Tài liệu Physics exercises_solution: Chapter 41 doc

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Tài liệu Physics exercises_solution: Chapter 41 doc

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41.1: 2 34 34 2 sJ10054.1 sJ104.716 )1()1(                       L llllL .40.20)1(      lll 41.2: a) .2so,2 maxmax  zl Lm b) .45.26)1(  ll c) The angle is arccos , 6 arccos              l z m L L and the angles are, for ,0.90,1.114,7.144,2to2  ll mm .3.35,9.65   The angle corresponding to lm l  will always be larger for larger .l 41.3: .)1(  llL The maximum orbital quantum number :ifSo.1   nl    5.199)200(199199200 49.19)20(191920 41.12)1(12    Lln Lln llLln The maximum angular momentum value gets closer to the Bohr model value the larger the value of . n 41.4: The ),( l ml combinations are (0, 0), (1, 0), )1,1(  , (2, 0), ),1,2(  ),2,2(  (3, 0), 4),(4,and),3,4(),2,4(),1,4(),0,4(),3,3(),2,3(),1,3(        a total of 25. b) Each state has the same energy ( n is the same), eV.544.0 25 eV60.13  41.5: J103.2 m100.1 C)1060.1( 4 1 . 4 1 18 10 219 0 21 0        πεr qq πε U eV.4.14 eVJ1060.1 J103.2 19 18       41.6: a) As in Example 41.3, the probability is                  2/ 0 2 0 2 322 3 22 1 422 4 4|| a a ar s e araar a dr πrψP .0803.0 2 5 1 1   e b) The difference in the probabilities is .243.0)2)(25())25(1()51( 2112   eeee 41.7: a) ))((|)(||)(||| 222  ll imim * AeAeθrRψψψ   ,|)(||)(| 222 θrRA  which is independent of  b)   ππ π AπAdAd 2 0 2 2 0 22 . 2 1 12|)(|  41.8: .)75.0( 22)4( 1 11 2 1 1212 22 4 2 0 EE E EEE n em πε E r n   a) kg1011.9If 31  mm r   2 29 234 49131 22 0 4 CNm10988.8 s)J10055.1(2 C)10kg)(1.602109.109 )4(       πε em r eV.59.13J10177.2 18   For 12  transition, the coefficient is (0.75)(13.59 eV)=10.19 eV. b) If , 2 m m r  using the result from part (a), eV.795.6 2 eV59.132 eV)59.13( )4( 22 0 4                m m πε em r  Similarly, the 12  transition, eV.095.5 2 eV19.10         c) If ,8.185 mm r  using the result from part (a), eV,2525 185.8 eV)59.13( )4( 22 0 4         m m πε em r  and the 12  transition gives  (10.19 eV)(185.8)=1893 eV. 41.9: a) 21931 234 0 2 2 0 1 C)10kg)(1.60210109.9( )sJ10626.6( :      π ε πme hε amm r m10293.5 11 1   a b) .m10059.12 2 10 12   aa m m r c) .m10849.2 185.8 1 m8.185 13 13   aam r 41.10: ),sin()cos(   ll im mime l  and to be periodic with period πmπ l 2,2 must be an integer multiple of l mπ so,2 must be an integer. 41.11:     a a ar s drπre πa dVψaP 0 0 22 3 1 )4( 1 )( .51)( 4422 4 422 44 )( 2 0 3 2 333 3 0 2 222 3 22 3                                         eaP e a e aaa a e araar a drer a aP a arar a o 41.12: a) ,2b),eV102.32T)400.0)(TVe1079.5( 55 B   l mBμE the lowest possible value of . l m c) 41.13: a) 4).3,2,1,,0(91)(2isstatesof#4state-           l mllg b) .J1056.5eV1047.3)T600.0)(TVe1079.5( 2455 B   BμU c) eV1078.2)T600.0)(TVe1079.5(88 45 B44    BμU .J1045.4 23  41.14: a) According to Fig. 41.8 there are three different transitions that are consistent with the selection rules. The initial l m values are 0, ;1  and the final l m value is 0. b) The transition from 0to0  ll mm produces the same wavelength (122 nm) that was seen without the magnetic field. c) The larger wavelength (smaller energy) is produced from the 0to1  ll mm transition. d) The shorter wavelength (greater energy) is produced from the 0to1  ll mm transition. 41.15: a) B B ,1,33 μ U BB μUlnp  .T468.0 T)Ve10(5.79 eV)1071.2( 5 5       B b) Three .1,0  l m 41.16: a) B m e U               22 )00232.2(  Bμ B 2 )00232.2(  .eV1078.2 )T480.0)(TVe10788.5( 2 )00232.2( 5 5     b) Since n = 1, l = 0 so there is no orbital magnetic dipole interaction. But if 0  n there could be since l < n allows for .0  l 41.17: BS m e BU zz 2 )00232.2(  . 