This preview shows page 1. Sign up to view the full content.
Unformatted text preview: CHAPTER 40 MORE ABOUT MATTER WAVES 1091 CHAPTER 40 Answer to Checkpoint Questions 1. 2. 3. 4. 5. (b), (a), (c) (a) all tie; (b) a, b, c (a), (b), (c), (d) (a) n = 1; (b) n = 3, n = 2, n = 1 (a) 5; (b) 7 Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. (a) 1=4; (b) same factor (a), (c), (b) (c) (a) 18; (b) 17 p p p (a) ( 1=L) sin(=2L)x; (b) ( 4=L) sin(2=L)x; (c) ( 2=L) cos(=L)x 12 eV (4 ! 2 in A) matches 1 ! 2 in C ; 9 eV (5 ! 4 in A) matches 1 ! 2 in D; 24 eV (5 ! 1 in A) matches 1 ! 3 in D; 15 eV (4 ! 1 in A) matches 1 ! 2 in E less equal (a) wider; (b) deeper (a) 3; (b) 4 n = 1, n = 2, n = 3 n = 3, n = 2, n = 1 (a) greater; (b) less; (c) less (b), (c), (d) same 1092 CHAPTER 40 MORE ABOUT MATTER WAVES 16. 17. (a) rst Lyman plus rst Balmer; (b) Lyman series minus Paschen series limit (a) n = 3; (b) n = 1; (c) n = 5 Solutions to Exercises & Problems 0 According to Eq. 404 En / L 2 . Thus the new energy level En satises 0 En = L0 En L 1E 2 which gives L0 = 2L, i.e., the width of the potential well must be multiplied by a factor p of 2.
2E p L = L0 2 1 = 2; (a) Use Eq. 404. For an electron h2 ( 2 (1240 eV nm)2 E1e = 8m L2 = 8(mhc) )L2 = 8(0:511 MeV)(0:100 nm)2 = 37:7 eV : e e c2
(b) For a proton
2 ( 2 (1240 eV E1p = 8(mhc) )L2 = 8(938 MeV)(0:nm) nm)2 = 2:05 10 2 eV : 100 p c2 3E For n = 3; let En = n2 h2 =(8mL2 ) = 4:7 eV and solve for L: L = p nh = p 3(1240 eV nm) = 0:85 nm : 8mEn 8(0:511 MeV)(4:7 eV)
2 4E (a) Use Eq. 404, with n = 1 and L equal to the atomic diameter, to estimate the energy: h2 10 34 2 E = n2 8mL2 = 8(9:11 (6:6331 kg)(1:4Js) 10 10 14 m)2 = 3:1 10 10 J = 1:9 103 MeV : CHAPTER 40 MORE ABOUT MATTER WAVES 1093 (b) The available binding energy is only a few MeV, so we cannot expect to nd an electron bound inside a nucleus.
5E Since En / L 2 , as L is doubled E1 becomes (2:6 eV)(2) 2 = 0:65 eV. The energy to be absorbed is
2 2) 2 )2 E = E4 E1 = (4 8m 1L2h = 3(15(hc)L2 me c2 e 2 15(1240 eV = 8(0:511 MeV)(0:nm)nm)2 = 90:3 eV : 250 6E By denition the probability density in a onedimensional potential distribution is the probability per unit length of nding a particle in a certain location. Its unit is therefore that of probility divided by length, or m 1 in SI units. Set n = 1 in Eq. 404 to obtain the zero point energy E1 :
8E 7E 10 h2 626 10 34 J 2 E1 = 8m L2 = (6:8(1:67 10 27 s) (1:60 10 kg)(100 p 19 eV/J) 12 m)2 = 2:08 10 2 eV : (An alternative, and perhaps easier, way to obtain this result is to use hc = 1240 eV nm and mp = 931 MeV=c2 .)
