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Physics 43 HW 11 Ch 39 E: 21, 22, 24, 27, 29, 30, 33 P: 47, 50, 51, 53, 56, 61, 63 39.21. Model: To conserve energy, the absorption spectrum must have exactly the energy gained by the atom in the quantum jumps. Visualize: Please refer to Figure EX39.20. Solve: (a) An electron with a kinetic energy of 2.00 eV can collide with an atom in the n = 1 state and raise its energy to the n = 2 state. This is possible because E2 – E1 = 1.50 eV is less than 2.00 eV. On the other hand, the atom cannot be excited to the n = 3 state. (b) The atom will absorb 1.50 eV of energy from the incoming electron, leaving the electron with 0.50 eV of kinetic energy. 39.22. Model: To conserve energy, the emission and the absorption spectra must have exactly the energy lost or gained by the atom in the appropriate quantum jumps. Solve: (a) (b) From Equation 39.4, the energy of a light quantum is E = hf = hc/λ. We can use this equation to find the emission and absorption wavelengths. The emission energies from the above energy-level diagram are: E2→1 = 4.00 eV, E3→1 = 6.00 eV, and E3→2 = 2.00 eV. The wavelength corresponding to the 2→1 transition is λ 2→1 = hc ( 4.14 × 10 = E21 −15 eV s )( 3.0 × 108 m/s ) 4.00 eV = 311 nm Likewise, λ 3→1 = hc E3→1 = 207 nm, and λ 3→2 = 622 nm. (c) Absorption transitions start from the n = 1 ground state. The energies = in the atom’s absorption spectrum are E1→2 4.00 eV and E1→3 = 6.00 eV. The corresponding wavelengths are λ 1→ 2 = hc E1→ 2 = 311 nm and λ1→3 = hc E1→3 = 207 nm. 39.24. Solve: The Bohr radius is defined as −12 −34 2 2 4πε 0h 2 4π ( 8.85 × 10 C /N m )(1.05 × 10 J s ) aB = = = 5.26 × 10−11 m = 0.0526 nm 2 −31 −19 me2 ( 9.11×10 kg )(1.60 × 10 C ) 2 This differs slightly from the accepted value of 0.0529 nm because of rounding error due to using constants accurate to only 3 significant figures. From Equation 39.29, the ground state energy level of hydrogen is − ( 9 × 109 N 2 m/C 2 )(1.60 × 10−19 C ) −e 2 E1 = = = −2.18 × 10 −18 J = − 13.61 eV 4πε 0 ( 2aB ) 2 ( 5.29 × 10−11 m ) 2 The slight difference from the accepted −13.60 eV is due to rounding error. 39.27. Solve: From Equation 39.30, the energy of the hydrogen atom in its first excited state is E2 = −13.60 eV ( 2) 2 = −3.40 eV The ionization energy of the hydrogen atom in its first excited state (n = 2) is thus 3.40 eV. 39.29. Solve: The units of h are the units of angular momentum, L = mvr. These units are J ⋅ s = kg ⋅ m2 m ⋅ s = kg ⋅ ⋅ m 2 s s 39.30. Solve: Photons emitted from the n = 4 state start in energy level n = 4 and undergo a quantum jump to a lower energy level with m < 4. The possibilities are 4 → 1, 4 → 2, and 4 → 3. According to Equation 39.36, the transition 4 → m emits a photon of wavelength. λ0 91.18 nm λ= = 1 1 1 ⎞ ⎛ ⎞ ⎛ 1 ⎜ 2− 2⎟ ⎜ 2− ⎟ n ⎠ ⎝ m 16 ⎠ ⎝m These values are given in the table below. Transition Wavelength 4→1 97.3 nm 4→2 486 nm 4→3 1876 nm 39.33. Solve: For hydrogen-like ions, Equation 39.37 is rn = n 2 aB ( rn )H = Z Z vn = Z v1 = Z ⋅ ( vn )H n ⎛ 13.60 eV ⎞ 2 En = − Z 2 ⎜ ⎟ = Z ( En ) H n2 ⎝ ⎠ Where (rn)H, (vn)H, and (En)H are the values of ordinary hydrogen. He+ has Z = 2. Using Table 39.2 for the values of hydrogen, we get n rn (nm) vn (m/s) En (eV) 1 2 3 4.38 × 106 2.19 × 106 1.46 × 106 0.026 0.106 0.238 −54.4 −13.6 −6.0 39.47. Model: The energy of a confined particle in a one-dimensional box is quantized. Solve: The energy of the nth quantum state of a particle in a box is En = n2E1, where E1 is the lowest energy level. The energies 12 eV, 27 eV, and 48 eV have the ratios 4:9:16. Thus, they are the n = 2, n = 3, and n = 4 states of an electron that has E1 = 3 eV = 4.8 × 10−19 J. The lowest energy level is ( 6.63 ×10−34 J s ) h2 h2 E1 = L ⇒ = = = 3.5 × 10−10 m = 0.35 nm 8mL2 8mE1 8 ( 9.11 × 10−31 kg )( 4.8 × 10−19 J ) 2 39.50. Model: Photons are emitted when an atom undergoes a quantum jump from a higher energy level to a lower energy level. On the other hand, photons are absorbed in a quantum jump from a lower energy level to a higher energy level. Because most of the atoms are in the n = 1 ground state, the only quantum