Tài liệu Physics exercises_solution: Chapter 38 doc

38.1: Hz105.77 m 10 5.20 sm103.00 λ 14 7 8      c f eV.2.40J103.84)sm10(3.00s)mkg10(1.28 smkg101.28 m105.20 sJ106.63 λ 19827 27 7 34          pcE h p 38.2: a) eV.107.49J0.0120s)10(20.0W)(0.600 163   Pt b) eV.1.90J103.05 λ 19   hc hf c) .1094.3 16  hf Pt 38.3: a) Hz.105.91 sJ106.63 eV)J10(1.60eV)10(2.45 20 34 196       h E fhfE b) m.105.08 Hz105.91 sm103.00 λ 13 20 8      f c c) λ is of the same magnitude as a nuclear radius. 38.4: hc W hc P hf P dNdE dtdE dt dN m)10(2.48)(12.0λ )( )( 7   sec.photons101.50 19  38.5:   hfmv 2 max 2 1 J 10 3.04 eVJ10(1.60eV)(5.1 m102.35 sJ10(6.63 20 19 7 34 ))                  c .sm102.58 kg109.11 J)102(3.04 5 31 20 max       v 38.6:          m102.72 Hz101.45 λ 7 15 0 c h hc hfhfE  eV.1.44J102.30 19   38.7: a) Hz105.0λ 14  cf b) Each photon has energy J.10.313 19  hfE Source emits sphotons102.3photons/s)10(3.31)sJ(75sosJ75 2019   c) No, they are different. The frequency depends on the energy of each photon and the number of photons per second depends on the power output of the source. 38.8: For red light nm700λ  eV1.77 J101.6 1eV J102.84 m)10(700 )sm10(3.00s)J10(6.626 λ 19 19 9 834                    hc hf  38.9: a) For a particle with mass, .4means2.2 1212 2 KKppmpK  b) For a photon, .2means2. 1212 EEpppcE    38.10:    hfK max Use the information given for :findtonm400λ   J 10 3.204 eV)J10(1.602eV)(1.10 m10400 )sm10(2.998s)J10(6.626 19 19 9 834 max           Khf  Now calculate :nm300λfor max K eV2.13J103.418 J103.204 m10300 )sm10(2.998s)J10(6.626 19 19 9 834 max           hfK 38.11: a) The work function 00 eV λ eV  hc hf  J.107.53 V)(0.181C)10(1.60 m102.54 s)m10(3.00s)J10(6.63 19 19 7 834           The threshold frequency implies   hchc hf th th  th λ λ m.102.64 J107.53 )sm10(3.00s)J10(6.63 λ 7 19 834 th        b) eV,4.70J107.53 19    as found in part (a), and this is the value from Table 38.1. 38.12: a) From Eq. (38.4), V.2.7V2.3 m)10(2.50 s)m10(3.00s)eV10(4.136 λ 1 7 815               hc e V b) The stopping potential, multiplied by the electron charge, is the maximum kinetic energy, 2.7 eV. c) s.m109.7 kg)10(9.11 V)(2.7C)102(1.6022 5 31 19       m eV m K v 38.13: a) )sm10(3.00)smkg10(8.24 828   pcE eV1.54 eVJ101.60 J102.47 J102.47 19 19 19        b) m.108.05 s)mkg10(8.24 s)J10(6.63 λ λ 7 28 34        p hh p This is infrared radiation. 38.14: a) The threshold frequency is found by setting V = 0 in Eq. (40.4), . 0 hf   b) eV.3.34105.35 m1072.3 λ 19 7 0      hchc hf  38.15: a) .eV1.44J102.31 m108.60 )sm10(3.00s)J10(6.63 λ 19 7 834        hc E  So the internal energy of the atom increases by     eV1.44eV6.52toeV44.1 E eV.5.08  b) eV.2.96J104.74 m 10 4.20 s)m10(3.00s)J10(6.63 λ 19 7 834        hc E  So the final internal energy of the atom decreases to eV.5.64eV2.96eV2.68      E 38.16: a) eV.20 1   E b) The system starts in the n = 4 state. If