9E The allowed energy values are given by En = n2 h2 =8mL2 . The dierence in energy between the states with quantum numbers n and (n + 1) is + 1) h E = En+1 En = (n + 1)2 n2 8mL2 = (2n mL2 h 8 2 2 E = (2n + 1)h2 8mL2 = 2n + 1 : E 8mL2 n2 h2 n2 As n becomes large 2n + 1 ! 2n and (2n + 1)=n2 ! 2n=n2 = 2=n. and 1094 CHAPTER 40 MORE ABOUT MATTER WAVES (a) Let the quantum numbers of the pair in question be n and n + 1, respectively. Then En+1 En = E1 (n + 1)2 E1 n2 = (2n + 1)E1 . Let 10P En+1 En = (2n + 1)E1 = 3(E4 E3 ) = 3(42 E1 32 E1 ) = 21E1 ;
we get 2n + 1 = 21, or n = 10. (b) Now let En+1 En = (2n + 1)E1 = 2(E4 E3 ) = 2(42 E1 32 E1 ) = 14E1 ;
we get 2n + 1 = 14, which does not have an integeral solution. So it is impossible to nd the pair of energy levels that ts the requirement. From Eq. 404
11P En+2
12P h2 En = 8mL2 (n + 2)2 h2 n2 = h2 (n + 2) : 8mL2 4mL2 (a) Let the quantum numbers of the pair in question be n and n + 1, respectively. Then from the result of 8P En+1 En = (2n + 1)E1 . Let En+1 En = (2n + 1)E1 = E5 = 52 E1 = 25E1 ;
which gives 2n + 1 = 25, or n = 12. (b) Now let En+1 En = (2n + 1)E1 = E6 = 62 E1 = 36E1 ; which gives 2n + 1 = 36, or n = 17:5, which is not an integer. So it is impossible to nd the pair that ts the requirement.
13P The fraction of time that the electron spends in any interval is equal to the probability that the electron is in the interval. This is given by P = j j2 dx ;
where the integral is over the interval. If the interval width x is small, the probability can be approximated by P = j j2 x ; Z CHAPTER 40 MORE ABOUT MATTER WAVES 1095 where the wave function is evaluated for the center of the interval, say. For an electron trapped in an innite well of width L the ground state probability density is j j2 = 2 sin2 x ; so L L (a) Take L = 100 pm, x = 25 pm, and x = 5:0 pm. Then (25 pm) pm) P = 2(5:0pm sin2 100 pm = 0:050 : 100 (b) Take L = 100 pm, x = 50 pm, and x = 5:0 pm. Then (50 pm) pm) P = 2(5:0pm sin2 100 pm = 0:10 : 100 (c) Take L = 100 pm, x = 90 pm, and x = 5:0 pm. Then (90 pm) pm) P = 2(5:0pm sin2 100 pm = 0:0095 : 100
14P 2 x sin2 x : P= L L (a) According to Sample Problem 401(a) Let m = me and solve for n: h E = En+1 En = 8mL2 (2n + 1) : 2 2 2 2 1 1 ec ) n = 8me2L 2E 2 = 8(m2(hcL2 E 2 h ) 2 (1:0 eV) 1 1:2 1019 : = 8(0:511 MeV)(3:0 m) 2(1240 eV nm)2 2 (b) (c) Since 2 2 1 n2 En = 8n hL2 = 2nE = 2 nE = 1 (1:2 1019 )(1:0 eV) = 6:0 1018 eV : me +1 2 the energy level is indeed in the relativistic range. En = 6:0 1018 eV 1 ; me c2 0:511 MeV 1096 CHAPTER 40 MORE ABOUT MATTER WAVES 15P (a) (b) P (0 < x < L=3) = Z L=3 0
Z 2 (x) dx = L 2Z
0 L=3 sin2 x dx = 0:196 : L P (L=3 < x < 2L=3) =
(c) 2L=3 L=3
Z L 2 (x) dx = L 2Z 2L=3 L=3 L sin2 x dx = 0:609 : L x dx = 0:196 : P (2L=3 < x < L) = L 2L=3 L 2L=3 Note that by symmetry P (0 < x < L=3) = P (2L=3 < x < L) and by normalization P (0 < x < L=3) + P (L=3 < x < 2L=3) + P (2L=3 < x < L) = P (0 < x < L) = 1.
2 (x) dx = 2Z sin2 16E From Fig. 406 the energy E of the n = 3 state is 5:5 eV. Thus from E = K + Epot we nd K = E Epot = 5:5 eV ( 30 eV) = 24:5 eV.
17E According to Fig. 406, the electron's initial energy is 11:6 eV. After the additional energy is absorbed, the total energy of the electron is 11:6 eV + 31:7 eV = 43:3 eV. Since it is in the region x > L, its potential energy is 30:0 eV, so its kinetic energy must be 43:3 eV 30:0 eV = 13:3 eV.
