Chapter 1 - Solutions 1. a. Find the quantum number of the initial energy state nf = 6 (fifth excited state), = 5910 nm c = 1 1 = R 2 2 = 3.29 1015 Hz n f n i 1 1 2 2 n f n i 1 1 c = 3.29 1015 Hz 2 2 λ n f n i 1 3.00 108 ms 109 nm 1 = 3.29 1015 Hz 2 2 5910 nm 1m n i 6 1 1 0.0154 2 6 n i2 1 0.0123 ; ni = 9, therefore the emitted photon corresponds to the third n i2 lowest-energy spectral line in the Humphreys series. b. B A Lymar (nf = 1) C Balmer (nf = 2) Paschen (nf = 3) D Humphreys (nf = 6) n=9 n=8 D n=7 n=6 n=5 C n=4 spectral line A spectral line B Eatom spectral line C n=3 A spectral line D B n=2 n=1 2 2 h 2 n 2x n y n Z 8m L2x L2y L2 Z 3. a. Enxnynz = (6.626 10 34 Js) 2 (1012 pm) 2 (8)(9.11 10 31 kg) 2 n 2y n 2Z nx (550) 2 (700) 2 (550) 2 n2 n 2y n 2Z x = 6.024 1014 J 302500 490000 302500 E211 = E112 = 1.12 1018 J 2nd excited state E121 = 8.90 1019 J 1st excited state E131 = 1.50 1018 J b. 112, 211 E 121 111 c. There is one node in the first excited state. d. There is one node in the second excited state. 5. Atom He N Na P Ground-state configuration 1a2 1a21b42a1 1a21b42a22b3 1a21b42a22b42c3 Number of unpaired electrons 0 1 1 3 7. a. Set (,) = 0 and solve for values of and that make this true. sin() = 0 when = 0 and a line, not a nodal plane cos(2) = 0 when 2 = /2, 3/2, 5/2, 7/2 and when = /4, 3/4, 5/4, 7/4 Hence, the nodal planes are defined by = /4 and 5/4 and = 3/4 and 7/4, but not the = 0 and line. b. Set R(r) = 0 and solve for values of r that make this true. R(r) = 0 = 2 (2 2) exp[/4] is true, when 2 2 = 0 The roots are = 0 (which is not a radial node) and = 2. Hence, r = 2a0 is the only radial node c. l = # of angular nodes = 2 n – l – 1 = # of radial nodes; 1 = n – 2 – 1; n = 4 9. a. S 999.6 kJ/mol Ca 589.8_______ Cl 1251______ Sc 631_________ Ar 1520______ Ti 658 kJ/mol___ b. S 2251 kJ/mol Ca 1145______ Cl 2297_____ Sc 1235______ Ar 2666_____ Ti 1310 kJ/mol K 418.8______ K 3051______ 11. a. Zeff = Force of attraction between nucleus and electron F = Q1 Q2 4or2 Where Q1 represents the charge of the nucleus Where Q2 represents the charge of the electron o is the permittivity of free space = 8.85419 1012 C2N1m2 r is the distance between charges (radius of the energy level) for Be F = Q1 Q2 = [4(1.6022 1019 C)] [1.6022 1019 C] 4or2 4(8.85419 1012 C2N1m2) 112 pm F = 8.24 1018 1m 11012 pm N For Mg F = Q1 Q2 = [12(1.6022 1019 C)] [1.6022 1019 C] 4or2 4(8.85419 1012 C2N1m2) 160 pm 1m 12 110 pm F = 1.73 1017 N For Ca F = Q1 Q2 = [20(1.6022 1019 C)] [1.6022 1019 C] 4or2 4(8.85419 1012 C2N1m2) 197 pm 1m 11012 pm F = 2.34 1017 N b. As the force of attraction between the nucleus and the outermost electron (Zeff) increases because of the increase in the number of protons in the nucleus, the number of electrons existing within the atom also increases. Electron-electron repulsion forces are also increased within the atom. These electron-electron repulsion forces work against the Zeff resulting in electrons in outer energy levels actually experiencing less attraction to the nucleus, so the amount of energy required to remove the electron from the atom (the ionization energy) is reduced as the number of electrons in the atom increases.