2 where)00232.2( )( 2 )00232.2( BB m e μBmμU Bm m e U s s     So the energy difference                2 1 2 1 )00232.2( B BμU )T45.1)(TVe10788.5)(00232.2( 5  U .eV1068.1 4  And the lower energy level is ).direction ˆ theinpointssince( 2 1 zBm s  41.18: The allowed ),( jl combinations are . 2 5 ,2and 2 3 ,2, 2 3 ,1, 2 1 ,1, 2 1 ,0                               41.19: j quantum numbers are either .4then,27and29ifSo, 2 1 or 2 1  ljll The letter used to describe is4  l “g” 41.20: a)         λ ,cm21 eV)10(5.9 )sm10s)(300eV10136.4( λ 6 815 c f E hc 9 8 104.1 0.21m )sm1000.3(   Hz, a short radio wave. b) As in Example 41.6, the effective field is for,T101.52 2 B   μEB smaller than that found in the example. 41.21: a) Classically 2 5 2 and, mRIIωL  for a uniform sphere. s.rad102.5 4 3 m)10(1.0kg)102(9.11 s)J105(1.054 4 3 2 5 4 3 5 2 30 21731 34 2 2        ω mR ω RmωL   b) .sm102.5)srad10(2.5m)10(1.0 133017   rωv Since this is faster than the speed of light this model is invalid. 41.22: For the outer electrons, there are more inner electrons to screen the nucleus. 41.23: Using Eq. (41.27) for the ionization energy: eV).(13.6 2 2 eff n Z E n   The s5 electron sees 5and771.2 eff  nZ eV.4.18eV)(13.6 5 (2.771) 2 2 5    E 41.24: However the number of electrons is obtained, the results must be consistent with Table (43-3); adding two more electrons to the zinc configuration gives 622 221 pss .43433 210262 pdsps 41.25: The ten lowest energy levels for electrons are in the n = 1 and n = 2 shells. states.6: 2 1 ,1,0,1,2 states.2: 2 1 ,0,0,2 states.2: 2 1 ,0,0,1    sl sl sl mmln mmln mmln 41.26: For the s4 state, .26.2)6.13()339.4(4andeV339.4 eff  ZE Similarly, 79.1 eff Z for the 4p state and 1.05 for the 4d state. The electrons in the states with higher l tend to be further away from the filled subshells and the screening is more complete. 41.27: a) Nitrogen is the seventh element (Z = 7). 2 N has two electrons removed, so there are 5 remaining electrons  electron configuration is .221 22 pss b) eV30.6eV)(13.6 2 4)(7 eV)(13.6 2 2 2 2 eff      n Z E c) Phosphorous is the fifteenth element (Z = 15). 2 P has 13 electrons, so the electron configuration is .33221 2622 pspss d) The least tightly held electron: eV.13.6eV)(13.6 3 12)(15 2 2    E 41.28: a) .26.1so, 4 e6.13 eff 2 eff2  ZZ V E b) Similarly, 2.26. eff Z c) eff Z becomes larger going down the columns in the periodic table. 41.29: a) Again using eV),(13.6 2 2 eff n Z E n   the outermost electron of the L  Be shell (n = 2) sees the inner two electrons shield two protons so 2. eff  Z eV.13.6eV)(13.6 2 2 2 2    2 E b) For ,Ca  outer shell has n = 4, so eV.3.4eV)(13.6 4 2 2 2 4    E 41.30: eV)(10.2)1( 2  ZE kx ,0.28 eV10.2 eV107.46 1 3    Z which corresponds to the element Nickel (Ni). 41.31: a) Hz108.951)Hz)(2010(2.48:20 17215  fZ m.103.35 Hz108.95 sm103.00 λ keV3.71Hz)10(8.95s)eV10(4.14 10 17 8 1715        f c hfE b) Z = 27: m.101.79λ keV6.96 Hz101.68 10 18     E f c) m.105.47λkeV,22.7Hz,105.48:48 1118   EfZ 41.32: See Example 41.3; )),2(2(, 22 2 2 /22 2 2 arrCe dr ψdr eCr ψr arar   and for a maximum, r = a, the distance of the electron from the nucleus in the Bohr model. 41.33: a)    2 4 2 0 1 2 0 2 4 2 0 1 2)4( 1 then),(. 4 1 )(and 2)4( 1  me πε rUEIf r e πε rU me πε E ss a me πε r r e πε 2 24 )4( 1 2 2 0 2 0    b) . 1 isand4)2( / 3 22 2 1 2 2 1 ar a s a s e a π ψ drrψπdVψarP     . 4 2 4 )2( 422 4 )2( 4 )2( 3 334 3 2 322 2 3 2 22 3                                a aae a arP araar e a arP drer a arP a ar a ar .238.013 4   e 41.34: a) For large values of n , the inner electrons will completely shield the nucleus, so 1 ff  e Z and the ionization energy would be 2 eV60.13 n . b) m106.48m)10(0.529(350))350(,eV101.11 350 eV13.60 6102 0 2 350 4 2   ar . c) Similarly for n = 650, 650 5 2 ,eV103.22 (650) eV13.60 r    (0.529(650) 2 m.102.24m)10 510   41.35: a) If normalized, then . 