18E (a) Apart from the common factor , the unit of the rst term on the LHS of Eq. 4012 is that of d2 =dx2 , which is length 2 (i.e., [L] 2 ); while the unit of the second term is that of mass times energy divided by the Plank's constant squared, or [M] [mass] [energy] = 2 2 ([M][L]2 [T] 2 ) ([T]2 ) = [L] ; ([energy] [time]) which is indeed the same as that of the rst term. (b) If we include the factor in determining the units, we must know whether represents a wave function in one, two, or threedimensions. In general, the unit of is dierent in dierent dimensions and can be determined from the fact that j j2 is the appropriate probability density. In onedimensional cases j j2 has the unit of [L] 1 so the SI unit of o is m 1=2 . Thus the common SI unit for each term in a onedimensional Schrdinger's 2 [m] 1=2 = [m] 5=2 . equation is given by [m] CHAPTER 40 MORE ABOUT MATTER WAVES 1097 19P (a) and (b) Schrdinger's equation for the region x > L is o d2 + 82 m [E E ] = 0 ; pot dx2 h2 where E Epot < 0. If 2 (x) = Ce 2kx , then (x) = C 0 e kx , where C 0 is another constant satisfying C 02 = C . Thus d2 =dx2 = 4k2 C 0 e kx = 4k2 and d2 + 82 m [E E ] = k2 + 82 m [E E ] : pot pot dx2 h2 h2
This is zero provided that
2 k2 = 8 2m [Epot E ] : h The quantity on the RHS is positive, so k is real and the proposed function satises Schrdinger's equation. If k is negative, however, the proposed function is physically unreo alistic. It increases exponentially with x and becomes large without bound. The integral of the probability density over the entire x axis must be unity. This is impossible if is the proposed function. Therefore we choose k = 2h 2m (Epot E ) > 0 :
q 20P Schrdinger's equation for the region x > L is o d2 + 82 m [E E (x)] = 0 : pot dx2 h2 If = De2kx , then d2 =dx2 = 4k2 De2kx = 4k2 and d2 + 82 m [E E (x)] = 4k2 + 82 m [E E (x)] : pot pot dx2 h2 h2 q2m (E k= h pot E ) : The quantity under the radical is positive, so k is real and the proposed function satises Schrdinger's equation. If k is positive, however, the proposed function is physically unreo alistic. It increases exponentially with x and becomes large without bound. The integral of the probability density over the entire x axis must be unity. This is impossible if is
the proposed function. This is zero provided that 1098 CHAPTER 40 MORE ABOUT MATTER WAVES 21P (a) and (b) In the region 0 < x < L Epot = 0, so Schrdinger's equation for the region is o d2 + 82 m E = 0 ; dx2 h2 where E > 0. If 2 (x) = B sin2 kx, then (x) = B 0 sin kx, where B 0 is another constant satisfying B 02 = B . Thus d2 =dx2 = k2 B 0 sin kx = k2 (x) and d2 + 82 m E = k2 + 82 m E : dx2 h2 h2
This is zero provided that
2 k2 = 8hmE : 2 Again, the quantity on the RHS is positive, so k is real and the proposed function satises Schrdinger's equation. In this case, however, there exists no physical restriction as to the o sign of k. It can assume either positive or negative values. Thus p k = 2h 2mE : 22E For n = 1 me e4 E1 = 82 h2 0 19 C)4 (9:11 31 kg)(1 6 = 8(8:85 10 12 F/m)210:63 10 :34 102 (1:60 10 (6 J s) = 13:6 eV : 19 J/eV) 23E E = hf = (4:14 10 15 eV s)(6:2 1014 Hz) = 2:6 eV : 24E The net energy absorbed is 1 E = hf = hc = (1240 eV nm) 3751nm 1 580 nm = 1:17 eV : CHAPTER 40 MORE ABOUT MATTER WAVES 1099 25E (a) The homebase energy level for the Balmer series is n = 2. Thus the transition with the least energetic photon is the one from the n = 3 level to the n = 2 level. The energy dierence for this transition is E = E3 E2 = (13:6 eV) 312 1 = 1:889 eV : 22 The corresponding wavelength is then hc :00 108 m/s) 626 34 = E = (6:(1:889 10 :J s)(310 19 J/eV) = 658 nm : eV)(1 60 (b) For the series limit the energy dirence is 1 E = E1 E2 = (13:6 eV) 12 The corresponding wavelength is then 1 = 3:40 eV : 22 108 m/s) 10 34 J hc = E = (6:62640 eV)(1:60s)(3:0019 J/eV) = 366 nm : (3: 10
26E (a) The energy level corresponding to the probability density distribution shown in Fig. 4015 is the n = 2 level. Its energy is given by 6 E2 = 13:22eV = 3:4 eV : (b) As the electron is removed from the hydrogen atom the nal energy of the protonelectron system is zero. Thus one needs to supply at least 3:4 eV of energy to the system in order to bring its energy up from E2 = 3:4 eV to zero. (If more energy is supplied then the electron will retain some kinetic energy after it is removed from the atom.)
