Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Question: Early on, the lanthanides were found to be quite uncooperative when attempts were made to chemically separate them from one another. One reason can be seen in Figure 8.16. Explain.

Here we consider adding two electrons to two "atoms," represented as finite wells. and investigate when the exclusion principle must be taken into account. In the accompanying figure, diagram (a) shows the four lowest-energy wave functions for a double finite well that represents atoms close together. To yield the lowest energy. the first electron added to this system must have wave function Aand is shared equally between the atoms. The second would al so have function Aand be equally shared. but it would have to be opposite spin. A third would have function B. Now consider atoms far a part diagram(b) shows, the bumps do not extend much beyond the atoms - they don't overlap-and functions Aand Bapproach equal energy, as do functions Cand D. Wave functionsAandBin diagram (b) describe essentially identical shapes in the right well. while being opposite in the left well. Because they are of equal energy. sums or differences ofandare now a valid alternative. An electron in a sum or difference would have the same energy as in either alone, so it would be just as "happy" inrole="math" localid="1659956864834" A,B,A+B, orA- B. Argue that in this spread-out situation, electrons can be put in one atom without violating the exclusion principle. no matter what states electrons occupy in the other atom.

Whether adding spins to get total spin, spin and orbit to get total angular momentum, or total angular momenta to get a "grand total" angular momentum, addition rules are always the same: Given J1=โ„j1(j1+1)andJ2=โ„j2(j2+1) . Where is an angular momentum (orbital. spin. or total) and a quantum number. the total isJT=โ„jT(jT+1) , where jTmay take on any value between |j1-j2|and j1+j2in integral steps: and for each value ofJฮ“JTz=mฯ€f . where mฯ€may take on any of2jr+I possible values in integral steps from-jT for +jTSince separately there would be 2j1+1possible values form11 and2j2+ I formฯ2 . the total number of stales should be(2j1+1)(2j2+1) . Prove it: that is, show that the sum of the2jT+1 values formit over all the allowed values forj7 is (2j1+1)(2j2+1). (Note: Here we prove in general what we verified in Example 8.5for the specialcase j1=3,j2=12.)

Were it to follow the standard pattern, what would be the electronic configuration of element 119.

Compare and contrast the angular momentum and magnetic moment related to orbital motion with those that are intrinsic.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free