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1061_PartUniversity Physics Solution

1061_PartUniversity Physics Solution - Photons Electrons...

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Photons, Electrons, and Atoms 38-5 38.20. (a) Equating initial kinetic energy and final potential energy and solving for the separation radius r , 19 14 6 0 0 1 (92 ) (2 ) 1 (184) (1.60 10 C) 5.54 10 m. 4 4 (4.78 10 J C) e e r K π π × = = = × × P P (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 6 19 14 (4.78 10 eV) (1.6 10 J ev) 13.8 N. (5.54 10 m) K F r × × = = = × 38.21. (a) I DENTIFY : If the particles are treated as point charges, 1 2 0 1 . 4 q q U r π = P S ET U P : 1 2 q e = (alpha particle); 2 82 q e = (gold nucleus); r is given so we can solve for U . E XECUTE : 19 2 9 2 2 13 14 (2)(82)(1.602 10 C) (8.987 10 N m /C ) 5.82 10 J 6.50 10 m U × = × = × × 13 19 6 5.82 10 J(1 eV/1.602 10 J) 3.63 10 eV 3.63 MeV U = × × = × = (b) I DENTIFY : Apply conservation of energy: 1 1 2 2 . K U K U + = + S ET U P : Let point 1 be the initial position of the alpha particle and point 2 be where the alpha particle momentarily comes to rest. Alpha particle is initially far from the lead nucleus implies 1 r ≈ ∞ and 1 0. U = Alpha particle stops implies 2 0. K = E XECUTE : Conservation of energy thus says 13 1 2 5.82 10 J 3.63 MeV. K U = = × = (c) 2 1 2 K mv = so 13 7 27 2 2(5.82 10 J) 1.32 10 m/s 6.64 10 kg K v m × = = = × × E VALUATE : / 0.044, v c = so it is ok to use the nonrelativistic expression to relate K and v . When the alpha particle stops, all its initial kinetic energy has been converted to electrostatic potential energy. 38.22. (a) , (b) For either atom, the magnitude of the angular momentum is 2 h π = 34 2 1.05 10 kg m s. × 38.23. I DENTIFY and S ET U P : Use the energy to calculate n for this state. Then use the Bohr equation, Eq.(38.10), to calculate L . E XECUTE : 2 (13.6 eV)/ , n E n = − so this state has 13.6/1.51 3. n = = In the Bohr model. L n = U so for this state 34 2 3 3.16 10 kg m /s. L = × U = E VALUATE : We will find in Section 41.1 that the modern quantum mechanical description gives a different result. 38.24. I DENTIFY and S ET U P : For a hydrogen atom 2 13.6 eV n E n = − . hc E λ Δ = , where E Δ is the magnitude of the energy change for the atom and λ is the wavelength of the photon that is absorbed or emitted. E XECUTE : 4 1 2 2 1 1 (13.6 eV) 12.75 eV 4 1 E E E Δ = = − = + . 15 8 (4.136 10 eV s)(3.00 10 m/s) 97.3 nm 12.75 eV hc E λ × × = = = Δ . 15 3.08 10 Hz c f λ = = × . 38.25. I DENTIFY : The force between the electron and the nucleus in 3+ Be is 2 2 0 1 , 4 Ze F r π = P where 4 Z = is the nuclear charge. All the equations for the hydrogen atom apply to 3+ Be if we replace 2 e by 2 . Ze (a) S ET U P : Modify Eq.(38.18). E XECUTE : 4 2 2 0 1 8 n me E n h = − P (hydrogen) becomes 2 2 4 2 2 3+ 2 2 2 2 2 0 0 1 ( ) 1 13.60 eV (for Be ) 8 8 n m Ze me E Z Z n h n h n = − = = P P The ground-level energy of 3+ Be is 1 2 13.60 eV 16 218 eV. 1 E = = − E VALUATE : The ground-level energy of 3+ Be is 2 16 Z = times the ground-level energy of H.
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