4 4 8 1 2 32 4 14 0 2 43 2 3 2 2 0 3 2 0 2 2 0 2 2 dre a r a r r a dre a r r πa π I drr ψπdVψ ar ar ss                          But recall      0 1 . ! n axn n dxex  So        5 2 4 3 )24( 1 )6( 4 )2(4 8 1 3 a a a a a a I   .normalizedisand124248 8 1 2 333 3 s ψaaa a I  b) We carry out the same calculation as part (a) except now the upper limit on the integral is 4a, not infinity. So .44 8 1 4 0 2 43 2 3 dre a r a r r a I ar a            Now the necessary integral formulas are:          )2424124( )663( )22( 54322344 332233 3222 araarararedrer araararedrer araaredrer arar arar arar All the integrals are evaluated at the limits 0  r and .4a After carefully plugging in the limits and collecting like terms we have: )]824568104()24248[( 8 1 43 3   ea a I ).4(Prob176.0)3608( 8 1 4 areI   41.36: a) Since the given .||real,is)( 2222 ψrψrrψ  The probability density will be an extreme when .022)( 2222                dr d ψ rψrψ dr ψ ψ rrψψr dr d This occurs at ,0  r a minimum, and when ,0  ψ also a minimum. A maximum must correspond to .0 dr d ψ rψ Within a multiplicative constant, ,)22( 1 ,)2()( 22 arar ear a dr dψ earrψ   and the condition for a maximum is .046or),22()()2( 22  ararararar The solutions to the quadratic are ).53(  ar The ratio of the probability densities at these radii is 3.68, with the larger density at .2at0)).53( arψbar  Parts (a) and (b) are consistent with Fig.(41.4); note the two relative maxima, one on each side of the minimum of zero at .2ar  41.37: a) .arccos        L L θ z L This is smaller for LL z and as large as possible. Thus 1and1  nlmnl l . )1( 1 arccos 1)(1)(and)1(             nn n θ nnllLnmL L lz  b) The largest angle implies )1(,1  nlmnl l   .)11(arccos )1( )1( arccos n nn n θ L             41.38: a) .)1(so)1( 2222222222  lyxlzyx mllLLmllLLLL  b) This is the magnitude of the component of angular momentum perpendicular to the z -axis. c) The maximum value is ,)( L1ll   when .0 l m That is, if the electron is known to have no z -component of angular momentum, the angular momentum must be perpendicular to the z -axis. The minimum is l when .lm l  41.39: ar er a rP 24 5 24 1 )(         ar e a r r a dr dP 2 4 3 5 )4( 24 1          ,4;04when0 4 3 ar a r r dr dP  In the Bohr model, ,4so 2 2 aranr n  which agrees. 41.40: The time required to transit the horizontal 50 cm region is ms.952.0 sm525 m500.0    x v x t The force required to deflect each spin component by 0.50 mm is N.1098.1 )s10952.0( )m1050.0(2 molatoms10022.6 molkg1079.02 22 23 3 232 zz                   t z mmaF According to Eq. 41.22, the value of z μ is .mA1028.9|| 224   z μ Thus, the required magnetic-field gradient is m.T3.21 TJ1028.9 N1098.1 24 22       z zz μ F dz dB 41.41: Decay from a d3 to p2 state in hydrogen means that     l mnn and23 .0,10,1,2  l m However selection rules limit the possibilities for decay. The emitted photon carries off one unit of angular momentum so l must change by 1 and hence l m must change by 0 or .1  The shift in the transition energy from the zero field value is just )( 2 )( 2323 B llll mm m Be B μmmU   where 3 l m is the l md3 value and 2 l m is the l mp2 value. Thus there are only three different energy shifts. They and the transitions that have them, labeled by , m are: 12,01,10: 2 11,00,11:0 10,01,12: 2    m Be m Be   [...]...  n      2   n 0 2   n 0   n 0  n 0 41. 53: The potential U ( x)   ( N  1)( N ) N   2w     w [ N 2  N  N ] 2 2  k  wN 2  N 2 m Here we used the hint from Problem 41. 51 to do the first sum, realizing that the first value of n is zero and the last value of n is N – 1, giving us a total of N energy levels filled 41. 