Homework #5 Chapter 12 Quantum Mechanics and Atomic Theory 13. Z is the atomic number or number of protons. Zeff is the effective nuclear charge or the nuclear charge after taking into account the shielding caused by electrons in the atom. 14. The ionization energy is the energy it takes to remove an electron from an atom. As one goes from left to right across the periodic table in one row, electrons are added to the same shell. As each electron is added a proton is also added to the nucleus. Because the electrons are added to the same shell the shielding of these electrons on the nucleus is minimal. This causes the effective nuclear charge to go up. The larger the positive effective nuclear charge the tighter the electrons are held to the nucleus causing a greater amount of energy to be needed to remove an electron. As you go down a period you are adding electrons to a new shell. Each new shell causes substantial shielding of the nucleus. In addition, the electrons are farther away from the nucleus causing less energy to be needed to remove a valence electron. 21. 1.0 0.010 2.998 10 3.010 0.010 Energy of 1 photon 6.626 10 ∙ 3.0 10 2.010 Energy of Avogadro’s number of photons 6.022 10 2.0 10 12 22. Wave a has a longer wavelength than wave b. Wave b has the higher frequency than wave a. Wave b has the higher energy than wave a. In 1.6×10‐3 m wave a completes 4 cycles. 1.6 10 4.010 4 2.998 10 7.510 4.0 10 6.626 10 ∙ 7.5 10 4.910 In 1.6×10‐3 m wave b completes 8 cycles. 1.6 10 2.010 8 2.998 10 1.510 2.0 10 6.626 10 ∙ 1.5 10 9.910 All electromagnetic radiation travels at the same velocity, the speed of light. Both waves a and b are infrared radiation. 1 27. a) 2.998 10 5.010 6.0 10 b) Infrared c) 6.626 10 ∙ 6.0 10 4.010 This is the energy per photon, calculate the energy per mol of photons 4.0 10 6.022 10 2.410 d) This bond absorbs radiation with a smaller frequency; therefore, the radiation is less energetic. 29. Calculate the energy needed to remove 1 electron 279.7 4.645 10 . Calculate the wavelength of the photon needed 6.626 10 ∙ 2.998 10 4.277 10 4.645 10 427.7 30. Need to determine is 225 nm light can ionize gold (890.1 ) 225 2.2510 Determine the energy of 225 nm light 6.626 10 ∙ 2.998 10 8.8310 2.25 10 Calculate the energy per mole of photons . 8.83 10 5.3210 532 Therefore, 225 nm light cannot ionize gold. 31. (total energy)=(energy to remove an e‐) + (kinetic energy of e‐) Calculate total energy 6.626 10∙2.998 10 7.8210 254 10 Calculate the energy needed to remove an e‐ from the surface 208.4 3.4610 . Calculate kinetic energy 7.82 10 3.46 10 4.36 10 32. Plank studied black body radiation and noted that short wavelength of light were not emitted from hot systems. This is inconsistent with classical theories because there should be no reason that small wavelength of light would not be given off. Therefore, electromagnetic radiation has both wave and matter like properties. In addition, the photoelectric effect showed that it took photons with a minimum frequency before electrons were seen to eject from metal. This held 2 true even if the intensity (number of photons) was increased but the frequency was below the threshold value. Matter was shown to have both particle and wave properties using the double slit experiment. When a beam of electrons is forced between two slits a diffraction pattern can be observed which is consistent with matter having wave properties. The wavelength of matter is calculated by therefore, the more massive the particle the shorter the wavelength and the less wave like the matter acts. You need to consider the wave properties of a system when the wavelength is greater than the particle size. . ∙ 34. a) 2.410 0.024 . .. . ∙ 3.410 3.410 b) . 