SolHome28 - 28.3 (a) From Coulombs law, ke q1q2 r2 99 ( 8....

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28.3 (a) From Coulomb’s law, () ( ) 2 92 2 1 9 12 8 2 2 10 8.99 10 N m C 1.60 10 C =2 31 0 N 1. 01 0 m e kqq F r ×⋅ × == × . × (b) The electrical potential energy is ( ) ( ) 2 1 9 1 9 18 -19 C C1 . 60 10 C 1 eV 2. 0 J1 4 eV 1.60 10 J e PE r −− × × × ⎛⎞ =− × = − ⎜⎟ × ⎝⎠ 28.6 Assuming a head-on collision, the -particle comes to rest momentarily at the point of closest approach. From conservation of energy, , or ffi KE PE +=+ i ( )( ) ( )( ) 27 9 9 0 ee i fi ke e rr += + With , this gives the distance of closest approach as i r → ∞ ( )( ) 2 2 1 9 2 -13 14 158 8.99 10 N m C C 158 5.0 MeV 1.60 10 J M eV 4. 51 m4 5 fm e f i ke r × × = 28.8 (a) With the electrical force supplying the centripetal acceleration, 22 2 en e nn mv ke = , giving 2 e n v mr = where ( ) 0 0.052 9 nm n rn an Thus, ( ) ( ) 2 2 1 9 2 6 1 9 1 C C 2.19 10 m s 9.11 10 kg 0.052 9 10 m e e v × = × ×× (b) ( ) 2 2 1 9 2 2 11 9 1 C C 1 20.052 9 10 m 2.18 10 3 . 6 eV e e m v r × = × = (c) ( ) 2 2 1 9 1 9 1 C C 0.052 9 10 m 4.35 10 J 27.2 eV e e PE r × × × 28.12 The change in the energy of the electron is
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22 11 13.6 eV fi if EE E nn ⎛⎞ Δ= − = ⎜⎟ ⎝⎠ Transition I: 2.86 eV 42 5 E Δ= = (absorption) Transition II: 0.967 eV 25 9 E − = (emission) Transition III: 0.572 eV 49 16 E = (emission) Transition IV: 16 49 E = (absorption) (a) Since hc EE γ λ == −Δ , tran sition II emits the shortest wavelength photon. (b) The atom gains the most energy in tran sition I (c) The atom loses energy in tran sitions II and III 28.22 (a) The time for one complete orbit is 2 r T v π = From Bohr’s quantization postulate, e Lmv rn = = h we see that e n v mr = h Thus, the orbital period becomes: () 2 2 2 2 0 3 0 2 e ee ma n Tn ππ = hh h or where 3 0 Tt n = ( ) 2 31 9 2 0 0 34 29 .
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This note was uploaded on 03/05/2009 for the course PHYS PHYS1C taught by Professor Douglassmith during the Summer '08 term at UCSD.

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SolHome28 - 28.3 (a) From Coulombs law, ke q1q2 r2 99 ( 8....

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