54: a) The radius is inversely proportional to Z,... configuration would be 1s 4 2s 4 2p 3 41. 46: a) Z 2 (13.6 eV)  (7) 2 (13.6 eV)  666 eV b) The negative of the result of part (a), 666 eV c) The radius of the ground state orbit is inversely proportional to the a nuclear charge, and  (0.529  10 10 m) 7  7.56  10 12 m Z hc hc d) λ   , where E 0 is the energy found in part (b), and λ  2.49 nm E E0 ( 112  212 ) 41. 47: a) The photon energy equals... photon wavelength increases due to the magnetic field 41. 48: The effective field is that which gives rise to the observed difference in the energy level transition, E hc  λ1  λ 2  2πmc  λ1  λ 2       B μ B  λ1λ 2  μB e  λ 1λ 2       λ  λ Substitution of numerical values gives B  3.64  10 3 T, much smaller than that for sodium 41. 49: a) The minimum wavelength means the largest...  4.72  1011 m c (3.00  108 m s) The maximum wavelength is λ   f (2.48  1015 Hz)(45  1) 2  λ  6.25  10 11 m e 2πmc hc B B 41. 50: a) E  (2.00232) μB BS Z  m λ λe 2π 9.11  10 31 kg (3.00  108 m s) b) B   0.307 T (0.0350 m)(1.60  10 19 C)   41. 51: a) To calculate the total number of states for the n th principle quantum number shell we must multiply all the possibilities The... (9.274 10  e ( 4.48210 3 c) B  5.00 T  n1 2 n1 2 n1 2 n1 2 n1 2 n1 2 J T)B (1.3811023 J K) (300 K) T 1 ) B a) B  5.00  10 5 T  b) B  0.500 T  24  0.9999998  0.9978  0.978 41. 44 Using Eq 41. 4 L  mvr  l (l  1), and the Bohr radius from Eq 38.15, we obtain the following value for v l (l  1)  2 (6.63  10 34 J  s)  v  7.74  10 5 m s 2π (9.11  10 31 kg) (4) (5.29  10.. .41. 42: a) The energy shift from zero field is U 0  ml  B B For ml  2, U 0  (2) (5.79  10 5 e V T ) (1.40 T)  1.62  10 4 eV For ml  1, U 0  (1)  (5.79  10 5 e V T) (1.40 T)  8.11  10... sodium 41. 49: a) The minimum wavelength means the largest transition energy If we assume that the electron makes a transition from a high shell, then using the screening approximation outlined in Section 41. 5, the transition energy is approximately the ionization energy of hydrogen Then E  E1  ( Z  1) 2 (13.6 eV) For vanadium, Z = 23  E  6.58  103 eV  1.05  10 15 J hc (6.63  10 34 J  s)(3.00... “moving” proton at the electrons position can be calculated from Eq 28.1 μ | q | v sin  (1.60  10 19 C) (7.74  105 m s) sin(90)  (10 7 A T  m A) B 0  0.277 T 4π (4) 2 (5.29  10 11 m) 2 r2 41. 45: m s can take on 4 different values: ms   3 ,  1 ,  1 ,  3 Each nlml state can 2 2 2 2 have 4 electrons, each with one of the four different m s values a) For a filled n  1 shell, the electron...  1)ml values So the total number of states is n 1 n 1 n 1 l 0 l 0 l 0 N  2 (2l  1)  2 1  4 l 4(n  1)(n)  2n   2n  2 n 2  2n 2 2 N  2n b) The n = 5 shell (O-shell) has 50 states 41. 52: a) Apply Coulomb’s law to the orbiting electron and set it equal to the centripetal force There is an attractive force with charge +2e a distance r away and a repulsive force a distance 2r away...  1.62  10 4 eV  8.11  10 5 eV  8.09  10 5 eV from part (a) Then, | λ | 2.81  10 11 m  0.0281 nm The wavelength corresponds to a larger energy change, and so the wavelength is smaller n 41. 43: From Section 38.6: 1  e ( E1  E0 ) kT We need to know the difference in energy n0 1 1 between the ms   and ms   states U   μ z B  2.00232 μB ms B 2 2 So U 1  U  1  2.00232 μB B 2  . this model is invalid. 41. 22: For the outer electrons, there are more inner electrons to screen the nucleus. 41. 23: Using Eq. (41. 27) for the ionization. larger for larger .l 41. 3: .)1(  llL The maximum orbital quantum number :ifSo.1   nl    5.199)200(199199200 49.19)20(191920 41. 12)1(12    Lln Lln llLln The

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