## Physics 334 Modern Physics

text “Modern Physics” by Tipler and Llewellyn. Others images are from a variety ... PowerPoint file. The PDF handouts are intended for my Modern Physics class.

Physics 334 Modern Physics

Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on the textbook “Modern Physics” by Thornton and Rex. I have replaced some images from the adopted text “Modern Physics” by Tipler and Llewellyn. Others images are from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc) and were part of original lectures. Contributions are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class.

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CHAPTER 7

Atomic Physics 7.1 Schrödinger Equation in 3D 7.2 Quantization of Angular Momentum and Energy in the Hydrogen Atom 7.3 Hydrogen Atom Wave Function 7.4 Electron Spin 7.5 Total Angular Momentum and the SpinOrbit Effect 7.6 The Schrödinger Equation for Two (or more) Particles 7.7 Ground States of Atoms: The Periodic Table 7.8 Excited States of Spectra and Atoms

Werner Heisenberg (1901-1976)

The atom of modern physics can be symbolized only through a partial differential equation in an abstract space of many dimensions. All its qualities are inferential; no material properties can be directly attributed to it. An understanding of the atomic world in that primary sensuous fashion…is impossible. - Werner Heisenberg

http://dir.niehs.nih.gov/ethics/images/photo-heisenberg.jpg

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7.1: The Schrödinger Equation in Three Dimension The wave function must be a function of all three spatial coordinates. Now consider momentum as an operator acting on the wave function. In this case, the operator must act twice on each dimension. Given:

So the three-dimensional Schrödinger wave equation is

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The 3D infinite potential well It’s easy to show that:

ψ ( x, y, z ) = A sin(k x x) sin(k y y) sin(k z z ) where:

k x = π nx / Lx k y = π n y / Ly k z = π nz / Lz

and:

π 2  2  nx2

nz2  E=  + +  2m  L2x L2y L2z  n y2

When the box is a cube:

E=

π 2 2 2mL2

(n

2 x

2 y

+n +n

2 z

)

Try 10, 4, 3 and 8, 6, 5

Note that more than one wave function can have the same energy.

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Degeneracy The Schrödinger wave equation in three dimensions introduces three quantum numbers that quantize the energy. And the same energy can be obtained by different sets of quantum numbers.

A quantum state is called degenerate when there is more than one wave function for a given energy.

Degeneracy results from particular properties of the potential energy function that describes the system. A perturbation of the potential energy can remove the degeneracy.

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Schrödinger Equation in Spherical Coordinates The potential energy of the electron-proton system is electrostatic:

Use the three-dimensional time-independent Schrödinger Equation.

For Hydrogen-like atoms (He+ or Li++) Replace e2 with Ze2 (Z is the atomic number). Replace m with the reduced mass, µ.

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Spherical Coordinates The potential (central force) V(r) depends on the distance r between the proton and electron. Transform to spherical polar coordinates because of the radial symmetry.

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The Schrödinger Equation in Spherical Coordinates

Transformed into spherical coordinates, the Schrödinger equation becomes:

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7-2 Quantization of Angular Momentum and Energy in the Hydrogen Atom The wave function ψ is a function of r, θ, φ. This is a potentially complicated function. Assume instead that ψ is separable, that is, a product of three functions, each of one variable only:

This would make life much simpler, and it turns out to work.

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Solution of the Schrödinger Equation for Hydrogen Separate the resulting equation into three equations: R(r), f(θ), and g(φ). The derivatives:

Substitute:

Multiply both sides by r2 sin2 θ / R f g:

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Solution of the Schrödinger Equation for H

r and θ appear only on the left side and φ appears only on the right side The left side of the equation cannot change as φ changes. The right side cannot change with either r or θ. Each side needs to be equal to a constant for the equation to be true. Set the constant to be −mℓ2

azimuthal equation

It is convenient to choose the solution to be

.

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Solution of the Schrödinger Equation for H satisfies the azimuthal equation for any value of mℓ. The solution must be single valued to be a valid solution for any φ:

mℓ must be an integer (positive or negative) for this to be true. Now set the left side equal to −mℓ2 and rearrange it [divide by sin2(θ)].

Now, the left side depends only on r, and the right side depends only on θ. We can use the same trick again!

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Solution of the Schrödinger Equation for H

Set each side equal to the constant ℓ(ℓ + 1).

Angular equation

We’ve separated the Schrödinger equation into three ordinary secondorder differential equations, each containing only one variable.

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Solution of the Radial Equation for H The radial equation is called the associated Laguerre equation and the solutions R are called associated Laguerre functions. There are infinitely many of them, for values of n = 1, 2, 3, … Assume that the ground state has n = 1 and ℓ = 0. Let’s find this solution. The radial equation becomes:

The derivative of

yields two terms.

This yields:

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Try a solution Equation for A is a normalization constant. a0 is a constant with the dimension of length. Take derivatives of R and insert them into the radial equation.

⇒ To satisfy this equation for any r, both expressions in parentheses must be zero. Set the second expression equal to zero and solve for a0: Set the first expression equal to zero and solve for E: Both are equal to the Bohr results!

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Principal Quantum Number n There are many solutions to the radial wave equation, one for each positive integer, n. The result for the quantized energy is:

A negative energy means that the electron and proton are bound together.

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Quantum Numbers The three quantum numbers: n: Principal quantum number ℓ: Orbital angular momentum quantum number mℓ: Magnetic (azimuthal) quantum number

The restrictions for the quantum numbers: n = 1, 2, 3, 4, . . . ℓ = 0, 1, 2, 3, . . . , n − 1 mℓ = −ℓ, −ℓ + 1, . . . , 0, 1, . . . , ℓ − 1, ℓ

Equivalently: n>0 ℓ