27E (a) Since energy is conserved the energy Eph of the photon is given by Eph = Ei Ef , where Ei is the initial energy of the electron and Ef is the nal energy. The electron energy is given by ( 13:6 eV)=n2 , where n is the principal quantum number. Thus Eph = Ei Ef = 13:62eV (3) 13:6 eV = 12:1 eV : (1)2 1100 CHAPTER 40 MORE ABOUT MATTER WAVES (b) The photon momentum is given by 19 ph p = Ec = (12:1 eV)(1:6010810 J/eV) = 6:45 10 3:00 m/s (c) The wavelength is given by eV nm hc = E = 1240:08 eV = 103 nm : 12 ph
28E 27 kg m/s : where According to Eq. 4015 the wavelength of the electromagnetic radiation emitted by hydrogen satises hc = Rhc 1 1 ; Eph = `2 u2 19 4 (1 m 9 (9: 10 31 kg)(1:6 = 8(8:85 11 12 F/m)2 (6:63 10 34 C) s)3 (3:=10nm) m/s) 10 10 J 00 108 = 0:01097 nm 1 is the Rydberg constant, ` is the principal quantum number associated with the lower electron state, and u is the principal quantum number associated with the upper electron state. A series limit is obtained if u = 1. For the Balmer series ` = 2 and B = 4=R. For the Lyman series ` = 1 and L = 1=R. The ratio is B = 4=R = 4 : L 1=R e4 R = 8m2eh3 c 0 If kinetic energy is not conserved some of the neutron's initial kinetic energy is used to excite the hydrogen atom. The least energy that the hydrogen atom can accept is the dierence between the rst excited state (n = 2) and the ground state (n = 1). Since the energy of a state with principal quantum number n is (13:6 eV)=n2 , the smallest excitation energy is 13:6 eV (13:6 eV)=(2)2 = 10:2 eV. The neutron does not have sucient kinetic energy to excite the hydrogen atom, so the hydrogen atom is left in its ground state and all the initial kinetic energy of the neutron ends up as the nal kinetic energies of the neutron and atom. The collision must be elastic.
30E 29E (a) E = (13:6 eV)(4 2 1 2 ) = 12:8 eV: (b) The various photon energies are depcited in the diagram in the next page. CHAPTER 40 MORE ABOUT MATTER WAVES 1101 4 3 E41 E31 E42 E32 E43 2 E21 n=1 The values of the photon energies are: E41 = E = 12:8 eV; E31 = (13:6 eV)(3 2 1 2 ) = 12:1 eV; E21 = (13:6 eV)(2 2 1 2 ) = 10:2 eV; E42 = (13:6 eV)(4 2 2 2 ) = 2:55 eV; E32 = (13:6 eV)(3 2 2 2 ) = 1:89 eV; and E43 = (13:6 eV)(4 2 3 2 ) = 0:66 eV:
31E Use Eq. 4020: (a) At r = 0 P (r) / r2 = 0. (b) At r = a P (r) = a43 r2 e
4e
2 2r=a : P (r) = a43 a2 e
(c) At r = 2a 2a=a = 4 2 = 5:29 e10 2 nm = 10:2 nm 1 : a
4 P (r) = a43 (2a)2 e
32E 4a=a = 16e a 16e 4 = 5:29 10 2 nm = 5:54 nm 1 : (a) Use Eq. 4016. At r = a
2 (r) = p 1 3=2 e
a a=a 2 1 = a3 e 2 = (5:29 1 2 nm)3 e 2 = 291 nm 3 : 10 4e 4 2 = 5:29 e10 2 nm = 10:2 nm 1 : a
2 (b) Use Eq. 4020. At r = a P (r) = a43 a2 e 2a=a = 1102 CHAPTER 40 MORE ABOUT MATTER WAVES (a) Take the electrostatic potential energy to be zero when the electron and proton are far removed from each other. Then the nal energy of the atom is zero and the work done in pulling it apart is W = Ei , where Ei is the energy of the initial state. The energy of the initial state is given by Ei = ( 13:6 eV)=n2 , where n is the principal quantum number of the state. For the ground state n = 1 and W = 13:6 eV. (b) For the state with n = 2, W = (13:6 eV)=(2)2 = 3:40 eV.
