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Chapter 8: Q12CQ (page 338)

Question: Solving (or attempting to solve!) a 4-electron problem is not twice as hard as solving a 2-electrons problem. Would you guess it to be more or less than twice as hard? Why?

Short Answer

Expert verified

Answer

Because there is an exact analytic solution for the two-electron problem, but not for the four-electron problem.

Step by step solution

01

Explanation.

A four-electron problem that involves only electrons and no nucleus would be more than twice as hard to solve as a two-electron problem" because there is an exact analytic solution for me two-electron problem, but not for the four-electron problem.

02

Reason.

However, what the question is probably asking about is finding the wave functions and energies for an atom with two electrons and an atom with four electrons. The two-electron problem is a three-body problem (with the nucleus as the third body).

The four-electron problem would be less man two times as difficult as the two-electron problem. This is because neither problem can be solved in analytic form like the hydrogen atom, but many of the approximations used in the approximate solution of the two-electron problem, such as regarding each electron to move in an average field of the other charges, would apply also in some form to the four-electron problem.

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Most popular questions from this chapter

Two particles in a box have a total energy 5π2ℏ2/2mL2.

(a) Which states are occupied?

(b) Make a sketch ofP5(x1,x2)versusx1for points along the linex2=x1.

(c) Make a similar sketch ofPA(x1,x2).

(d) Repeat parts (b) and (c) but for points on the linex2=L−x1

Imagine two indistinguishable particles that share an attraction. All other things being equal, would you expect their multiparticle spatial state to be symmetric, ant symmetric, or neither? Explain.

The 21cm Line: One of the most important windows to the mysteries of the cosmos is the 21cm line. With it astronomers map hydrogen throughout the universe. An important trait is that it involves a highly forbidden transition that is, accordingly, quite long-lived. But it is also an excellent example of the coupling of angular momentum. Hydrogen's ground state has no spin-orbit interaction—forl=0there is no orbit. However, the proton and electron magnetic moments do interact. Consider the following simple model.

(a) The proton seesitself surrounded by a spherically symmetric cloud of 1s electron, which has an intrinsic magnetic dipole moment/spin that of course, has a direction. For the purpose of investigating its effect the proton, treat this dispersed magnetic moment as behaving effectively like a single loop of current whose radius isa0then find the magnetic field at the middle of the loop in terms of e,ℏ,me , μ0anda0.

(b) The proton sits right in the middle of the electron's magnetic moment. Like the electron the proton is a spin−12particle, with only two possible orientations in a magnetic field. Noting however, that its spin and magnetic moment are parallel rather than opposite, would the interaction energy be lower with the proton's spin aligned or anti-aligned with that of the electron?

(c) For the protongρ.is 5.6. Obtain a rough value for the energy difference between the two orientations.

(d) What would be the wavelength of a photon that carries away this energy difference?

Figureshows the Stern-Gerlach apparatus. It reveals that spin-12particles have just two possible spin states. Assume that when these two beams are separated inside the channel (though still near its centreline). we can choose to block one or the other for study. Now a second such apparatus is added after the first. Their channels are aligned. But the second one is rotated about the-axis by an angle \(\phi\) from the first. Suppose we block the spin-down beam in the first apparatus, allowing only the spin-up beam into the second. There is no wave function for spin. but we can still talk of a probability amplitude, which we square to give a probability. After the first apparatus' spin-up beam passes through the second apparatus, the probability amplitude iscos(ϕ/2)↑2nd+sin(ϕ/2)↓2ndwhere the arrows indicate the two possible findings for spin in the second apparatus.

(a) What is the probability of finding the particle spin up in the second apparatus? Of finding it spin down? Argue that these probabilities make sense individually for representative values ofϕand their sum is also sensible.

(b) By contrasting this spin probability amplitude with a spatial probability amplitude. Such asψ(x)=Ae−te2. Argue that although the arbitrariness ofϕgives the spin cases an infinite number of solves. it is still justified to refer to it as a "two-state system," while the spatial case is an infinite-state system.

Determine the electronic configuration for phosphorus, germanium and cesium.
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