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 00 1 (92 ) (2 ) 1 (184) (1.60 10 C) 5.54 10 m. 44 ( 4 . 7 8 1 0 J C ) ee r K ππ × == = × × PP (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 61 9 14 (4.78 10 eV) (1.6 10 J ev) 13.8 N. (5.54 10 m) K F r ×× = × 38.21. (a) IDENTIFY: If the particles are treated as point charges, 12 0 1 . 4 qq U r π = P SET UP: 1 2 qe = (alpha particle); 2 82 = (gold nucleus); r is given so we can solve for U . EXECUTE: 19 2 92 2 1 3 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) IDENTIFY: Apply conservation of energy: 11 2 2 . KU KU += + SET UP: 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 = EXECUTE: Conservation of energy thus says 13 5.82 10 J 3.63 MeV. KU × = (c) 2 1 2 Km v = so 13 7 27 22 ( 5 . 8 2 1 0 J ) 1.32 10 m/s 6.64 10 kg K v m × = × × EVALUATE: / 0.044, vc = 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 = 34 2 1.05 10 kg m s. ×⋅ 38.23. IDENTIFY and SET UP: Use the energy to calculate n for this state. Then use the Bohr equation, Eq.(38.10), to calculate L . EXECUTE: 2 (13.6 eV)/ , n En =− so this state has 13.6/1.51 3. n In the Bohr model. Ln = U so for this state 34 2 33 . 1 6 1 0 k g m / s . L U= EVALUATE: We will find in Section 41.1 that the modern quantum mechanical description gives a different result. 38.24. IDENTIFY and SET UP: 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 is the wavelength of the photon that is absorbed or emitted. EXECUTE: 41 (13.6 eV) 12.75 eV EE E ⎛⎞ − = = + ⎜⎟ ⎝⎠ . 15 8 (4.136 10 eV s)(3.00 10 m/s) 97.3 nm 12.75 eV hc E × = Δ . 15 3.08 10 Hz c f × . 38.25. IDENTIFY: The force between the electron and the nucleus in 3+ Be is 2 2 0 1 , 4 Ze F r = 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) SET UP: Modify Eq.(38.18). EXECUTE: 4 0 1 8 n me E nh P (hydrogen) becomes 4 3 + 2 1( ) 1 1 3 . 6 0 e V (for Be ) 88 n mZe me EZ Z n = = The ground-level energy of 3+ Be is 1 2 13.60 eV 16 218 eV. 1 E = EVALUATE: The ground-level energy of 3+ Be is 2 16 Z = times the ground-level energy of H. (b) SET UP: The ionization energy is the energy difference between the n →∞ level energy and the 1 n = level energy.

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38-6 Chapter 38 EXECUTE: The n →∞ level energy is zero, so the ionization energy of 3+ Be is 218 eV.
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1061_PartUniversity Physics Solution - Photons, Electrons,...

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