34P 33P From the formula obtained in 28E 1 hf = hc = hcR n2 1 1 : n2
2 Set n2 = 1 to ontain and set n2 = n1 + 1 to obtain 1 =R 1 n2
max 1 1 =R; min n2 1 n2 2 1 1 = R n2
1 1 (n1 + 1)2 : (a) For the Lyman series n1 = 1, so = max 1 = R 112 1 = R 212 f = c min So for the Lyman series
8 01097 nm 1 ) = 8:23 1014 Hz ; f = (3:00 10 m/s)(0:2 (1 + 1) 1 1 min = R n2 1 22 1 1 n2 1 1 (n1 + 1)2 R 1 1 12 = 1 = R 3R 3(0:01097 nm 1 ) = 30:4 nm : (b) For the Balmer series n1 = 2, so 1 32 1 6 22 = 36 = R 5R 5(0:01097 nm 1 ) = 292 nm : (c) Since f = c= the frequency interval is given by c = cR cR 1 max n2 n2 1 1 1 cR 2 = (n1 + 1)2 : (n1 + 1) CHAPTER 40 MORE ABOUT MATTER WAVES 1103 and for the Balmer series 8 01097 nm 1 ) = 3:66 1014 Hz : f = (3:00 10 m/s)(0:2 (2 + 1)
35P (a) The kinetic energy K of the electron is related to Epot and E by K + Epot = E . Here 1 2 Epot = 4 er : 0 As the electron moves around the proton, the proton exerts a Coulomb force FC which serves as the centripetal force, Fcen , which is necessary for the electron to maintain its orbital motion. This means 2 K 1 e2 FC = 4 r2 = Epot = Fcen = me v = 2r ; r r 0 which gives Epot = 2K: Thus K + Epot = K 2K = E , or K = E = ( 13:6 eV) = 13:6 eV :
(b) From the discussion above Epot = 2K = 2( E ) = 2E = 2(13:6 eV) = 27:2 eV :
36P Conservation of linear momentum of the atomphoton system requires that precoil = pphoton , or mp vrecoil = Ephoton =c. Thus
2) ( 13:6 eV)(4 2 vrecoil = E4m cE1 = 938 MeV=(3:00 101 m/s) = 4:1 m/s : 8 p (a) Let or 1 37P 1 Eph = hc = hcR n2 1 1 ; n2
2 1 = 1 = 1 3 1 1 2 R (486:1 nm)(0:01097 nm 1 ) 16 = 22 42 : n2 Thus n1 = 2 and n2 = 4. So the transition is from the n2 = 4 to the n1 = 2 state. n2 2 1104 CHAPTER 40 MORE ABOUT MATTER WAVES (b) The Balmer series (since n1 = 2).
