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Unformatted text preview: 1110 CHAPTER 41 ALL ABOUT ATOMS CHAPTER 41 Answer to Checkpoint Questions 1. 2. 3. 4. 7 (a) decrease; (b) { (c) same less A, C , B Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 0, 2, and 3 same number (10) 6p 2, 1, 0, and 1 (a) 2, 8; (b) 5, 50 (a) bromine; (b) rubidium; (c) hydrogen (a) n; (b) n and f all true (a), (c), (e), (f) (a) rubidium; (b) krypton (a) unchanged; (b) decrease; (c) decrease (a) 2; (b) 3 (a) and (b) CHAPTER 41 ALL ABOUT ATOMS 1111 14. In addition to the quantized energy, a helium atom has kinetic energy; its total energy can equal 20:66 eV Solutions to Exercises & Problems 1E 34 h = 2h = 6:62607542 10 J s = 1:0546 10 34 J s = 1::0546 10 J/eV = 6:59 10 16 eV s : 1 60 10 19 34 J s For a given quantum number l there are (2l + 1) dierent values of ml . For each given ml the electron can also have two dierent spin orientations. Thus the total number of electron states for a given l is given by Nl = 2(2l + 1). (a) Now l = 3 so Nl = 2(2 3 + 1) = 14. (b) Now l = 1 so Nl = 2(2 1 + 1) = 6. (c) Now l = 1 so Nl = 2(2 1 + 1) = 6. (d) Now l = 0 so Nl = 2(2 0 + 1) = 2. For a given quantum number n there are n possible values of l, ranging from 0 to n 1. For each l the number of possible electron states is Nl = 2(2l + 1) (see 2E). Thus the total number of possible electron states for a given n is
3E 2E Nn = n 1 X l=0 Nl = 2 n 1 X l=0 (2l + 1) = 2n2 : (a) Now n = 4 so Nn = 2(42 ) = 32. (a) Now n = 1 so Nn = 2(12 ) = 2. (a) Now n = 3 so Nn = 2(32 ) = 18. (a) Now n = 2 so Nn = 2(22 ) = 8. (a) Use Eq. 412: L = l(l + 1) h = 3(3 + 1) (1:055 10
p p
4E 34 J s) = 3:653 10 34 J s : 1112 CHAPTER 41 ALL ABOUT ATOMS (b) Use Eq. 415: Lz = ml h. For the maximum value of Lz set ml = l. Thus [Lz ]max = lh = 3(1:055 10
5E 34 J s) = 3:165 10 34 J s : (a) For a given value of the principal quantum number n, the orbital quantum number l ranges from 0 to n 1. For n = 3 there are 3 possible values: 0, 1, and 2. (b) For a given value of l, the magnetic quantum number ml ranges from l to +l. For l = 1, there are 3 possible values: 1, 0, and +1. The minimum angle min satises max =p 5 = 5: cos min = Lz,L = (ml h)max = p lh L 6 l(l + 1)h 5(5 + 1)) The angle is min = cos 1 ( 5=6) = 24:1 :
p r
6E n = 4; l = 3; ml = +3; +2; +1; 0; 1; 2; 3; ms = 1 . 2
8E 7E The principal quantum number n must be greater than 3. The magnetic quantum number ml can have any of the values 3, 2, 1, 0, +1, +2, or +3. The spin quantum number can have either of the values 1 or + 1 . 2 2 l = [ml ]max = 4; n = lmax + 1 l + 1 = 5; and ms = 1 . 2
The total number of electron states is Nn = 2n2 = 2(5)2 = 50:
11E 10E 9E (a) For n = 3 there are 3 possible values of l: 0, 1, and 2. (b) For each l there are 2l + 1 possible values of ml . Thus the number of possible m0l s for n = 3 is 3 1 X (2l + 1) = 32 = 9 :
l=0 CHAPTER 41 ALL ABOUT ATOMS 1113 (c) Regardless of the values of n, l and ml , for an electron there are always two possible values of ms : 1 . 2 (d) The population in the n = 3 shell is equal to the number of electron states in the shell, or 2n2 = 2(32 ) = 18. (e) Each subshell has its own value of l. Since there are three dierent values of l for n = 3, there are three subshells in the n = 3 shell. Use Lz = ml h to calculate the z component of the angular momentum, z = ml B p to calculate the z component of the magnetic dipole moment, and cos = ml = l(l + 1) to calculate the angle between the angular momentum vector and the z axis. Here h is the Planck constant divided by 2 (h = 1:055 10 34 J s) and B is the Bohr magneton (B = 9:274 10 24 J/T). Spin was ignored in computing the dipole moment. For l = 3 the magnetic quantum number ml can take on the values 3, 2, 1, 0, +1, +2, +3. The results are tabulated below.
