Ch. 41----4, 8, 14, 16, 17, 18, 19, 20, 21, 25, 27, 28, 30, 32, 33
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SAMPLE EXAMS  #2
REAL EXAM 3
REAL EXAM 4
DISCUSSION
4.(a) ml = - l, -(l-1), ...., -1,0,1, ....., (l-1), l    Note: Lz -= ml*hbar  . (b) Equations 41. 4 and 41.5.   (c) See example 41.2.
8. Note the integral from r = 0 to r = infinity is ONE  for the normalized wave function. 
(a)  Example 41.3 for guidance. Integration needed but you may use  the  indefinite integral formula or  perform yourself using integration  by parts. Evaluate the indefinite integral formula  at r = 0 and r = a/2 and compute the difference  to get definite integral.
(b) It should be clear the integral from 0 to a/2 + integral  from a/2 to a equals 0.323. Thus,
(integral  from a/2 to a)= 0.323 - (integral from 0 to a/2).
14. (a) d state means  l =  2 and ml = - l, -(l-1), ...., -1,0,1, ....., (l-1), l. See figure 41.0.  For each value of ml,  you add UB = ml*uB*B (formula 41.18) to the energy for various ml values, where uB = e*hbar/(2m).  Note: When  ml > 0 , the magnetic dipole moment  has a z-component opposite  the upward magnetic field since  the electron is negative. Thus UB is positive when ml > 0 using the definition of UB for  dipole moments in Ch. 27. 
(b) See previous comments; the lowest energy is for a specific value of   ml  < 0, i.e. , when the z -component of the dipole moment points in the same direction as the magnetic field 
(c)  See figures in section 41.2 referenced above.

Note the selection rules: In  transitions, l  must change by 1 and    ml changes by  - l,,0, or 1. See Figure 41.11 and the following problem.

16. Note the selection rules: In  all  Zeeman effect  transitions, l  must change by 1 and    ml changes by  - l,,0, or 1. See Figure 41.11 only as a guide. Note in that figure: l  must change by 1 and    ml changes by  - l,,0, or 1.

(a) In this current situation , the starting value of l is 1 with ml = -1, 0, 1and  the final l is 0 with ml = 0. FInd the wavelength of photon for   transitions from  l = 1 with  ml = -1, 0 or 1 to l = 0 with ml = 0. THERE ARE THREE TRANSITIONS; one of them ( from l = 1,  ml  = 0 to  l = 0,  ml = 0) will have   wavelength  = 122 nm, the same as without a magnetic field.
(b) See comment that   one of the transitions( from l = 1,  ml  = 0 to  l = 0,  ml = 0) will have   wavelength  = 122 nm, the same as without a magnetic field.
(c)  and (d) : See figure 41.11 as a guide only; The largest energy changes have the smallest wavelength and the largest changes have the smallest lambda.
17. 3p means n = 3 and l =1; for

 l = 1,    ml = - l,,0,1..
(a) Find B such that level difference uB*B = 0.0000271 eV.
(b) THREE by inspection, each separated from one another by uB*B,  where uB = e*hbar/(2m). 

18. (a) See example 41.5. FIND -uz*B, ms = -1/2. Note U = -uz*B, where uz is given by formula 41.22. Note the negative sign.  When  ms > 0 , the magnetic dipole moment  has a z-component opposite  the upward magnetic field since  the electron is negative. Thus UB is positive when ms > 0 using the definition of  UB for  dipole moments in Ch. 27.  In this case,  UB is negative  since ms < 0.
(b) Since, n = 1, l = 0, and we have an s state in which case ml  = 0, and thus   no orbital magnetic interaction splittings given by formula  41.18.  
19. Find the difference between -uz*B  for  ms = -1/2 and  -uz*B  for ms = 1/2. NOTE:  BECAUSE THE MAGNETIC FIELD IS IN THE NEGATIVE  DIRECTION, UB is positive when ms < 0 using the definition of  UB for  dipole moments in Ch. 27. UB is NEGATIVE  when ms >  0.
20. See Formula 41. 24. Note: j = | l + 1/2| or j =   | l - 1/2|  . For n = 3, l = 2, 1, or 0. As you can see , for each of  three possible l values there are two  possibilities for   j  and thus 6 combinations of j and l.
21.  Read page 1417 carefully. Read about the notation 2P1/2.  This applies when l = 1, and thus j is either 1/2 or 3/2; Notations 2P1/2 means l =1 (hence we use upper case "P"), there are 2 states of spin,  ms = -1/2 or ms = 1/2 (hence the superscript "2"); and j = l - 1/2 = 1 - 1/2 = 1/2.  In the case of this problem, j = l - 1/2 = 7/2 or j = l + 1/2 = 9/2. Both of  these equations give the same result for l, from which you can label the states s, p, d, f, g, , h, etc. Which letter is it?   
25. Ok to refer to Table 41.2 or 41.3, despite what the book says. Neon = 1s22s22p6.  See my notes and ICQ  from FRIDAY 4-27-12 when we wrote the spectrographic notation for NEON and also sketched the diagram showing the filled orbitals. To see a similar diagram, go here and scroll down to Neon: http://www.brazosport.edu/sites/CurrentStudents/Faculty/JudyChu/Tutorials/elecconf3.htm .
At the web page,  we see the 10 electrons shown with spin up or down as they occupy the lowest possible  energy states. Neon is thus said to be in it "ground state." For each of the 10 electrons write the value of n, l, ml and ms. .
27. Ionization energy = 0 -( -Zeff2*13.6 eV/n2), where n and  Zeff are given.  
28.  This problem expands comments in book  asserting energy can depend on n and l rather than only n as with hydrogen---see pages 1418 and 1421,  under  "Central Field Approximation" and "Screening, " respectively.

When you visit example 41.8, you see the results of the energy's  dependence on n and l.
First off, visit Table 41.53. Potassium (K) has 16 electrons;,  see  potassium's row: 1s22s22p6 3s23p6 4s. Note 4s fills before  3d (for l = 2) ; the additional electron goes into the 4s energy state rather than 3d state because 4s  has a lower energy. See comments page 1421 under  "The Periodic Table." For reinforcement, see figure 41.4 and note 4s radial function has two peaks closer to  nucleus than a  third outer peak. The inner most peak is closer to nucleus than  3d's spike, hence  electrons spend some time closer to center, reducing the  energy.

Solve:  energy =  -Zeff2*13.6 eV/n2) for  Zeff in each of the three cases---4s, 4p and 4d.  Does  Zeff increase with the magnitude of the energy?   Does it show that as the electron spends a bit more time within the inner shells, screening is lower and Zeff is larger? Note as the angular  momentum index  increases (from s to p to d) , energy magnitude decreases.
30. (a) and (b), solve:   energy =  -Zeff2*13.6 eV/n2) for  Zeff
(c) Compare energy of the sodium 3s state from example 41.7 with  those of 2s and  4s states from parts (a) and (b), respectively. How does Zeff vary with atomic number? Explain
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