1.0 10 1.010 36. 6.626 10 ∙ 7,300 9.109 10 1.0 10 1.0 1.0 10 6.626 10 ∙ 7.310 9.109 10 1.0 10 37. 3.31 10 3.3110 6.626 10 ∙ 6.6810 3.31 10 0.01002.998 10 Determine molar mass 6.68 10 6.022 10 0.0402 40.2 The element is calcium 42. ∆ 2.178 10 a) ∆ 2.178 10 1.059 10 Therefore, the photon emitted would carry 1.059×10‐19 J and 6.626 10 ∙ 2.998 10 1.876 10 1.059 10 Infrared 3 b) ∆ 2.178 10 4.901 10 Therefore, the photon emitted would carry 4.901×10‐20 J 6.626 10 ∙ 2.998 10 4.053 10 4.901 10 Infrared c) ∆ 2.178 10 1.549 10 Therefore, the photon emitted would carry 1.549×10‐19 J 6.626 10 ∙ 2.998 10 1.282 10 1.549 10 Infrared 40. The final energy state will be n=∞, which corresponds to an electron not being associated with the atom anymore, and will result is Ef=0. n 1 ∞ 1 1 ∆ 2.178 10 2.178 10 2.178 10 ∞ 1 and 6.626 10 ∙ 2.998 10 9.121 10 91.21 2.178 10 n 3 ∞ 1 1 ∆ 2.178 10 2.178 10 2.420 10 ∞ 3 6.626 10 ∙ 2.998 10 8.209 10 820.9 2.420 10 45. a) This statement is false. The higher the energy orbital the electron is in, the farther away from the nucleus and the easier (the less energy) it is to remove the electron. b) This statement is true. The greater the value of n, the higher the energy level, and the farther it is away from the nucleus. c) This statement is false. n=3 n=1 is a larger jump than n=3 n=2. Therefore, more energy will be released for the n=3 n=1 jump. The larger the energy, the greater the frequency of the photon; the greater the frequency of the photon, the smaller the wavelength. d) This statement is true. In order for an electron transitions from the ground state to the n=3 state the electron must absorb a specific amount of energy. When the electron falls back down to the ground state (n=3 n=1) the electron emits the same amount of energy that it absorbed. e) This statement is false. The ground state of the hydrogen atoms is n=1 and the first excited state is n=2. 46. n=1 n=5 1 1 ∆ 2.178 10 2.178 10 5 1 2.091 10 and 4 6.626 10 ∙ 2.998 10 9.500 10 95.00 2.091 10 No visible light does not have enough energy to promote an e‐ from n=1 n=5 state. n=2 n=6 1 1 ∆ 2.178 10 2.178 10 6 2 4.840 10 6.626 10 ∙ 2.998 10 4.104 10 410.4 4.840 10 Yes visible light does have enough energy to promote and e‐ from n=2 n=6 state. 48. 6.626 10 ∙ 6.90 10 4.5710 ∆ 2.178 10 Light was emitted therefore ΔE is negative 1 1 4.57 10 2.178 10 5 2 50. When an electron is removed from an atom the final n value is n=∞ a) ∆ 2.178 10 2.178 10 2.178 10 You need 2.178×10‐18 J in order to remove the e‐ from the ground state of H. Calculate the energy per mol 6.022 10 2.178 10 1,312,000 1,312 b) ∆ 2.178 10 2.178 10 8.712 10 You need 8.712×10 ‐18 J in order to remove the e‐ from the ground state of He+. Calculate the energy per mol 6.022 10 8.712 10 5,246,000 5,246 c) ∆ 2.178 10 2.178 10 1.960 10 You need 1.960×10 ‐18 J in order to remove the e‐ from the ground state of Li2+. Calculate the energy per mol 6.022 10 1.960 10 1.180 10 1.180 10 d) ∆ 2.178 10 2.178 10 7.841 10 You need 7.841×10‐17 J in order to remove the e‐ from the ground state of C5+. Calculate the energy per mol 6.022 10 7.841 10 4.722 10 4.722 10 5 e) ∆ 2.178 10 2.178 10 1.472 10 You need 1.472×10‐15 J in order to remove the e‐ from the ground state of Fe25+. Calculate the energy per mol 6.022 10 1.472 10 8.866 10 8.866 10 253.4 2.534 10 51. and 6.626 10 ∙ 2.998 10 7.839 10 2.534 10 ∆ 2.178 10 Light was emitted therefore ΔE is negative 7.839 10 2.178 10 4 53. ∆∆ 1 ∆∆ 2 6.626 10 ∙ 1.0510 ∙ 2 2 a) ∆∆ 1.05 10 ∙ ∆9.109 10 0.100 2 ∆ 5.76 10 This is the minimum value of Δx b) ∆∆ 1.05 10 ∙ ∆0.145 0.100 2 ∆ 3.62 10 This is the minimum value of Δx The diameter of a hydrogen atom is 74 pm= 7.4×10‐11m (pg.