jumps in the absorption spectrum start from the n = 1 state. hc = ΔEatom , the 3 wavelengths in the absorption spectrum give 2.49 eV, 4.14 eV, Solve: (a) Using Ephoton = λ and 6.21 eV as the energies of the n = 2, 3, and 4 energy levels. (b) The emission spectrum of the atom will contain the following wavelengths: hc hc hc hc λ 31 = = = 300 nm λ 41 = = = 200 nm E3 − E1 4.14 eV E4 − E1 6.21 eV λ 21 = hc hc = = 500 nm E2 − E1 2.49 eV λ 42 = hc hc = = 334 nm E4 − E2 3.72 eV λ 43 = hc hc = = 601 nm E4 − E3 2.07 eV λ 32 = hc hc = = 753 nm E3 − E2 1.65 eV 39.51. Model: Photons are emitted when an atom undergoes a quantum jump from a higher energy level to a lower energy level. On the other hand, photons are absorbed in a quantum jump from a lower energy level to a higher energy level. Because most of the atoms are in the n = 1 ground state, the only quantum jumps in the absorption spectrum start from the n = 1 state. Solve: (a) The ionization energy is E1 = 6.5 eV. (b) The absorption spectrum consists of the transitions 1 → 2 and 1 → 3 from the ground state to excited states. According to the Bohr model, the required photon frequency and wavelength are f = c hc ΔE ⇒ λ= = f ΔE h where ΔE = Ef – Ei is the energy change of the atom. Using the energies given in the figure, we calculated the values in the table below. Ei (eV) Δ E (eV) λ (nm) Transition Ef (eV) 1→2 1→3 −3.0 –2.0 −6.5 −6.5 3.5 4.5 355 276 (c) Both wavelengths are ultraviolet (λ < 400 nm). (d) A photon with wavelength λ = 1240 nm has an energy Ephoton = hf = hc/λ = 1.0 eV. Because Ephoton must exactly match Δ E of the atom, a 1240 nm photon can be emitted only in a 3 → 2 transition. So, after the collision the atom was in the n = 3 state. Before the collision, the atom was in its ground state (n = 1). Thus, an electron with vi = 1.4 × 106 m/s collided with the atom in the n = 1 state. The atom gained 4.5 eV in the collision as it is was excited from the n = 1 to n = 3, so the electron lost 4.5 eV = 7.20 × 10−19 J of kinetic energy. Initially, the kinetic energy of the electron was K i = 12 melecvi2 = 1 2 ( 9.11×10 kg )(1.40 × 106 m/s ) = 8.93 × 10 −19 J 2 −31 After losing 7.20 × 10−19 J in the collision, the kinetic energy is K f = K i − 7.20 × 10−19 J = 1.73 × 10−19 J = 12 melecvf2 ⇒ vf = 2 (1.73 × 10−19 J ) 2Kf = = 6.16 × 105 m/s melec 9.11 × 10−31 kg 39.53. Solve: The radius of the orbit in state n is rn = n2aB. The quantum number is found as follows: B 1 r ( 5.18 nm ) = 48.96 ≈ 49 ⇒ n = 7 n2 = n = 2 aB 0.0529 nm Thus the energy is En = − 13.60 eV 13.60 eV ⇒ E7 = − = −0.278 eV 2 n 49 39.56. Solve: (a) From Equation 39.36, the wavelengths of the emission spectrum are λ n→m = 91.18 nm m −2 − n −2 m = 1, 2, 3, … n = m + 1, m + 2, … For the 200 → 199 transition, λ 200 →199 = 91.18 nm (199 ) −2 − ( 200 ) −2 = 0.362 m (b) For the 2 → 199 transition, λ 2 →199 = 91.18 nm ( 2) −2 − (199 ) −2 = 4.000404071 (91.18 nm) Likewise, λ 2 → 200 = 4.00040004(91.18 nm). The difference in the wavelengths of these two transitions is 0.0000041 × (91.18 nm) = 0.000368 nm. 39.61. Solve: (a) Nearly all atoms spend nearly all of their time in the ground state (n = 1). To cause an emission from the n = 3 state, the electrons must excite hydrogen atoms from the n = 1 state to the n = 3 state. The energy gained by each atom is E3 – E1 = −1.51 eV – (−13.60 eV) = 12.09 eV This means the electrons must each lose 12.09 eV of kinetic energy. Thus the electrons must have at least Kmin = 12.09 eV of kinetic energy to cause the emission of 656 nm light. The minimum speed of the electrons is vmin = 2 (12.09 eV ) 1.6 × 10−19 J 2 K min = × = 2.06 × 106 m/s melec 9.11 × 10−31 kg 1 eV (b) An electron gains 12.09 eV of kinetic energy by being accelerated through a potential difference of 12.09 V. This is simply the definition of electron volt. 39.63. Solve: The energy of the ultraviolet light photon is E= hc λ = ( 6.63 × 10 −34 J s )( 3.0 × 108 m/s ) 70 × 10 −9 m × 1 eV = 17.76 eV 1.60 × 10−19 J To ionize a hydrogen atom, a minimum energy of 13.60 eV is required. Thus, the kinetic energy of the freed electrons is K = 12 mv 2 = E − 13.60 eV = 17.76 eV − 13.60 eV = 4.16 eV