38P (a) The energy of the photon of wavelength 121:66 nm emitted in the transition is given by Thus the energy dierence between the initial and the nal states must be 10:2 eV. A direct inspection of the energy level diagram of Fig. 4011 tells us that the only two states with this energy dierence are the n = 1 and the n = 2 states. Thus the transition is from the n = 2 to the n = 1 state. (b) The transition belongs to the Lyman series, since n1 = 1. (a) Since E2 = 0:85 eV and E1 = 13:6 eV + 10:2 eV = 3:4 eV, the photon energy is Ephoton = E2 E1 = 0:85 eV ( 3:4 eV) = 2:6 eV : (b) From 1 1 = 2:6 eV E2 E1 = ( 13:6 eV) n2 n2 1 = 2:6 eV 3 = 1 1 : n2 13:6 eV 16 42 22 1 Thus n2 = 4 and n1 = 2. So the transition is from the n = 4 state to the n = 2 state. You can easily verify this by inspecting the energy level diagram of Fig. 4011. we obtain
2 1 nm E = hc = 1240:eVnm = 10:2 eV : 121 66 39P n2 2 1 40P where n = 3; 4; 5; . . . Solve for : The wavelength of the photon emitted in a transition belonging to the Balmer series satises hc = E E = (13:6 eV) 1 1 ; Eph = n 2 n2 22
2 eV 4hcn2 = (13:6 eV)(n2 4) = 4(1240:6 eVnm) n2n 4 : 13 Plug in the various values of n to obtain these values of the wavelength: = 656 nm (for n = 3), = 486 nm (for n = 4), = 434 nm (for n = 5), = 410 nm (for n = 6), = 397 nm (for n = 7), = 389 nm (for n = 8), etc. Finally for n = 1 = 365 nm. These values agree well with the data found in Fig. 4012. [You can also nd beyond three signicant gures by using the more accurate values for me , e and h listed in Appendix B when calculating En in Eq. 4015. Another factor that contributes to the error is the CHAPTER 40 MORE ABOUT MATTER WAVES 1105 motion of the atomic nucleus. It can be shown that this eect can be accounted for by replacing the mass of the electron me by me mp =(mp + me ) in Eq. 4015, where mp is the mass of the proton. Since mp me this is not a major eet.]
41P This problem is similar to 37P and 38P. Let 1 Eph = hc = hcR n2 1 1 ; n2
2 1 = 1 = 1 8 1 1 2 R (102:6 nm)(0:01097 nm 1 ) 9 = 12 32 : n2 Thus n1 = 1 and n2 = 3. So the transition is from the n2 = 3 to the n1 = 1 state. or n2 1 1 42P From Sample Problem 407 we know that the probability of nding the electron in the ground state of the hydrogen atom inside a sphere of radius r is given by p(r) = 1 e 2x (1 + 2x + 2x2 ) ; where x = r=a. Since the total probability of nding the electron anywhere in space is unity, the probability of nding the electron outside the sphere, p0 (r), satises p0 (r) = 1 p(r) :
Now set r = a to obtain the desired probability p0 (a): p0 (a) = 1 p(a) = e
43P 2x (1 + 2x + 2x2 ) x=1 = 0:68 : Again from Sample Problem 407 we know that the probability of nding the electron in the ground state of the hydrogen atom inside a sphere of radius r is given by p(r) = 1 e 2x (1 + 2x + 2x2 ) ; where x = r=a. Thus the probability of nding the electron in between the two shells is given by p(a < r < 2a) = p(2a) p(a) = 1 e 2x (1 + 2x + 2x2 ) x=2 = 0:44 : 1 e 2x (1 + 2x + 2x2 ) x=1 1106 CHAPTER 40 MORE ABOUT MATTER WAVES 44P The proposed wave function is = p 1 3=2 e where a is the Bohr radius. Substitute this into the right side of Schrdinger's equation o and show that the result is zero. The derivative is d = p 1 e r=a ; a r=a ; dr a5=2
2 so and r r2 d = pa5=2 e dr d r2 d = p 1 dr dr a5=2 r=a a r a : Now the energy of the ground state is given by E = me4 =82 h2 and the Bohr radius is 0 given by a = h2 0 =me2 , so E = e2 =80 a. The potential energy is given by Epot = e2 =40 r, so
r=a 2+1 e r a =1 2+1 82 m [E E ] = 82 m pot h2 h2 = me h2 2 0 a 1+2 e2 + e2 80 a 40 r r 1 =a a 2 e2 = 8 2m 8 h 0 1+2 : a 1+2 r r The two terms in Schrdinger's equation obviously cancel and the proposed function o satises that equation. The radial probability function for the ground state of hydrogenRis P (r) = (4r2 =a3 )e 2r=a , where a is the Bohr radius. You want to evaluate the integral 01 P (r) dr. Let = 2=a. Then 2 d2 P (r) = 4ar3 e r = a43 d 2 e r and the integral is