ml 3 2 1 0 1 2 3 Lx (kg.m2/s) 3.16 x 1034 2.10 x 1034 1.05 x 1034 0 1.05 x 1034 2.10 x 1034 3.16 x 1034
p
12P x (J/T)
2.78 x 1023 1.85 x 10 23 9.27x 10 24 0 9.27 x 1024 1.85 x 1023 2.78 x 1023 (degrees)
150.0o 125.3o 106.8o 90.0o 73.2o 54.7o 30.0o kg The magnitude of L is l(l + 1) h = 3 4 (1:055 10 34 J s) = 3:655 10 34p m2 /s. p 24 J/T) 3 4 = The magnitude of the dipole moment is B l(l + 1) = (9:274 10 3:21 10 23 J/T. (a) The value of l satises l(l + 1)h l2 h = lh = L, so l ' L=h ' 3 1074 . 13P p p p 1114 CHAPTER 41 ALL ABOUT ATOMS (b) The number is 2l + 1 2(3 1074 ) = 6 1074 : (c) Since 1 m cos min = p l max h = p l 1 2l = 1 2(3 11074 ) l(l + 1)h l(l + 1)
2 or cos min ' 1 min =2 1 10 74 =6; we have min ' 10 74 =3 = 6 10 38 rad: The correspondance principle requires that all the quantum eects vanish as h ! 0. In this case h=L is extremely small so the quantization eects are barely existent, with min ' 38 rad ' 0. 10
14P p ml h to obtain Since L2 = L2 + L2 + L2 ,
x y z q L2 + L2 = L2 L2 . Replace L2 with l(l + 1)h2 and Lz with x y z
q p L2 + L2 x y = h l(l + 1) m2 : l
q q For a given value of l, the greatest that ml can be is l, so the smallest that L2 + L2 can x y p p 2 = h l. The smallest possible magnitude of ml is zero, so the largest be is h l(l + 1) l q p L2 + L2 can be is h l(l + 1). Thus x y h l L2 + L2 h l(l + 1) : x y p q p 15E The magnitude of the spin angular momentum is S = s(s + 1) h = ( 3=2)h, where s = 1 2 was used. The z component is either Sz = h=2 or h=2. If Sz = +h=2 the angle between the spin angular momentum vector and the positive z axis is p p = cos 1 Sz = cos S 1 1 p = 54:7 : 3 If Sz = h=2 the angle is = 180 54:7 = 125 .
16E The acceleration is where m is the mass of a silver atom, is its magnetic dipole moment, B is the magnetic eld, and is the angle between the dipole moment and the magnetic eld. Take the F )( a = m = ( cos m dB=dz) ; CHAPTER 41 ALL ABOUT ATOMS 1115 moment and the eld to be parallel (cos = 1) and use the data given in Sample Problem 411 to obtain 24 J/T)(1: 3 a = (9:27 10 1:8 10 254kg 10 T/m) = 7:2 104 m/s2 : (a)
17E which is in the short radio wave region. (a)
18E 24 E = 2B B = 2(9:27:60 1010 J/T)(0:50 T) = 58 eV : 1 19 J/eV) (b) From E = hf we get :27 10 24 f = E = 6963 10 34 JJs = 1:4 1010 Hz = 14 GHz : h : (c) The wavelength is c : 108 = f = 3100 1010m/s = 2:1 cm ; :4 Hz (b) The vertical displacement is F = B @B = (9:27 10 @z 24 J/T)(1:6 102 T/m) = 1:5 10 21 N : 1 at2 = 1 F l 2 x = 2 2 m v 2 1 1:5 10 21 N 0:80 m = 2 1:67 10 27 kg 1:2 105 m/s = 2:0 10 5 m : Let Eph = hf = hc= = 2B B ms and solve for : hc 1240 nm eV = 2 B m = 2(5:788 10 5 eV/T)(0:200 T)(1) = 5:35 cm : B s Let E = 2B Be and solve for Be : hc 1240 nm eV Be = 2E = 2 = 2(21 10 7 nm)(5:788 10 5 eV/T) = 51 mT : B B
20E 19E 1116 CHAPTER 41 ALL ABOUT ATOMS The total magnetic eld, B = Blocal + Bext ; satises E = hf = 2B . So hf B = (6:63 10 34 J s)(34 106 Hz) 0:78 T = 19 mT : Blocal = 2 ext 2(1:41 10 26 J/T)
