Physics exercises_solution: Chapter 41

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Physics exercises_solution: Chapter 41

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 41

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  1. 2 2  L   4.716  10 J  s  34 41.1: L  l (l  1)  l (l  1)       34       1.054  10 J  s   l (l  1)  20.0  l  4. 41.2: a) mlmax  2, so Lzmax  2. b) l (l  1)  6  2.45. c) The angle is arccos  Lz  m     arccos l , and the angles are, for ml  2 to ml  2, 144.7, 114.1, 90.0,  L  6 65.9, 35.3. The angle corresponding to ml  l will always be larger for larger l . 41.3: L  l (l  1). The maximum orbital quantum number l  n  1. So if : n2 l 1  L  l (l  1)   2  1.41 n  20 l  19  L  19(20)   19.49 n  200 l  199  L  199(200)   199.5 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), (2,  1), (2,  2), (3, 0), (3,  1), (3,  2), (3,  3), (4, 0), (4,  1), (4,  2), (4,  3), and (4,  4), a total of 25. 13.60 eV b) Each state has the same energy ( n is the same),   0.544 eV. 25 1 q1q 2  1 (1.60  10 19 C) 2 41.5: U  .   2.3  10 18 J 4πε0 r 4πε0 1.0  10 10 m  2.3  10 18 J   14.4 eV. 1.60  10 19 J eV 41.6: a) As in Example 41.3, the probability is a 2 a/2 4  ar 2 a 2 r a 3   2 r a  P | ψ1s | 4πr dr  3 2 2    2  2  4 e   0 a   0 1 5e  1  0.0803. 2 b) The difference in the probabilities is (1  5e 2 )  (1  (5 2)e 1 )  (5 2)(e 1  2e 2 )  0.243.
  2. 41.7: a) | ψ |2  ψ * ψ | R(r ) |2 | (θ ) |2 ( Ae iml )( Ae  iml )  A2 | R(r ) |2 | (θ ) |2 , which is independent of  2π 2π 1 b)  |  ( ) |2 d  A2  d  2πA2  1  A  . 0 0 2π 1 mr e 4 E1 41.8: En   E12  E2  E1  2  E1  (0.75) E1 . (4πε0 ) 2n  2 2 2 2 a) If mr  m  9.11  10 31 kg 9.109  10 31kg)(1.602  10 19 C) 4 mr e 4 (4πε0 ) 2 2  2(1.055  10 34 J  s) 2  8.988  109 Nm 2 C  2  2.177  10 18 J  13.59 eV. For 2  1 transition, the coefficient is (0.75)(13.59 eV)=10.19 eV. m b) If mr  , using the result from part (a), 2 mr e 4  m 2   13.59 eV   (13.59 eV)    6.795 eV. (4πε0 )  2 2  m   2   10.19 eV  Similarly, the 2  1 transition,     5.095 eV.  2  c) If mr  185.8m, using the result from part (a), mr e 4  185.8m   (13.59 eV)   2525 eV, (4πε0 )  2 2  m  and the 2  1 transition gives  (10.19 eV)(185.8)=1893 eV. ε0 h 2 ε0 (6.626  10 34 J  s) 2 41.9: a) mr  m : a1   πme 2 π (9.109  10 31 kg)(1.602  10 19 C) 2  a1  5.293  10 11 m m b) mr   a2  2a1  1.059  10 10 m. 2 1 c) mr  185.8 m  a3  a1  2.849  10 13 m. 185.8 41.10: eiml  cos(ml )  i sin( ml ), and to be periodic with period 2π , ml 2π must be an integer multiple of 2π , so ml must be an integer.