we look at all paths to n = 1 we find the 4-3, 4-2, 4-1, 3-2, 3-1, and 2-1 transitions are possible (the last three are possible in combination with the others), with energies 3 eV, 8 eV, 18 eV, 5 eV, 15 eV, and 10 eV, respectively. c) There is no energy level 8 eV above the ground state energy, so the photon will not be absorbed. d) The work function must be more than 3 eV, but not larger than 5 eV. 38.17: a)        22 1 2 1 λ 1 n R (Balmer series implies final state is n = 2) nm433m104.33m )1021(1.10 100 21 100 λ 100 21 25 1 4 1 λ 1 :5 7 7             R RRnH  b) Hz106.93 m104.33 sm103.00 λ 14 7 8      c f c) eV.2.87   hfE 38.18: Lyman: largest is nm,122 )m10(1.097 )34()34( λ,2 17     R n in the ultraviolet. Smallest is nm,91.2 1 λ,  R n also ultraviolet. Paschen: largest is nm,1875 )7144( λ,4  R n in the infrared. Smallest is nm,820 9 λ,  R n also infrared. 38.19: m105.890 )sm10(3.000s)J10(6.626 λ 7 834 1 2 3       hc E g eV.102.00 .eV2.107J103.371 m105.896 )sm10(3.000s)J10(6.626 λ eV.2.109J103.375 3 19 7 834 2 19 2 1 2 3 2 1              E hc E g 38.20: a) Equating initial kinetic energy and final potential energy and solving for the separation radius r, m.105.54 C)J10(4.78 C)10(1.60(184) 4 1 )2()92( 4 1 14 6 19 0 0        ε K ee ε r   b) The above result may be substituted into Coulomb’s law, or, the relation between the magnitude of the force and the magnitude of the potential energy in a Coulombic field is N.13.8 m)10(5.54 )evJ10(1.6eV)10(4.78 14 196       r K F 38.21: a) )m1050.6(4 C)1060.1(164 4 )82()2( 4 14 0 219 00 21      εrε ee r ε qq U  MeV.3.63eV103.63J105.81 613   U b) MeV.3.63J105.81 13 212211   UKUKUK c) .sm101.32 kg106.64 J)102(5.812 2 1 7 27 13 2       m K vmvK 38.22: Hz.103.09 λ andnm97.0λso, )4( 1 1 λ 1 15 2           c fR 38.23: a) Following the derivation for the hydrogen atom we see that for 3 Be all we need do is replace Then.4by 22 ee . eV13.60 16)(Be)H(16 8 )4(1 )(Be 2 3 22 22 2 0 3          n EE hn em ε E nnn So for the ground state, eV.218)(Be 3 1   E b) The ionization energy is the energy difference between the 1and    nn levels. So it is just 218 eV for 3 Be , which is 16 times that of hydrogen. c) . 11 )m10(1.74 11 8 )4( λ 1 2 2 2 1 18 2 2 2 1 32 0 22                    nnnnch em  So for m101.31 4 1 1)m10(1.74 λ 1 ,1to2 818          nn m.107.63λ 9  This is 16 times shorter than that from the hydrogen atom. d) (H). 4 1 )4( )(Be 2 22 0 3 nn r em hn ε r    38.24: a), b) For either atom, the magnitude of the angular momentum is   2 h s.mkg101.05 234   38.25: ,)eV6.13( 2 nE n  so this state has .351.16.13 n In the Bohr model.  nL  so for this state .smkg103.163 234   L 38.26: a) We can find the photon’s energy from Eq. 38.8 J.104.58 5 1 2 1 )m10(1.097)sm10(3.00s)J10(6.63 1 2 1 19 22 17834 22                  n hcRE The corresponding wavelength is nm.434λ  hc E b) In the Bohr model, the angular momentum of an electron with principal quantum number n is given by Eq. 38.10 . 