22E 21E (a) 1 = (1240 eV nm) 588:995 nm 1 2 (b) From E = 2B B we get 1 E = hc 1 1 589:592 nm = 2:13 meV : E = 2:13 10 3 eV = 18 T : B = 2 B 2(5:788 10 5 eV/T) For a given shell with quantum number n the total number of available electron states is 2n2 . Thus for the rst four shells (n = 1 through 4) the number of available states are 2, 8, 18, and 32 (see Appendix G). Since 2 + 8 + 18 + 32 = 60 < 63, according to the \logical" sequence the rst four shells would be completely lled in an europium atom, leaving 63 60 = 3 electrons to partially occupy the n = 5 shell. Two of these three electrons would ll up the 5s subshell, leaving only one remaining electron in the only partially lled subshell (the 5p subshell). In chemical reactions this electron would have the tendency to be transferred to another element, leaving the remaining 62 electrons in chemically stable, completely lled subshells. This situation is very similar to the case of sodium, which also has only one electron in a partially lled shell (the 3s shell). The rst three shells (n = 1 through 3), which can accomodate a total of 2 + 8 + 18 = 28 electrons, are completely lled. For selenium (Z = 34) there are still 34 28 = 6 electrons left. Two of them go to the 4s subshell, leaving the remaining four in the highest occupied subshell, the 4p subshell. Similarly, for bromine (Z = 35) the highest occupied subshell is also the 4p subshell, which contains ve electrons; and for krypton (Z = 36) the highest occupied subshell is also the 4p subshell, which now accomodates six electrons; Without the spin degree of freedom the number of available electron states for each shell would be reduced by half. So the values of Z for the noble gas elements would be half as
25P 24P 23P CHAPTER 41 ALL ABOUT ATOMS 1117 what they are now, i.e., Z = 1, 5, 9, 18, 27, and 43. Of this set of numbers the only one which coincides with one of the numbers in the actual set (Z = 2, 10, 18, 36, 54, and 86) is 18. So argon would be the only one that would remain a nobel gas. When a helium atom is in its ground state both of its electrons are in the 1s state. Thus for each of them n = 1, l = 0, and ml = 0. One of the electrons is spin up (ms = + 1 ), 2 while the other is spin down (ms = 1 ). 2 (a) All states with principal quantum number n = 1 are lled. The next lowest states have n = 2. The orbital quantum number can have the values l = 0 or 1 and of these the l = 0 states have the lowest energy. The magnetic quantum number must be ml = 0 since this is the only possibility if l = 0. The spin quantum number can have either of the values ms = 1 or + 1 . Since there is no external magnetic eld the energies of these two states 2 2
27P 26P are the same. Thus in the ground state the quantum numbers of the third electron are either n = 2, l = 0, ml = 0, ms = 1 or n = 2, l = 0, ml = 0, ms = + 1 . 2 2 (b) The next lowest state in energy is an n = 2, l = 1 state. All n = 3 states are higher in energy. The magnetic quantum number can be ml = 1, 0, or +1; the spin quantum number can be ms = 1 or + 1 . If both external and internal magnetic elds can be 2 2 neglected, all these states have the same energy. (a) The number of dierent ml 0 s is 2l + 1 = 3 and the number of dierent ms 0 s is 2. Thus the number of combinations is N = (3 2)2 =2 = 18. (b) There are six states disallowed by the exclustion principle, in which both electrons share the quantum numbers (n; l; ml ; ms ) = (2; 1; 1; 1 ); (2; 1; 1; 1 ); (2; 1; 0; 1 ); (2; 1; 0; 1 ); 2 2 2 2 (2; 1; 1; 1 ); (2; 1; 1; 1 ): 2 2 For a given value of the principal quantum number n there are n possible values of the orbital quantum number l, ranging from 0 to n 1. For any value of l there are 2l + 1 possible values of the magnetic quantum number ml , ranging from l to +l. Finally, for each set of values of l and ml there are two states, one corresponding to the spin quantum number ms = 1 and the other corresponding to ms = + 1 . Hence the total number of 2 2 states with principal quantum number n is
29P 28P N =2 n 1 X
0 (2l + 1) : 1118 CHAPTER 41 ALL ABOUT ATOMS Now n 1 X
0 2l = 2 n 1 X
0 l = 2 n(n2 1) = n(n 1) ;
n 1 X
0 since there are n terms in the sum and the average term is (n 1)=2. Furthermore, 1 = n: Thus N = 2 [n(n 1) + n] = 2n2 : Let eV = hc=min , we get where V is measured in kV. Use eV = hc=min :
31E 30E nm hc min = eV = 1240eV eV = 1240 pm keV = 1240 pm ; eV V h = eV min = (1:60 10 c
32E 19 C)(40:0 103 eV)(31:1 10 12 m) 3:00 108 m/s = 6:63 10 34 J s : The kinetic energy of the electron is eV , where V is the accelerating potential. A photon with the minimum wavelength is produced when all of the electron's kinetic energy goes to a single photon in a bremsstrahlung event. Thus eV = hc = 1240 eV nm = 1:24 104 eV : 0:10 nm min The accelerating potential is V = 1:24 104 V = 12:4 kV. (a) and (b) Let the wavelength of the two photons be 1 and 2 = 1 + , then
33P hc hc eV = + + ; 1 1 CHAPTER 41 ALL ABOUT ATOMS 1119 2 1 = (=0 2)2 (=0 ) + 4 : = Here = 130 pm and 0 = hc=eV = 1240 keVpm=20 keV = 62 pm: Choose the plus sign in the expression for 1 (since 1 > 0) to obtain 2 + (130 1 = (130 pm=62 pm 2) 62 pm pm=62 pm) + 4 = 87 pm 2=
p or p and The energy of the electron after its rst deceleration is hc K = Ki = 20 keV 1240 keV pm = 5:7 keV: 87 pm 1 The energies of the two photons are hc E1 = = 1240 keV pm = 14 keV 87 pm 1 and hc E2 = = 1240 keV pm = 5:7 keV : 130 pm
2
34P 2 = 1 + = 87 pm + 130 pm = 2:2 102 pm : The initial kinetic energy of the electron is 50:0 keV. After the rst collision the kinetic energy is 25 keV, after the second it is 12:5 keV, and after the third it is zero. The energy of the photon produced in the rst collision is 50:0 keV 25:0 keV = 25:0 keV. The wavelength associated with the photon is 1240 nm = 25:0 eV 3 eV = 4:96 10 2 nm = 49:6 pm : 10 The energies of the photons produced in the second and third collisions are each 12:5 keV and their wavelengths are 1240 nm = 12:5 eV 3 eV = 9:92 10 2 nm = 99:2 pm : 10
35P Suppose an electron with total energy E and momentum p spontaneously changes into a photon. If energy is conserved, the energy of the photon is E and its momentum has 1120 CHAPTER 41 ALL ABOUT ATOMS magnitude E=c. Now the energy and momentum of the electronp related by E 2 = are 2 + (me c2 )2 , where me is the rest mass of the el;ectron. So pc = E 2 (me c2 )2 . Since (pc) the electron has nonzero mass E=c and p cannot have the same value. Hence momentum cannot be conserved. A third particle must participate in the interaction, primarily to conserve momentum. It does, however, carry o some energy. (a)
36E hc min = eV = 1240:keV pm = 24:8 pm : 50 0 keV (b) and (c) The values of for the K and K lines do not depend on the extenal potential
and are therefore unchanged.
