Lecture 16 Hydrogen Atom. PHYS 302: Modern Physics
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1 Lecture 16 Hydrogen Atom PHYS 302: Modern Physics
2 Summary of Central Force Analysis Starting from the assumption of a central force potential U(r) we first separated the stationary- state wavefunction into spherical coordinate components: The spherical form of the Laplacian was then inserted into the (time- independent) Schrodinger equation, which after some tidying up gave us: We then do some more work on this to get all the phi terms on one side: Now we equate the LHS to a constant, which we call, and so get: which is easy to solve:
3 Summary (cont.) Equating the RHS to gives (after more rearranging): and we have again separated the variables. We call the constant of separation, and when we equate this to the RHS we get: solutions exist only if is a nonegative integer AND has absolute value that is not larger than. Solutions are associated Legendre polynomials:
4 Summary (cont.) The angular part of the wavefunction is and if we combine the two solutions we got so far, we get the spherical harmonics where:
5 Summary (cont.) The separation constants relate to sharp observables central forces. We do not derive these relations: for Note that the limitation on values corresponds to the fact that the z component of L cannot be greater than the magnitude of L. Notice, also, that the angular wavefunctions are not connected to E or U. Only the radial function R(r) contains E and U. To find R(r) we return to the last equation and equate LHS to to get: This is the radial wave equation. It contains orbital quantum number so different give different energies. But it does not contain the magnetic quantum number so different values are all same energy states.
6 Space Quantization The orbital and magnetic quantum numbers and specify the magnitude of the angular momentum vector and its z component. Recall: The fact that the direction of L is quantized with respect to an arbitrary axis is referred to as space quantization. For a given value of corresponding to a different value., the L vector is constrained to certain cones, each It is as if the L vector is precessing around the z axis. (not really, though!).
7 L and L z make a right triangle, so the angle between them (θ) can be found using: Space Quantization (cont.) Consider an atomic electron in the l = 3 state. Calculate the magnitude of the total angular momentum L and the allowed values of L z and θ. Solution: Use the equation with l = 3. Thus: Allowed values of L z are given by where = 0, ±1, ±2, ±3 so Now we can compute cosθ from the above equation: Inserting allowed values for gives:
8 Atomic Hydrogen Now try to include an actual central force: the coulomb attraction between electron and one- proton nucleus. We will look at atomic number Z=1 (hydrogen) as well as singly ionized Helium He +, and doubly ionized Lithium, Li 2+. All of these have one electron, so they are "hydrogen- like". Assume that nucleus is fixed (because M>>m). The potential for this force is: Previous analysis tells us that because this is a central force, the stationary states are: where is a spherical harmonic. The one thing left to find is the radial wave function, which is a solution of:
9 Solving the Radial Wave Function This looks a bit like kinetic energy + potential energy = total energy. The term with in it is orbital contribution to K.E. Explanation. If particle is moving in circular orbit, all K.E. is orbital. In that case And we can combine these to eliminate v: But so this is the the above term. The derivative terms are radial contribution to K.E. The first one is K.E. for matter wave traveling in r direction - - recall one of the operators derived in earlier lecture:
10 Now what about the term? If we look hard enough, we notice that if the effective one- dimensional matter wave was not but then everything would make sense, because: So if we call the "effective" radial matter wave g(r) = rr(r), and if we also create an "effective" potential energy function U eff that is the sum of the real potential energy and the orbital kinetic energy: Then we get a kind of 1- dimensional Schrodinger equation:
11 Solving Schrodinger Equation for H- like atom Deriving the solution is beyond our scope, so we just write it: For solutions only exist for E values that satisfy: where and is the Bohr radius Å is the Rydberg energy, 13.6 ev The integer n is the principal quantum number. n can have any value, but l is now restricted to This makes sense: for a given energy L cannot become arbitrarily large.
12 Radial Wave Solution What is the form of the R(r) solutions? R depends on n and l so the solutions are labeled R nl (r) All states with same n are said to form a shell. Shells are labeled: K, L, M (n = 1, 2, 3 ) States with same n and l are said to form a subshell. Subshells are labeled s, p, d, f, (l = 0, 1, 2, 3, ) This is called spectroscopic notation.
13 These are some of the energy levels for atomic hydrogen: Electron Transitions When electron drops a photon can carry away the energy However, not all transitions can be optical transitions. Photons also carry angular momentum, so to conserve total angular momentum, the ang. mom. of the electron must change by one unit: This is a selection rule that must be obeyed in optical transitions. So, 3p 1s and 2p 1s are allowed but 3p 2p is forbidden.
14 Homework Homework is from chapter 8:
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