to specify the value of only one component of the orbital angular momentum, what is that possible value for this atom?

Picture the Problem   The picture shows the Bohr orbit around the nucleus of the electron in a hydrogen atom.

Strategy   We must deduce the values of the relevant quantum numbers and use the above expressions to calculate the desired quantities.

Solution

Part (a)

1. The ground state of hydrogen has Z = n = 1. So the total energy is:
Part (b)
1. The range of the orbital angular momentum quantum number is:
2. The magnitude of the orbital angular momentum then is:
Part (c)
1. The range of the magnetic quantum number is:
2. The "z-component" of the orbital angular momentum is:

Insight   In part (b), L = 0 may seem odd for an orbiting electron, but keep in mind that the electron spends time in all directions at all distances. In part (c) the word "z-component" is in quotes because it could really be any component. Since the atom is not in a magnetic field, no one component, that we can call "z," is picked out.

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Practice Quiz

How many possible values are there for when n = 4? 0 1 2 3 4

How many possible values are there for ms when n = 4? 0 1 2 3 4

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