2  h nL  Thus, when an electron makes a transition from n = 5 to n = 2 orbital, there is the following loss in angular momentum (which we would assume is transferred to the photon): s.J103.17 2 s)J103(6.63 2 )52( 34 34      π h L  However, this prediction of the Bohr model is wrong (as shown in Chapter 41). 38.27: a) m/s1018.2 )sJ1063.6(2 C)1060.1( 1: 2 1 6 34 0 219 1 2 0         vn nh e v n .sm1027.7 3 3 sm1009.1 2 2 5 1 3 6 1 2   v vh v vh b) Orbital period 4 332 0 2 0 222 0 4 21 22 me hn ε nhe mehn v r n n      .s1013.4)3(:3 s1022.1)2(:2 s1053.1 C)1060.1(kg)1011.9( )sJ1063.6(4 1 153 13 153 12 16 41931 3342 0 1            TTn TTn Tn  c) number of orbits .102.8 s1022.1 s100.1 6 15 8       38.28: a) Using the values from Appendix F, keeping eight significant figures, gives .m100973731.1 17  R (Note: On some standard calculators, intermediate values in the calculation may have exponents that exceed 100 in magnitude. If this is the case, the numbers must be manipulated in a different order.) b) Using the eight-figure value for 18 101798741.2 λ gives   hcR hc ER eV.605670.13J  c) Using the value for the proton mass as given in Appendix F gives .m5970so,m100967758.1 117   RR 38.29: kTEE p s ps e n n )( 3 5 35    But J10306.3eV66.20 18 5   s E J103.14 J10992.2eV70.18 19 18 3     E E p a) .1015.1 33)K300KJ1038.1()J1014.3( 3 5 2319    e n n p s b) .1039.3 17)K600KJ1038.1()J1014.3( 2319    e c) .1082.5 9)K1200KJ1038.1()J1014.3( 2319    e d) The s5 state is not highly populated compared to the p3 state, so very few atoms are able to make the required energy jump to produce the 632.8 nm light. 38.30: . )( 2 2 212232 2/1 23 KTEE p p pp e n n   From the diagram .J10375.3 m10890.5 s)m10000.3()J10626.6( λ 19 7 834 1 2/3         hc E g .J1000.4J10371.3J10375.3so .J10371.3 m10896.5 s)m10000.3()J10626.6( λ 221919 2/12/3 19 7 834 2 21            E hc E g .944.0 K).500K/1038.1()J1000.4( 2 2 2322 2/1 2/3    J p p e n n So more atoms are in the 21 2 p state. 38.31: .J1088.1 m1006.1 s)m1000.3()sJ1063.6( λ 20 5 834        hc E  Total energy in 1 second from the laser .J1050.7 3  PtE So the number of photons emitted per second is .1000.4 J101.88 J1050.7 17 20 3        E E 38.32: 20.66 nm,632λandJ,103.14eV1.96eV18.70eV 19   E hc f c in good agreement. 38.33: min max λ hc hfeV AC  m1011.3 )V4000)(C1060.1( )sm10s)(3.00J1063.6( λ 10 19 834 min        AC eV hc This is the same answer as would be obtained if electrons of this energy were used. Electron beams are much more easily produced and accelerated than proton beams. 