Ch. 42----2, 3, 4, 5, 8, 9, 10, 12, 13, 15, 18, 19, 22, 24, 26, 27, 28, 29, 30, 32
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SAMPLE EXAMS  #2
REAL EXAM 3
REAL EXAM 4
2. (a) Find U.  See example 42.1.
(b) See page 1434 about the additional "net investment " in pulling molecule apart and returning electron to the sodium (Na) , which "donated" it to Cl when the two elements formed an ionic bond. Total binding energy is 
|U + 3.5 eV - 4.3 eV |; see page 1434. This is how much work you have to do to pull the molecule  apart and separate the neutral atoms. - 
3. This transition is not consistent  with photon selection rules.  See Figure 42.5.  5*hbar2/I = hc/lambda; find I and find inter-atomic distance from I formula.
4. This transition is more consistent  with photon selection rules.  hc/lambda = hbar2/I . Oven frequency should  be near f you found  for maximum absorption of heat.
5. (a) Eqn 42.5 (b) They should agree.
8. This transition is consistent  with photon selection rules: change in n = 1. See Eqn 42.8. Set hc/lambda = expression for change in E. Find k'.
9.(a)  4*hbar2/I  Compute I to complete  problem.  (b) hc/lambda = 4*hbar2/I  .
10.  (i) 2*hbar2/I  = hc/lambda = 8.841x10 -4 eV  and   (ii) hc/lambda' = 0.2560 eV =  expression for change in E. Find k'.

From (i) get mass & use in (ii) for k'.

12.  Now E = E(l, n) is a function of two variables, l and n,  given by equation 42.9. (a) Set hc/lambda =  E(1, 2)  - E(0,1)   Use similar substitutions for part (b) and (c).
13. (a) Equation 42.8. (b) Equation 42.8. (c ) E = E(n). Set hc/lambda =  E(n + 1)  - E(n) = right side of equation 42. 8 
15. Use simple cubic: (atom mass/L3,  where  L is given. Use average of Na  and Cl atom  for atom  mass
18. (a), (b) , (c) based on   hc/lambda = Eg.
19. hc/lambda =  n*(1.12 eV); find n.
22. Review Ch. 18, equation 18.19,  to get vrms then compare with  v in example 42.9.
24. Eliminate nrs by taking the square root of both sides of 42.13. Solve for nrs  in terms of E. The substitute into equation 42.12 to get 42.14..
26. EF is derived on page 1450. It is given by the total number of electrons N .  On page 1450 we integrate 42.18 to get N at T = 0 and   get the fermi energy EF0  at absolute zero.

(a) Now that you know EF0    is use it to evaluate  equation 42.22.
(b) Use classic definition (1/2)mv2  for kinetic energy to get v in terms of Eav .

(c) Set kT = EF.

27. Evaluate equation 42.17. Remember T is in Kelvin.
DISCUSSIONS TO COME