  3. a a 1 2r a 41.11: P(a)   ψ1s dV   e (4πr 2 dr ) 0 0 πa 3 a 4 a 4   ar 2 a 2 r a 2   2 r a     2  2  4 e 2 2 r a  P(a)  3 r e dr  3   a o a   0   a 3 a 3 a 3  2 a 3 0  4    2  2  4 e  4 e   a3    2  P(a )  1  5e . 41.12: a) E  μB B  (5.79  10 5 e V T )(0.400 T)  2.32  10 5 eV, b)ml  2, the lowest possible value of ml . c) 41.13: a) g - state  l  4  # of states is (2l  1)  9 (ml  0 ,  1,  2,  3, 4). b) U  μB B  (5.79  10 5 e V T)(0.600 T)  3.47  10 5 eV  5.56  10 24 J. c) U 4 4  8 μB B  8(5.79  10 5 e V T)(0.600 T)  2.78  10 4 eV  4.45  10 23 J. 41.14: a) According to Fig. 41.8 there are three different transitions that are consistent with the selection rules. The initial ml values are 0,  1; and the final ml value is 0. b) The transition from ml  0 to ml  0 produces the same wavelength (122 nm) that was seen without the magnetic field. c) The larger wavelength (smaller energy) is produced from the ml  1 to ml  0 transition. d) The shorter wavelength (greater energy) is produced from the ml  1 to ml  0 transition. U 41.15: a) 3 p  n  3, l  1, U  μB B  B  μB (2.71  10 5 eV) B  0.468 T. (5.79  10 5 e V T) b) Three ml  0,  1.
  4.  e     41.16: a) U  (2.00232)   B  2m   2  (2.00232)  μB B 2 (2.00232)  (5.788  10 5 e V T )(0.480 T) 2  2.78  10 5 eV. b) Since n = 1, l = 0 so there is no orbital magnetic dipole interaction. But if n  0 there could be since l < n allows for l  0. e 41.17: U    z B  (2.00232) Sz B 2m e  U  (2.00232) (ms ) B 2m e  U  (2.00232) μB ms B where μB  . 2m So the energy difference  1  1  U  (2.00232) μB B      2    2   U  (2.00232)(5.788  10 5 e V T)(1.45 T)  1.68  10 4 eV. 1 And the lower energy level is ms   (since B points in the  z direction). ˆ 2  1  1  3  3  5 41.18: The allowed (l , j ) combinations are  0, , 1, , 1, ,  2,  and  2, .  2  2  2  2  2 1 1 41.19: j quantum numbers are either l  or l  , So if j  9 2 and 7 2 , then l  4. 2 2 The letter used to describe l  4 is “g” hc (4.136  10 15 eV  s)(300  108 m s) c 41.20: a) λ   6  21 cm, f   E (5.9  10 eV) λ (3.00  108 m s)  1.4  109 Hz, a short radio wave. 0.21m b) As in Example 41.6, the effective field is B   E 2 μB  5.1  10 2 T, for smaller than that found in the example.
  5. 2 41.21: a) Classically L  Iω, and I  mR 2 for a uniform sphere. 5 2 3 L mω R 2   5 4 5 3 5(1.054  10 34 J  s) 3 ω  2mR 2 4 2(9.11  10 31 kg) (1.0  10 17 m) 2 4  ω  2.5  1030 rad s. b) v  rω  (1.0  10 17 m) (2.5  1030 rad s)  2.5  1013 m s . 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.  Z eff 2 41.23: Using Eq. (41.27) for the ionization energy: En  (13.6 eV). The 5s n2 electron sees Z eff  2.771 and n  5  (2.771) 2  E5  (13.6 eV)  4.18 eV. 52 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 1s 2 2 s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p 2 . 41.25: The ten lowest energy levels for electrons are in the n = 1 and n = 2 shells. 1 n  1, l  0, ml  0, ms   : 2 states. 2 1 n  2, l  0, ml  0, ms   : 2 states. 2 1 n  2, l  1, ml  0,  1, ms   : 6 states. 2 41.26: For the 4 s state, E  4.339 eV and Z eff  4 (4.339) (13.6)  2.26. Similarly, Z eff  1.79 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.