37E (a) The cuto wavelength min is characteristic of the incident electrons, not of the target material. This wavelength is the wavelength of a photon with energy equal to the kinetic energy of an incident electron. Thus 1240 nm min = 35:0 eV 3 eV = 3:54 10 2 nm = 35:4 pm : 10 (b) A K photon results when an electron in a target atom jumps from the Lshell to the K shell. The energy of this photon is 25:51 keV 3:56 keV = 21:95 keV and its wavelength is K = (1240 eV nm)=(21:95 103 eV) = 5:65 10 2 nm = 56:5 pm. (c) A K photon results when an electron in a target atom jumps from the M shell to the K shell. The energy of this photon is 25:51 keV 0:53 keV = 24:98 keV and its wavelength is K = (1240 eV nm)=(24:98 103 eV) = 4:96 10 2 nm = 49:6 pm. For the K line from iron E = hc = 1240 pm keV = 6:42 keV : 193 pm For the hydrogen atom the corresponding energy dierence is E12 = (13:6 eV) 212 38E 1 = 10:2 eV: 11 The dierence is much greater in iron because its atomic nucleus contains 26 protons, exerting a much greater force on the K  and Lshell electrons than that provided by the singleproton nucleus of hydrogen. CHAPTER 41 ALL ABOUT ATOMS 1121 The energy dierence EL EM for the xray atomic energy levels of molybdenum is 39E hc E = EL EM = L hc = 1240 pm keV 1240 pm keV = 2:2 keV : M 63:0 pm 71:0 pm As shown in Sample Problem 415, the ratio of the wavelength Nb for the K line of niobium to the wavelength Ga for the K line of gallium is given by 40E Nb = ZGa 1 Ga ZNb 1 2 ; where ZNb is the atomic number of niobium (41) and the ZGa is the atomic number of gallium (31). Thus Nb = 30 2 = 9 : 40 16
Ga
41P The slope s of the Moseley plot can be found directly from Fig. 4117:
9 1=2 = 5:0 107 Hz 1=2 ; a = (1:95 0:50)10 Hz 40 11 which is in agreement with the result of Sample Problem 411. (a) An electron must be removed from the K shell, so that an electron from a higher energy shell can drop. This requires an energy of 69:5 keV. The accelerating potential must be at least 69:5 kV. (b) After it is accelerated the kinetic energy of the bombarding electron is 69:5 keV. The energy of a photon associated with the minimum wavelength is 69:5 keV, so its wavelength is 1240 nm min = 69:5 eV 3 eV = 1:78 10 2 nm = 17:8 pm : 10 (c) The energy of a photon associated with the K line is 69:5 keV 11:3 keV = 58:2 keV and its wavelength is K = (1240 eV nm)=(58:2 103 eV) = 2:13 10 2 nm = 21:3 pm. The energy of a photon associated with the K line is 69:5 keV 2:30 keV = 67:2 keV and its wavelength is K = (1240 eV nm)=(67:2 103 eV) = 1:85 10 2 nm = 18:5 pm.
42P 1122 CHAPTER 41 ALL ABOUT ATOMS (a) From Fig. 4114 we nd the wavelengths corresponding to K and K lines to be = 70:0 pm and = 6:30 pm, respectively. So E = hc= = (1240 keV pm)=(70:0 pm) = 17:7 keV and E = (1240 keV nm)=(63:0 pm) = 19:7 keV: (b) Both Zr and Nb can be used, since E < 18:00 eV < E and E < 18:99 eV < E . According to the hint given in the problem statement Zr is the better choice. Let 2d sin = m = mhc=E , where = 74:1 . Solve for d: (1)(1240 keV nm) d = 2mhc = 2(8:979 keV 0:951 keV)(sin 74:1 ) = 80:3 pm : E sin (a) According to Eq. 4119 f / (Z 1)2 , so f=f 0 = [(Z 1)=(Z 0 1)]2 : (b) E = f = Z 1 2 = 92 1 2 = 57:5 ( uranium vs aluminum) : E0 f 0 Z0 1 13 1 (c)
45P 44P 43P E = 92 1 E0 13 1 2 = 2070 ( uranium vs lithium) : (a) and (b) The theoretical value for the energy of the K xrays is given by me4 (Z 1)2 1 1 = (10:2 eV)(Z 1)2 : E = 82 h2 12 22 0 The theoretical vs measured values of E are listed in the table below.