38.34: a) kV.8.29V1029.8 λ 3  e hc b) The shortest wavelength would correspond to the maximum electron energy, ,eV and so .nm0414.0λ  eV hc 38.35: An electron’s energy after being accelerated by a voltage V is just .eVE  The most energetic photon able to be produced by the electron is just: V λ e hc E hc  .nm0829.0m1029.8 V)105.1)(C1060.1( )sm1000.3)(sJ1063.6( λ 11 419 834        38.36: a) From Eq. (38.23), , )( λ 1cos mch    and so a) nm,0.0500nm0542.0λ    .137and,731.0 nm002426.0 nm0042.0 1cos    b) .3.82.134.0 nm002426.0 nm0021.0 1cos.nm0500.0nm0521.0 λ    c) ,0λ   the photon is undeflected, 1cos   and .0   38.37: )cos1(λλ    mc h .)λmaximizetois180))(1(1(λλ max     chosen mc h  mc h2 λ  .m1014.7m)102(2.426m1065.6λ 111211 max     38.38: a)     m)10426.2()cos)(1λλb)nm.0691.0λ 12  mc(h eV hc .nm0698.0λsom,1011.7)0.45cos1( 13     c) .keV8.17 λ    hc E 38.39: )cos1(λλ    mc h mc h λ 2 λ;180,λλ      6 1070.9 λ 2 λ λ   mc hΔ 38.40: The change in wavelength of the scattered photon is given by Eq. 38.23   ).cos1(λ)cos1( λλ Δλ λ Δλ   mc h mc h Thus, .m1065.2)11( )100.0)(m/s1000.3)(kg1067.1( )sJ1063.6( λ 14 827 34        38.41: The derivation of Eq. (38.23) is explicitly shown in Equations (38.24) through (38.27) with the final substitution of ).cos1(λλyieldingλandλ        mc h hphp 38.42: .K1025.7 m10400 Km1090.2m1090.2 3 9 33         m λ K T 38.43: mm.06.1m1006.1 K728.2 Km1090.2Km1090.2 λ 3 33 m        T This is in the microwave part of the electromagnetic spectrum. 38.44: From Eq. (38.30), a)     m c f λ and,mm966.0 K00.3 Km10898.2 λ 3 m .Hz1010.3 11  Note that a more precise value of the Wien displacement law constant has been used. b) A factor of 100 increase in the temperature lowers m λ by a factor of 100 to 9.66 m  and raises the frequency by the same factor, to .Hz1010.3 13  c) Similarly, nm966λ m  .Hz1010.3and 14 f [...]... to the visible luminosity Rstar Rsun  38. 47: Eq (38. 32): I ( λ)  x2 2hc 2 but e x  1  x    1  x for 2 λ 5 (e hc λkT  1) 2hc 2 2ckT x  1  I ( λ)  5   Eq (38. 31), which is Rayleigh’s distribution λ (hc λkT ) λ4 38. 48: a) As in Example 38. 10, using four-place values for the physical constants, hc  95.80, from which λkT I (λ)λ  6.44  10 38 4 σT b) With T  2000 K and the same... s)(3.00  108 m s) hc b) λ mT    2.90  10 3 m  K (4.965)k (4.965)(1 .38  10 23 J K)  Solve 5  x  5e x where x  38. 50: Combining Equations (38. 28) and (38. 30), 2.90  10 3 m  K λm  ( I σ)1/4 (2.90  10 3 m  K)  (6.94  10 6 W/m 2 5.67  10 8 W m 2  K 4 )1/4  8.72  10 7 m  872 nm E (mole) 1.00  10 5 J 38. 51: a) Energy to dissociate an AgBr molecule is just E    1 mole 6.02...  6 4 2 (4.00  10 m )(5.67  10 8 W m 2  K 4 )(473 K) 4 AT 38. 78:  8.81  103 s  2.45 hrs 38. 79: a) The period was found in Exercise 38. 27b: T  2 4ε 0 n 3 h 3 and frequency is me 4 just f  1 me 4  2 3 3 T 4 0 n h b) Eq (38. 6) tells us that f  1 me 4  1 1 ( E 2  E1 ) So f  2 3  2  2 n h 8 0 h  2 n1   (from Eq   (38. 18)) If n2  n and n1  n  1, then 1 1 1 1  2  2  2 n2 n1... given by 38. 