  6. 41.27: a) Nitrogen is the seventh element (Z = 7). N 2 has two electrons removed, so there are 5 remaining electrons  electron configuration is 1s 2 2s 2 2 p.  Z eff 2  (7  4) 2 b) E  (13.6 eV)  (13.6 eV)  30.6 eV n2 22 c) Phosphorous is the fifteenth element (Z = 15). P 2 has 13 electrons, so the electron configuration is 1s 2 2s 2 2 p 6 3s 2 3 p. (15  12) 2 d) The least tightly held electron: E   (13.6 eV)  13.6 eV. 32 13.6 eV 2 41.28: a) E2   Z eff , so Z eff  1.26. b) Similarly, Z eff  2.26. c) Z eff 4 becomes larger going down the columns in the periodic table.  Z eff 2 41.29: a) Again using En  2 (13.6 eV), the outermost electron of the Be  L shell n (n = 2) sees the inner two electrons shield two protons so Z eff  2.  22  E2  (13.6 eV)  13.6 eV. 22  22 b) For Ca  , outer shell has n = 4, so E 4  2 (13.6 eV)  3.4 eV. 4 41.30: Ekx  ( Z  1) 2 (10.2 eV) 7.46  10 3 eV Z  1  28.0, 10.2 eV which corresponds to the element Nickel (Ni). 41.31: a) Z  20 : f  (2.48  1015 Hz)(20  1) 2  8.95  1017 Hz E  hf  (4.14  10 15 eV  s) (8.95  1017 Hz)  3.71 keV c 3.00  108 m s λ   3.35  10 10 m. f 8.95  10 Hz 17 b) Z = 27: f  1.68  1018 Hz E  6.96 keV λ  1.79  10 10 m. c) Z  48 : f  5.48  1018 Hz, E  22.7 keV, λ  5.47  10 11 m.
  7. 2 2 2 2 r / a dr 2 ψ 41.32: See Example 41.3; r ψ  Cr e 2 ,  Ce  2 r a (2r  (2r 2 a)), and for a dr maximum, r = a, the distance of the electron from the nucleus in the Bohr model. 1 me 4  1 e2 1 me 4 41.33: a) E1s   and U (r )  . If E1s  U (r ), then  (4πε0 ) 2 22 4πε0 r (4πε0 ) 2 22  1 e2 4πε0 22 r  2a (4πε0 ) r me 2   1 b) P (r  2a )   ψ1s dV  4π  ψ1s r 2 dr and ψ is  e r / a . 2 2 3 2a 2a πa  4 r e 2 2 r a  P ( r  2a )  dr a3 2a  4  2 r a  ar 2 a 2 r a 3   P ( r  2a )   3 e   2  2  4   a    2 a 4 4  3 a3   P ( r  2a )  e  2a  a 3  .  a3  4  4  13e  0.238. 41.34: a) For large values of n, the inner electrons will completely shield the nucleus, so 13.60 eV Z eff  1 and the ionization energy would be . n2 13.60 eV b) 2  1.11  10 4 eV, r350  (350) 2 a0  (350) 2 (0.529  10 10 m)  6.48  10 6 m . 350 13.60 eV c) Similarly for n = 650, 2  3.22  10 5 eV, r650  (650) 2 (0.529  (650) 10 10 m)  2.24  10 5 m.
  8. 41.35: a) If normalized, then    ψ 2 s dV  4π  2 2 ψ 2 s r 2 dr  1 0 0 2  4π 2  r I  3 r  2   e  r a dr 0 32πa  a 1  2 4r 3 r 4   r a   3 8a 0  4r  a  a 2  e dr.      n! But recall  x n e ax dx  n1 . 0  1  3 4 1  So I  3  4(2)a  (6)a 4  2 (24)a 5  8a  a a  1    I  3 8a 3  24a 3  24a 3  1 and ψ 2 s is normalized. 8a b) We carry out the same calculation as part (a) except now the upper limit on the integral is 4a, not infinity. 1 4a  r3 r4  So I  3   4r 2  4  2  e r a dr. 8a 0   a a  Now the necessary integral formulas are:  r e dr  e (r a  2ra  2a ) 2 r a r a 2 2 3  r 3e r a dr  e r a (r 3 a  3r 2 a 2  6ra 3  6a 3 )  r 4 e r a dr   e r a (r 4 a  4r 3 a 2  12r 2 a 3  24ra 4  24a 5 ) All the integrals are evaluated at the limits r  0 and 4a. After carefully plugging in the limits and collecting like terms we have: 1 I  3  a 3 [(8  24  24)  e 4 (104  568  824)] 8a 1  I  (8  360e 4 )  0.176  Prob(r  4a ). 8
  9. 41.36: a) Since the given ψ (r ) is real, r 2 | ψ |2  r 2ψ 2 . The probability density will be an extreme when d 2 2  ψ  dψ  (r ψ )  2  rψ 2  r 2ψ   2rψ  ψ  r   0. dr  dr   dr  This occurs at r  0, a minimum, and when ψ  0, also a minimum. A maximum must correspond to ψ  r dψ dr  0. Within a multiplicative constant, dψ 1 ψ ( r )  ( 2  r a )e r 2 a ,   ( 2  r 2 a )e  r 2 a , dr a and the condition for a maximum is (2  r a )  (r a ) (2  r 2a ), or r 2  6ra  4a 2  0. The solutions to the quadratic are r  a(3  5 ). The ratio of the probability densities at these radii is 3.68, with the larger density at r  a(3  5 ). b) ψ  0 at r  2a. 