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SolHome28 - 28.3(a From Coulombs law ke q1q2 r2 99 8 10 = 9...

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28.3 (a) From Coulomb’s law, ( )( ) ( ) 2 9 2 2 19 1 2 8 2 2 10 8.99 10 N m C 1.60 10 C = 23 10 N 1.0 10 m e k q q F r × × = = × . × (b) The electrical potential energy is ( )( )( ) 9 2 2 19 19 1 2 10 18 -19 8.99 10 N m C 1.60 10 C 1.60 10 C 1.0 10 m 1 eV 2.3 10 J 14 eV 1.60 10 J e k q q 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 f f i KE PE KE PE + = + i ( )( ) ( )( ) 2 79 2 79 0 e e i f i k e e k e e KE r r + = + With , this gives the distance of closest approach as i r ( )( ) ( ) 2 9 2 2 19 2 -13 14 158 8.99 10 N m C 1.60 10 C 158 5.0 MeV 1.60 10 JMeV 4.5 10 m 45 fm e f i k e r KE × × = = × = × = 28.8 (a) With the electrical force supplying the centripetal acceleration, 2 2 2 e n e n n m v k e r r = , giving 2 e n e n k e v m r = where ( ) 2 2 0 0.0529 nm n r n a n = = Thus, ( )( ) ( )( ) 2 9 2 2 19 2 6 1 31 9 1 8.99 10 N m C 1.60 10 C 2.19 10 m s 9.11 10 kg 0.0529 10 m e e k e v m r × × = = = × × × (b) ( )( ) ( ) 2 9 2 2 19 2 2 1 1 9 1 18 8.99 10 N m C 1.60 10 C 1 2 2 2 0.0529 10 m 2.18 10 J 13.6 eV e e k e KE m v r × × = = = × = × = (c) ( ) ( )( ) ( ) 2 9 2 2 19 1 9 1 18 8.99 10 N m C 1.60 10 C 0.0529 10 m 4.35 10 J 27.2 eV e k e e PE r × × = = − × = − × = − 28.12 The change in the energy of the electron is
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2 2 1 1 13.6 eV f i i f E E E n n Δ = = Transition I: 1 1 13.6 eV 2.86 eV 4 25 E Δ = = (absorption) Transition II: 1 1 13.6 eV 0.967 eV 25 9 E Δ = = − (emission) Transition III: 1 1 13.6 eV 0.572 eV 49 16 E Δ = = − (emission) Transition IV: 1 1 13.6 eV 0.572 eV 16 49 E Δ = = (absorption) (a) Since hc hc E E γ λ = = −Δ , transition II emits the shortest wavelength photon.
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