Z 45P 1 0 4 d2 Z 1 e P (r) dr = a3 d 2 = 4 d2 1 a3 d 2 0 8 4 2 = a3 3 = a3 8 = 1 ; 2 r dr = 4 d a3 d 2 a3 1 e 1 r 0 where the substitution = 2=a was made in the next to last step. CHAPTER 40 MORE ABOUT MATTER WAVES 1107 (a) The allowed values of l for a given n are 0; 1; 2; ; n 1. Thus there are n dierent values of l. (b) The allowed values of ml for a given l are l, l + 1, , l. Thus there are 2l + 1 dierent values of ml . (c) According to part (a) above, for a given n there are n dierent values of l. Also, each of these l's can have 2l + 1 dierent values of ml [see part (b) above]. Thus the total number of ml 's is n 1 X (2l + 1) = n2 :
l=0 46P 47P According to Fig. 4018 the quantum number n in question satises r = n2 a. Let r = 1:0 mm and solve for n: r r r = 1:0 10 3 m 4:3 103 : n= a 5:29 10 11 m Since r is small we may calculate the probability using p = P (r) r, where P (r) is the radial probability density. The radial probability density for the ground state of hydrogen is given by Eq. 4020: 2 P (r) = 4ar3 e 2r=a ; where a is the Bohr radius. (a) Put r = 0:500a and r = 0:010a. Then 2 4r r e 2r=a = 4(0:500)2 (0:010) e 1 = 3:68 10 3 : p = a3 48P (b) Put r = 1:00a and r = 0:010a. Then 2 4r r e 2r=a = 4(1:00)2 (0:010) e 2 = 5:41 10 3 : p = a3
49P* The radial probability function for the ground state of hydrogen is P (r) = (4r2 =a3 )e where a is the Bohr radius. Let = 2=a. Then 3 3 rP (r) = 4r e 2r=a = 4 d e r 2r=a , a3 a3 d 3 1108 CHAPTER 40 MORE ABOUT MATTER WAVES d3 1 r = a3 d 3 e r dr = a43 d 3 e r 0 0 3 1 24 1 4 a 4 d = a3 d 3 = a3 4 = 24 16 = 1:5a ; 3 a where the substitution = 2=a was made in the next to last step. and 4 d3 Z 1 1 50P* (a) 300 250 200  200(r)2 in (nm)3 150 100 50 0 0 0.5 1 1.5 2 2.5 3 r/a The plot above for j 200 (r)j2 is consistent with the dot plot of Fig. 4015. In the plot above there is a high central peak between r = 0 and r 2a, corresponding to the densely dotted region around the center of the dot plot of Fig. 4015. Outside this peak is a region of nearzero values centered at r = 2a, where 200 = 0. This is represented in the dot plot by the empty ring surrounding the central peak. Further outside is a broader, atter, lower peak which reaches its maximum value at r = 4a. This corresponds to the outer ring with nearuniform dot density which is lower than that of the central peak. (b) The maxima of 2 (r) and (r) occur at the same value of r. So to nd the value of r that maximizes 2 (r), set d (r)=dr = 0. Disregarding the unimportant prefactor, we get d h2 r e r=2a i = 1 r 2 e r=2a = 0 ; dr a a 2a a which gives r = 4a. In fact you can verify directly from the plot above that r = 4a is 2 indeed a local maximum of 200 (r). CHAPTER 40 MORE ABOUT MATTER WAVES 1109 (c) The radial probability is given by P200 (r) = 4r2
(d) Let x = r=a. Then
Z 2 200 (r) =
Z r2 2 r 2 e 8a3 a r=a : 1 0 P200 (r) dr = r 2 2 a e r=a dr 0Z 8a3 1 1 x2 (2 x)2 e x dx =8 0 Z 1 = (x4 4x3 + 4x2 )e x dx 0 1 [4! 4(3!) + 4(2!)] =8 = 1; xn e x dx = n! 1 r2 where the integral formula was used.
51P* 210 is real. Z 1 0 Simply square it to obtain the probability density: P210 = j Each of the other functions is multiplied by its complex conjugate, obtained by replacing i with i in the function. Since ei e i = e0 = 1, the result is the square of the function without the exponential factor: r 2 210 j = 32a5 e 2 r=a cos2 : P211 = j
and 211 j2 = r2 e 64a5
2 r=a sin2 r 2 r=a sin2 : 1 = j 2 1 1 j = 64a5 e The last two functions lead to the same probability density. The total probability density for the three states is the sum: r2 e r=a cos2 + 1 sin2 + 1 sin2 = r2 e r=a : total probability density = 32a5 2 2 32a5 Note that the total probability density does not depend on or . It is spherically symmetric. P2 1 ...
View
Full
Document
 Fall '07
 GROUPTEST

Click to edit the document details