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 r  2a. L  41.37: a) θ L  arccos  z . This is smaller for Lz and L as large as possible. Thus  L l  n  1 and ml  l  n  1  Lz  ml   (n  1) and L  l (l  1)  (n  1) n  n 1   θ L  arccos  .  n(n  1)    b) The largest angle implies l  n  1, ml  l  (n  1)   (n  1)   θ L  arccos    n(n  1)      arccos  (1  1 n) .  41.38: a) L2  L2  L2  L2  l (l  1)2  ml2 2 so L2  L2  l (l  1)  ml2 . x y z x y b) This is the magnitude of the component of angular momentum perpendicular to the z -axis. c) The maximum value is l (l  1)  L, when ml  0. 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 ml  l.
  10.  1  4 r 2 a 41.39: P(r )   5  r e  24a  dP  1  r4   (4r 3  )e  r 2 a dr  24a 5  a dP r4  0 when 4r 3   0; r  4a, dr a In the Bohr model, rn  n 2 a so r2  4a, which agrees. 41.40: The time required to transit the horizontal 50 cm region is x 0.500 m t   0.952 ms. v x 525 m s The force required to deflect each spin component by 0.50 mm is 2z  0.1079 kg mol  2(0.50  10 3 m)  22 Fz  maz   m 2     6.022  10 23 atoms mol  (0.952  10 3 s) 2  1.98  10 N.  t   According to Eq. 41.22, the value of μ z is | μ z | 9.28  10 24 A  m 2 . Thus, the required magnetic-field gradient is dBz F 1.98  10 22 N  z   21.3 T m. dz μ z 9.28  10 24 J T 41.41: Decay from a 3d to 2 p state in hydrogen means that n  3  n  2 and ml   2,  1, 0  ml  1, 0. 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 ml must change by 0 or  1. The shift in the transition energy from the zero field value is just eB U  (ml3  ml2 ) μB B  (ml3  ml2 ) 2m where ml3 is the 3d ml value and ml2 is the 2 p ml value. Thus there are only three different energy shifts. They and the transitions that have them, labeled by m, are: eB : 2  1, 1  0, 0  1 2m 0 : 1  1, 0  0,  1  1 eB  : 0  1,  1  0,  2  1 2m
  11. 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 5 eV. b) | λ | λ 0 | E | E0 , where E0  (13.6 eV((1 4)  (1 9)), λ 0   36  R  5 1 6.563  10 7 m and E  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 2 n1 2   e ( 2.00232 ) μB B kT n1 2 24 J T)B (1.3811023 J K) (300 K)  e (2.00232) (9.274 10 3 T 1 ) B  e ( 4.48210 n1 2 a) B  5.00  10 5 T   0.9999998. n1 2 n1 2 b) B  0.500 T   0.9978 n1 2 n1 2 c) B  5.00 T   0.978 n1 2 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. m ( n 2 a0 ) 2π (9.11  10 31 kg) (4) (5.29  10 11 m) The magnetic field generated by the “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) B 0  (10 7 A T  m A)  0.277 T. 4π r2 (4) 2 (5.29  10 11 m) 2
  12. 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 configuration would be 1s 4 ; four electrons and Z  4. For a filled n  2 shell, the electron configuration would be 1s 4 2s 4 2p12 ; twenty electrons and Z  20. b) Sodium has Z  11; 11 electrons. The ground-state electron 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 the atom’s transition energy. The hydrogen atom decays from n  2 to n  1, so:  1 1  E  13.60 eV  2  2   (10.2 eV) (1.60  10 19 J eV)  (2)  (1)   18  1.63  10 J hc (6.33  10 -34 J  s)(3.00  108 m s) λ  18  1.22  10 7 m. E 1.63  10 J b) The change in an energy level due to an external magnetic field is just U  ml μB B. The ground state has ml  0, and it is not shifted. The n = 2 state has ml  1, so it is shifted by U  (1)(9.274  10 24 J T)(2.20 T)  2.04  10  23 J and since λ E  λ E E  2.04  10 23 J   λ  λ  (1.22  10 7 m)   1.63  10 18 J   1.53  10 m.  12 E   Since the n = 2 level is lowered in energy (brought closer to the n = 1 level) the change in energy is less, and the 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     . μB μ B  λ1λ 2    e  λ 1λ 2    Substitution of numerical values gives B  3.64  10 3 T, much smaller than that for sodium.
