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Home , Ground state, Quantum Jump

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 , the absorption spectrum must have exactly the energy gained by the in the quantum jumps. Visualize: Please refer to Figure EX39.20. Solve: (a) An 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 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

−15 8 hc ()4.14×× 10 eV s() 3.0 10 m/s λ 21→ == =311 nm E21 4.00 eV

Likewise, λ 31→→==hc E 31 207 nm, and λ3→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

λ12→→==hc E 12 311 nm and λ1→3 = hc E13→ = 207 nm.

39.24. Solve: The is defined as 2 2 48.8510π ××−−12 C/N 2 m1.0510 2 34 J s 4πε 0h ()()−11 aB ==2 2 =×=5.26 10 m 0.0526 nm me ()9.11×× 10−−31 kg() 1.60 1019 C 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 of is

92 2 −19 2 −e2 −×()910 N m/C1.6010( × C) E == =−×=−2.18 10−18 J 13.61 eV 1 −11 42πε 0B()a 25.2910()× m 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 in its first is −13.60 eV E2 ==−3.40 eV ()2 2 The 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 mm2 Js⋅ =⋅ kg ⋅=⋅⋅ s kg m ss2

39.30. Solve: 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 of wavelength. λ 91.18 nm λ ==0 ⎛⎞⎛11 11⎞ ⎜⎟⎜22−− 2 ⎟ ⎝⎠⎝mn m16 ⎠ 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

2 a (rn ) v ⎛⎞13.60 eV rn==B H vZ==⋅1 Zv 22 n nn()H EZnn=− ⎜⎟2 = ZE() Z Z n ⎝⎠n H + 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 0.026 4.38 × 106 −54.4 2 0.106 2.19 × 106 −13.6 3 0.238 1.46 × 106 −6.0

39.47. Model: The energy of a confined particle in a one-dimensional box is quantized. 2 Solve: The energy of the nth of a is En = n E1, 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 −19 states of an electron that has E1 = 3 eV = 4.8 × 10 J. The lowest energy level is

−34 2 hh22()6.63× 10 J s EL=⇒== =×=3.5 10−10 m 0.35 nm 1 2 −−31 19 88mL mE1 89.1110()×× kg4.810() J

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 are in the n = 1 ground state, the only quantum jumps in the absorption spectrum start from the n = 1 state. hc Solve: (a) Using E==ΔE, the 3 wavelengths in the absorption spectrum give 2.49 eV, 4.14 eV, photon λ atom 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 41 == =200 nm λ 31 == =300 nm EE41− 6.21 eV EE31− 4.14 eV hc hc hc hc λ 21 == =500 nm λ 42 == =334 nm EE21− 2.49 eV EE42− 3.72 eV hc hc hc hc λ 43 == =601 nm λ 32 == =753 nm EE43− 2.07 eV EE32− 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 =1ground 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 ΔE chc f = ⇒ λ == h f ΔE 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.

Transition Ef (eV) Ei (eV) ΔE (eV) λ (nm) 1 → 2 −3.0 −6.5 3.5 355 1 → 3 –2.0 −6.5 4.5 276 (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 ΔE 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 6 n = 3 state. Before the collision, the atom was in its ground state (n = 1). Thus, an electron with vi = 1.4 × 10 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

2 1123−−16 19 Kmvieleci==×22(9.11 10 kg)( 1.40 × 10 m/s) =× 8.93 10 J After losing 7.20 × 10−19 J in the collision, the kinetic energy is

21.7310× −19 J −−19 19 1 2 2Kf ( ) 5 KKfi=−7.20 × 10 J = 1.73 × 10 J =2 mvvelecff ⇒= = −31 =6.16 × 10 m/s melec 9.11× 10 kg

2 39.53. Solve: The radius of the orbit in state n is rn = n aB.B The is found as follows: r 1 (5.18 nm) n2 ==n 2 =48.96 ≈ 49 ⇒ n = 7 aB 0.0529 nm Thus the energy is 13.60 eV 13.60 eV EE=− ⇒ =− =−0.278 eV n n2 7 49

39.56. Solve: (a) From Equation 39.36, the wavelengths of the emission spectrum are 91.18 nm λ = m = 1, 2, 3, … n = m + 1, m + 2, … nm→ mn−−2− 2 For the 200 → 199 transition, 91.18 nm λ 200→ 199 ==0.362 m ()()199−−22− 200 (b) For the 2 → 199 transition, 91.18 nm λ 2→ 199 ==4.000404071 (91.18 nm) ()2−−22− ( 199 )

Likewise, λ 2200→ = 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 in the ground state (n = 1). To cause an emission from the n = 3 state, the 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

−19 2Kmin 2( 12.09 eV) 1.6× 10 J 6 vmin ==−31 × =×2.06 10 m/s melec 9.11× 10 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

−34 8 hc (6.63×× 10 J s)( 3.0 10 m/s) 1 eV E == × =17.76 eV λ 70×× 10−−91 m 1.60 109 J To ionize a hydrogen atom, a minimum energy of 13.60 eV is required. Thus, the kinetic energy of the freed electrons is

1 2 KmvE==−=−=2 13.60 eV 17.76 eV 13.60 eV 4.16 eV