element Li Be B C N O F Ne Na Mg Z 3 4 5 6 7 8 9 10 11 12 E theory (eV) 40.8 91.8 163.2 255 367.2 499.8 653.0 826.0 1020 1234 Emeasured (eV) 54.3 108.5 183.3 277 392.4 524.9 676.8 848.6 1041 1254 percentage deviation (%) 25 15 11 7.9 6.4 4.8 3.5 2.7 2.0 1.6 46P CHAPTER 41 ALL ABOUT ATOMS 1123 (c)
percentage deviation (%) 25 Z
3 The percentage deviation decreases as Z increases. The theory is more precise for large Z 0 s. (a) The length of the pulse's wave train is given by L = ct = (3:00 108 m/s)(10 10 15 s) = 3:0 10 6 m. Thus the number of wavelengths contained in the pulse is L = 3:0 10 6 m = 6:0 : N = 500 10 9 m (b) Solve for X from 10 fm=1 m = 1 s=X : 1 X = 10(1 s)(1 m)m = (10 10 15 )(3s15 107 s/y) = 3:2 106 y : 15 10 : According to Sample Problem 417 Nx =N0 = 1:3 10 38 . Let the number of moles of the lasing material needed be n, then N0 = nNA , where NA is the Avogadro constant. Also Nx = 10. Solve for n: 10 N n = (1:3 10x 38 ) N = (1:3 10 38 )(6:02 1023 /mol) = 1:3 1015 mol : A (a) The distance from the Earth to the Moon is dem = 3:82 108 m. Thus the times required is given by 8 m) t = 2dcem = 2(3:82 10m/s = 2:55 s : 3:00 108
49E 48E 47E 1124 CHAPTER 41 ALL ABOUT ATOMS (b) Denote the uncertainty in time measurement as t and let 2des = 15 cm. Then since dem / t, t=t = dem =dem . Solve for t: (2: :15 t = tdem = 2(355 s)(0108 m) = 5:0 10 10 s : dem :82 m) The number of atoms in a state with energy E is proportional to e E=kT , where T is the temperature on the Kelvin scale and k is the Boltzmann constant. Thus the ratio of the number of atoms in the thirteenth excited state to the number in the eleventh excited state is n13 = e E=kT ; n11 where E is the dierence in the energies: E = E13 E11 = 2(1:2 eV) = 2:4 eV. For the given temperature kT = (8:62 10 2 eV/K)(2000 K) = 0:1724 eV. Hence n13 = e 2:4=0:1724 = 9:0 10 7 : n
11
50E From Eq. 4121 n2 =n1 = e (E2 E1 )=kT . Solve for T : E T = k E2 n =n1 ) = (1:38 10 23 J/K) 3:2 eV 1015 =6:1 1013 ) = 10; 000 K : ln( 1 2 ln(2:5 Consider two levels, labeled 1 and 2, with E2 > E1 . Since T = jT j < 0, n2 = e (E2 E1 )=kT = e jE2 E1 j=( kjT j) = ejE2 E1 j=kjT j > 1 ; n
1
52E 51E i.e., n2 > n1 . This is population inversion. Solve for T : 2:26 eV E T = jT j = k E2 n =n1 ) = (8:62 10 5 eV/K) ln(1 + 10:0%) = 2:75 105 K : ln( 2 1 Let the power of the laser beam be P and the energy of each photon emitted be E . Then the rate of photon emission is 3 P 2: P R = E = hc= = (1240 eV nm=63238 10 :W 10 19 J/eV) = 7:3 1015 s 1 : : nm)(1 60 (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.
54E 53E CHAPTER 41 ALL ABOUT ATOMS 1125 (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 Ep = (0:150 J)=(1:602 10 19 J/eV) = 9:36 1017 eV and the energy of a single photon is E = (1240 eV nm=694:4 nm) = 1:786 eV. Hence N = Ep = 9:36: 10eV eV = 5:24 1017 photons : E 1 786
17 Similar to 53E, the rate is given by 55E P = P = (5:0 10 3 W)(0:80 103 nm) = 2:0 1016 s 1 : R = E hc= (1240 eV nm)(1:60 10 19 J/eV)
56E Let the range of frequency of the microwave be f . Then the number of channels that could be accomodated is 8 1 (650 nm) 1 ] N = 10f = (3:00 10 m/s)[(450 nm) = 2:1 107 : MHz 10 MHz The higher frequencies of the visible light would allow many more channels to be carried compared with using the microwave. Refer to the gure to the right. The diameter of the spot on the Moon is
57E D D = 2R tan + d 2R tan 2R sin 1:22 = 2R d 2(3:82 108 m)(1:22)(600 10 9 m) = 0:12 m = 4:7 103 m = 4:7 km : R d (a) If both mirrors are perfectly re