  13. 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  108 m s) λ  15  1.89  10 10 m. E 1.05  10 J For the longest wavelength, we need the smallest transition energy, so this is the n  2  n  1 transition ( K  ). So we use Moseley’s Law: f  (2.48  1015 Hz)(23  1) 2  1.20  1018 Hz c  λ   2.50  10 10 m. f b) The rhenium, Z = 45, the minimum wavelength is hc (6.63  10 34 J  s)(3.00  108 m s) λ  ( Z  1) 2 (13.6 eV) (44) 2 (13.6 eV)(1.60  10 19 J eV)  λ  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 hc 2πmc 41.50: a) E  (2.00232) μB BS Z  B B . m λ λe b) B    2π 9.11  10 31 kg (3.00  108 m s)  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 spin states multiply everything by 2. The maximum l value is (n –1), and each l value has (2l  1)ml values. So the total number of states is n 1 n 1 n 1 N  2 (2l  1)  2 1  4 l l 0 l 0 l 0 4(n  1)(n)  2n   2n  2 n 2  2n 2 N  2n . 2 b) The n = 5 shell (O-shell) has 50 states.
  14. 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. So, (2e)(e) (e)(e)  mv 2   . 4 0 r 2 4 0 (2r ) 2 r But, from the quantization of angular momentum in the first Bohr orbit, L  mvr    v  mv 2  m mr  2  2e 2 e2  2 So     3 4πε0 r 2 4πε0 (4r ) 2 r r mr  7 e2 4πε0 2 4  4πε0 2    r   4 r2 mr 3 7  me 2    4 4  a 0  (0.529  10 10 m)  3.02  10 11 m. 7 7  7  7 (1.054  10 34 J  s) And v     3.83  10 6 m s . mr 4 ma0 4 (9.11  10 31 kg)(0.529  10 10 m) 1  b) KE  2 mv 2   9.11  10 31 kg (3.83  106 m s) 2  1.34  10 17 J  83.5 eV. 2    2e  2 e2  4e 2 e2  c) PE  2      4πε0 r  4πε0 (2r ) 4πε0 r 4πE0 (2r )  e2  7    4πε r   2.67  10 J  166.9 eV.  17 2 0  d) E   [166.9 eV  83.5 eV]  83.4 eV, which is only off by about 5% from the real value of 79.0 eV.
  15. 1 2 41.53: The potential U ( x)  k x is that of a simple harmonic oscillator. Treated 2 quantum mechanically (see Section 40.4) each energy state has energy En  ω (n  1 ). 2 Since electrons obey the exclusion principle, this allows us to put two electrons (one for each m s   1 ) for every value of neach quantum state is then defined by the ordered 2 pair of quantum numbers (n, ms ). By placing two electrons in each energy level the lowest energy is then  N 1   N 1  1   N 1 N 1 1 2   En   2   w  n     2w  n       n 0   n 0  2   n 0 n 0 2   ( 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, so the classical turning radius is 2a Z . b) The normalized wave function is 1 ψ1s (r )  e Z r a 3 3 πa Z and the probability of the electron being found outside the classical turning point is  4  P   ψ1s 4πr 2 dr  3 3  2 e 2 Zr a r 2 dr. 2a Z a Z 2a Z Making the change of variable u  Zr a , dr  (a Z )du changes the integral to  P  4  e 2u u 2 du , 2 which is independent of Z. The probability is that found in Problem 41.33, 0.238, independent of Z.
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