Chapter 40
2. For a given quantum number n there are n possible values of , ranging from 0 to n 1. For each the number of possible electron states is N = 2(2 + 1). Thus the total number of possible electron states for a given n is
nn11 2 Nn N 2 2 1 2 n . 00
2 2 Thus, in this problem, the total number of electron states is Nn = 2n = 2(5) = 50.
6. For a given quantum number there are (2 + 1) different values of m . For each given the electron can also have two different spin orientations. Thus, the total number of electron states for a given is given by N = 2(2 + 1).
(a) Now = 3, so = 2(2 3 + 1) = 14.
(b) In this case, = 1, which means = 2(2 1 + 1) = 6.
(c) Here = 1, so = 2(2 1 + 1) = 6.
(d) Now = 0, so = 2(2 0 + 1) = 2.
28. For a given value of the principal quantum number n, there are n possible values of the orbital quantum number , ranging from 0 to n – 1. For any value of , there are
21 possible values of the magnetic quantum number m , ranging from to .
Finally, for each set of values of and m , there are two states, one corresponding to the 1 1 spin quantum number ms 2 and the other corresponding to ms 2 . Hence, the total number of states with principal quantum number n is
n1 N 2 (2 1). 0 Now nn11n 2 2 2 (n 1) n ( n 1), 002 since there are n terms in the sum and the average term is (n – 1)/2. Furthermore,
1515 1516 CHAPTER 40
n1 1. n 0 Thus N 2 nbn 1g n 2n2 .
30. When a helium atom is in its ground state, both of its electrons are in the 1s state. Thus, for each of the electrons, n = 1, = 0, and m = 0. One of the electrons is spin up 1 1 bms 2 g while the other is spin down bms 2 g. Thus,
(a) the quantum numbers (,,,)n m ms for the spin-up electron are (1,0,0,+1/2), and
(b) the quantum numbers for the spin-down electron are (1,0,0,1/2).
52. The energy of the laser pulse is
66 Ep P t (2.80 10 J/s)(0.500 10 s) 1.400 J .
Since the energy carried by each photon is
hc (6.63 1034 J s)(2.998 10 8 m/s) E 4.69 1019 J , 424 109 m the number of photons emitted in each pulse is
E 1.400J N p 3.0 1018 photons. E 4.69 1019 J
With each atom undergoing stimulated emission only once, the number of atoms contributed to the pulse is also 3.0 1018 .
55. (a) If t is the time interval over which the pulse is emitted, the length of the pulse is
L = ct = (3.00 108 m/s)(1.20 10– 11 s) = 3.60 10– 3 m.
(b) If Ep is the energy of the pulse, E is the energy of a single photon in the pulse, and N is the number of photons in the pulse, then Ep = NE. The energy of the pulse is
– 19 17 Ep = (0.150 J)/(1.602 10 J/eV) = 9.36 10 eV and the energy of a single photon is E = (1240 eV·nm)/(694.4 nm) = 1.786 eV. Hence, 1517
E 9.36 1017 eV N p 5.24 1017 photons. E 1.786 eV
65. (a) Using hc = 1240 eV·nm,
1 1 1 1 E hc 1240eV nm 2.13meV . 12 588.995nm 589.592nm
(b) From E = 2BB (see Fig. 40-10 and Eq. 40-18), we get
E 2.13103 eV B 18T . 5 2 B 2c5.788 10 eV Th
71. The principal quantum number n must be greater than 3. The magnetic quantum number m can have any of the values – 3, – 2, – 1, 0, +1, +2, or +3. The spin quantum 1 1 number can have either of the values 2 or 2 .