ecting there is a node at each end of the crystal. With one end partially silvered there is a node very close to that end. Assume nodes at both 58P 1126 CHAPTER 41 ALL ABOUT ATOMS ends, so there are an integer number of halfwavelengths in the length of the crystal. The wavelength in the crystal is c = =n, where is the wavelength in vacuum and n is the index of refraction of ruby. Thus N (=2n) = L, where N is the number of standing wave nodes, so N = 2nL = 2(1:75)(0:0600 m) = 3:03 105 : 694 10 9 m (b) Since = c=f , where f is the frequency, N = 2nLf=c and N = (2nL=c) f . Hence 108 m/s)(1) f = c2N = (3:0075)(0:0600 m) = 1:43 109 Hz : nL 2(1:
The speed of light in the crystal is c=n and the roundtrip distance is 2L, so the roundtrip travel time is 2nL=c. This is the same as the reciprocal of the change in frequency. (c) The frequency is f = c= = (3:00 108 m/s)=(694 10 9 m) = 4:32 1014 Hz and the fractional change in the frequency is f=f = (1:43 109 Hz)=(4:32 1014 Hz) = 3:31 10 6 . For the nth harmonic of the standing wave of wavelength in the cavity of width L we have n = 2L, so n + n = 0. Let n = 1 and use = 2L=n to obtain
59P jnj = = = (533 nm)2 = 1:8 10 jj = n n 2L 2(8:0 107 nm) 60P 12 m = 1:8 pm : (a) Denote the upper level as level 1 and the lower one as level 2. From n1 =n2 = e (E1 E2 )=kT we get n1 = n2 e (E1 E2 )=kT = n2 e hc=kT = (4:0 1020 )e (1240 eVnm)=[(580 nm)(8:6210 = 5:0 10 16 1 ;
so practically no electron occupies the upper level. (b) The energy released is
20 E = n1 Eph = n1hc = (3:0 10 )(1240 eV nm)(1:6 10 580 nm 5 eV/K)(300 K)] 19 J/eV) = 1:0 102 J : (a) 61P : R = 1:22f = (1:22)(3350 cm)(515 nm) = 7:33 m : d :00 mm CHAPTER 41 ALL ABOUT ATOMS 1127 (b) The average power ux density in the incident beam is P = 4(5:00 W) = 707 kW/m2 ; d2 =4 (3:00 mm)2
and the average power ux density in the central disk is (84%)P = (84%)(5:00 W) = 24:9 GW/m2 : R2 (7:33 m)2
62P (a) The intensity at the target is given by I = P=A, where P is the power output of the source and A is the area of the beam at the target. You want to compute I and compare the result with 108 W/m2 . The beam spreads because diraction occurs at the aperture of the laser. Consider the part of the beam that is within the central diraction maximum. The angular position of the edge is given by sin = 1:22=d, where is the wavelength and d is the diameter of the aperture. At the target, a distance D away, the radius of the beam is r = D tan . Since is small we may approximate both sin and tan by , in radians. Then r = D = 1:22D=d and 2 (5:0 106 W)(4:0 m)2 P I = r2 = (1:PdD)2 = 22 [(1:22(3000 103 m)(3:0 10 6 m)]2 = 2:1 105 W/m2 ; not great enough to destroy the missile. (b) Solve for the wavelength in terms of the intensity and substitute I = 1:0 108 W/m2 : P 5:0 106 W 4:0 m d = 1:22D I = 1:22(3000 103 m) (1:0 108 W/m2 ) = 1:4 10 7 m = 140 nm :
63 r s (a) In the lasing action the molecules are excited from energy level E0 to energy level E2 . Thus the wavelength of the sunlight that causes this excitation satises which gives E = E2 E0 = hc ; 1240 = E hc E = 0:289eV nm = 4:29 103 nm = 4:29 m : eV 0 2 0 1128 CHAPTER 41 ALL ABOUT ATOMS E1 . Thus the lasing wavelength 0 satises (b) Lasing occurs as electrons jump down from the higher energy level E2 to the lower level E 0 = E2 E1 = hc ; 0 which gives 1240 = E hc E = 0:289 eV eV0:nm eV = 1:00 104 nm = 10:0 m : 165 2 1 (c) Both and 0 belong to the infrared region of the electromagnetic spectrum.
64 (a) The energy dierence between the two states 1 and 2 was equal to the energy of the photon emitted. Since the photon frequency was f = 1666 MHz its energy was given by hf = (4:14 10 15 eV s)(1666 MHz) = 6:90 10 6 eV. Thus E2 E1 = hf = 6:90 10 6 eV = 6:90 eV :
(b) The emission was in the radiowave region of the electromagnetic spectrum. ...
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.
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