68:  ( Eex  Eg ) n2  ( E  E ) kT  e ex g T  n1  n2  kln  n   1  13.6 eV Eex  E 2   3.4 eV 4 E g  13.6 eV E ex  E g  10.2 eV  1.63  10 18 J a) n2  10 12 n1  (1.63  10 18 J) T  4275 K (1 .38  10  23 J K ) ln(10 12 ) b) n2  10 8 n1  (1.63  10 18 J) T  6412 K (1 .38  10 23 J K ) ln(10 8 ) c) n2  10  4 n1  (1.63  10 18 J)  12824 K (1 .38  10...  10 34 J  s)(3.00  108 m s)  1 1  V0  11  2.65  10 7 m  2.95  10 7 m   0.477 V  (1.60  10 C)   So the change in the stopping potential is an increase of 0.739 V 38. 57: a) Recall eV0  38. 58: From Eq (38. 13), the speed in the ground state is v1  Z (2.19  10 6 m s) Setting c v1  gives Z = 13.7, or 14 as an integer b) The ionization energy is E  Z 2 (13.6 10 mc 2 eV), and the rest... λ  mc p  h λ  6.99  10  24 kg  m s b) E  E   Ee ; hc λ  hc λ  Ee λ  λ 1 1  Ee  hc    (hc)  1.129  10 16 J  705 eV λλ   λ λ  38. 59: λ  λ  38. 60: a) The change in wavelength of the scattered photon is given by Eq 38. 23 h h (1  cos  )  λ  λ  (1  cos  )  λ  λ  mc mc (6.63  10 34 J  s) (0.0830  10 9 m)  (1  1)  0.0781 nm (9.11  10 31 kg)(3.00  108... wavelength interval would decrease From the Wien displacement law, the temperature that has the peak of the corresponding distribution in this wavelength interval is 5800 K (see Example 38. 10), close to that used in part (c) 38. 49: a) To find the maximum in the Planck distribution: dI d  2hc 2  (2hc 2 ) 2hc 2 (  λ 2 )  5    0  5 5     5 λ dλ dλ  λ (e  1)  λ (e  1) λ (e  1) 2  .. .38. 45: a) H  Aeσ T 4 ; A  r 2l 14 14   100 W    (0.20  10 3 m) 2 (0.30 m)(0.26)(5.671  10 8 W m 2  K 4 )     4 T  2.06  10 K b) λ mT  2.90  10 3 m  K; λ m  141 nm Much of the emitted radiation is in the ultraviolet  H  T    Aeσ  38. 46: (a) Wien’s law: λ m  k T 2.90  10 3 K  m  9.7  10 8 m ... 23  23  1.65  10 kg  m s  2  10 kg  m s c) Since the electron is non relativistic (   0.06), pe2 Ke   1.49  10 16 J  10 16 J 2m 38. 61: a) mr  207me m p m1m2   1.69  10 28 kg m1  m2 207me  m p b) The new energy levels are given by Eq (38. 18) with me replaced by mr En   1 mr e 4  mr    13.60 eV      02 8n 2 h 2  me   n2     1.69  10 28 kg    13.60 eV ... E2 )  λ hc (6.63  10 34 J  s) (3.00  108 m s)  1.53  10 9 m 1  λ  6.55  10 10 m 38. 62: a) The levels are E 4  1.0 eV, E3  5.0 eV, E 2  8.0 eV, and E1  10.0 eV b) We can go from 4-3(4 eV), 4-2(7 eV), and 4-1(9 eV) directly, but also 3-2(3 eV), 3-1(5 eV), and 2-1(2 eV) after starting from 4 38. 63: a) The maximum energy available to be deposited in the atom is hc (6.63  10 34 J  .  ).cos1(λ)cos1( λλ Δλ λ Δλ   mc h mc h Thus, .m1065.2)11( )100.0)(m/s1000.3)(kg1067.1( )sJ1063.6( λ 14 827 34        38. 41: The derivation of Eq. (38. 23) is explicitly shown in Equations (38. 24) through (38. 27) with the final substitution. b) .Km1090.2 K)J1 038( 4.965)(1. )sm1000.3)(sJ1063.6( )965.4( λ 3 23 834 m        k hc T 38. 50: Combining Equations (38. 28) and (38. 30), nm.872m1072.8 )KmW1067.5W/m10(6.94 K)m1090.2( ( Km1090.2 λ 7 1/442826 3 1/4 3 m           σ)I 38. 51: