Practice Modern Physics II, W018, Set 1 Question 1 Energy Level Diagram of Boron ion B + For neutral B, Z = 5 (A) Draw the fine-structure diagram of B + that includes all n = 3 states Label the states in spectroscopic notation(b) Using the selection rules of equation 88 draw all the allowed transitions of B + Such diagrams are called Grotian diagrams The ground state B + has three electrons, electronic configuration is 1s s 1 As shown in class, only the electron in the unfilled subshell s 1 contributes to the angular momentum of the atom We show in class that S = s 1 = 1/,L = l 1 = 0, J = L+ S = S = 1/, with term symbol S 1/ The first excited states electronic configuration is 1s p 1 There is a single electron in the unfilledp 1 ( l 1 = 1) subshell So S = 1/ and L = 1 The possible values of J = L+1/,L 1/ gives two states J = 3/,1/ The states are: 1) S = 1/,L = 1, J = 1/giving P 1/ ; ) S = 1/,L = 1, J = 3/giving P 3/ The next excited state electronic configuration is 1s 3s 1 It should be clear that that S = s 1 = 1/,L = l 1 = 0, J = L+ S = S = 1/, with term symbol 3 S 1/ The next excited states electronic configuration is 1s 3p 1 The 3p 1 electron gives S = 1/and L = 1 The possible values of J = L+1/,L 1/ gives two states J = 3/,1/ The states are: 1) S = 1/,L = 1, J = 1/giving3 P 1/ ; ) S = 1/,L = 1, J = 3/giving3 P 3/ The final excited states for the n = 3 shell has electronic configuration is 1s 3d 1 The ( ) electron gives S = 1/ and L = l 1 = The possible values of 3d 1 l 1 = J = L+1/,L 1/ gives J = 5/,3/ The states are: 1) S =1/, L =, J = 3/ giving 3 D 3/ ; ) S = 1/,L =, J = 5/giving3 D 5/ The schematic energy diagram is below: Just like the p states, Hund s rule 3 suggests has lower energy than Using Hund s rule 3, since the p (3p) subshell is less than half filled P 1/ ( 3 P 1/ ) has lower energy than P 3/ ( 3 P 1/ ) s 1 S 1/ In the diagram, a thick line indicates a state of B +, with the term symbol, n L J, on the right of the line The electronic configurations are on the left of the lines The
double-arrow lines that connect two states indicate allowed transitions that obey the transition rules of equation 88 Question Atomic Physics (Fine Structure and Zeeman Effects) Excited states of atoms, fine-structure, and Zeeman effects The ground-state configuration of SODIUM (Z = 11) is1s s p 6 3s 1, in term symbol is 3 S 1/ A) Write is the first excited state electronic configuration of Sodium Find the spectroscopic notation (term symbol) of the two states associated with this excited state Hint: find all the total orbital Angular Momentum (L) and total spin (S), and all the possible total angular momentum (J) Also see equation sheet for energy order!! The electronic configuration of the first excited state is1s s p 6 3p 1 For this configuration, only the 3p 1 electron contribute to the angular momentum, and hence S = 1 /, L = 1 The total angular momentum is J = L + S, L S There are two states: i) J = L + S = 3 / with S = 1 / and L = 1 The term symbol is 3 P 3/ ii) J = L S = 1 / with S = 1 / and L = 1 The term symbol is 3 P 1/ B) Write the second excited state electronic configuration of SODIUM Find the spectroscopic notation (term symbol) of the one state associated with this excited state Briefly explain why there is only one state Look at the energy order list in back! The electronic configuration of the second excited state is1s s p 6 4s 1 For this configuration, only the 4s 1 electron contribute to the angular momentum, and hence S = 1 /, L = 0 There is only value of total angular momentum is J = S = 1 / For J = 1 /, S = 1 / and L = 0 The term symbol is 4 S 1/ C) The first excited state of LITHIUM (Z = 3) has electronic configuration 1s p 1, and is split into two states (doublet) P 1/ and P 3/ It is estimated that the internal magnetic field for this state is B int = 036T Find the fine-structure energy difference between the P 1/ and P 3/ states Fine-structure splitting is ΔV fs B int = 5788 10 5 ev / T ( ) 036T ( ) = 417 10 5 ev D) Suppose the LITHIUM atom is placed in an external magnetic field of magnitude B = 05T How many energy levels do the P 3/ have? Draw a schematic diagram to illustrate these energy levels Find the energy spacing between adjacent energy levels HINT: This is the Zeeman Effects! Lande factor is relevant! The Zeeman energy of these levels are given by V z B ext g For P 3/, J = 3 / and = 3, 1, 1, 3 giving four distinct energy levels
E P 3/ = 1 / The energy separation between adjacent levels (for example m j = 3 / and m j = 1 /, or m j = 0 and m j = 1 / ) is simply, ΔV z B ext gδm j B ext g, where Δm j = 1 for adjacent levels For P 3/ state J = 3, L = 1, and S = ½ ( ) + 05( 05 +1) 1 ( 1+1 ) ( )( 15 +1) 15 15 +1 g = 1+ 15 = 133 For B ext = 050T, we use µ B = 5788 10 5 ev / T and g = 133 to obtain for the energy difference between adjacent Zeeman P 3/ levels, ΔV z gb ext = 5788 10 5 ev / T 05T = 3 / ( )( 133) = 384 10 5 ev = 1 / = 3 / Question 3 Molecular Spectroscopy: It is known that the hydrogen molecule H has a vibration absorption (emission) frequency of ν 0 = 13 10 14 Hz A) Model the H molecule as two H atoms connected by a spring Based on the data given calculate the spring constant k Use mh = 1u Reduced mass µ = m m H H = 1 m H + m H m = H 833 10 8 kg ω = k µ ν = 1 π k µ k = 4π ν µ = 4π ( 13 10 14 s 1 ) ( 833 10 8 kg) = 573 N m B) Now consider a deuterium molecule D, where D is a heavy hydrogen with nucleus of one proton and one neutron with a mass md = amu Use the data of part A to calculate the vibration frequency of this molecule µ = m Dm D m D + m D = 1 m D = 167 10 7 kg ν = 1 π k µ = 1 π 573N / m 167 10 7 kg = 93 1013 Hz C) Are H and D infrared active? Briefly explain your answer No Only vibration modes that induce a change in the dipole moment of the molecule are infrared active Since homonuclear diatomic molecules such as H and D do not even have dipole moments, they cannot be infrared active
Question 4 Vibrational Energy Level of oxygen molecule O A) Assume that the O molecule behaves like a harmonic oscillator with a force constant of 10 N/m Find the energy (in ev) of its ground (n = 0) and first excited (n = 1) vibrational states For 16 O, m O = 16 u, where 1 u = 166 10 7 kg For 16 O, m O = 1 6u, where 1 u = 166 10 7 kg First m O = 1 6u 166 10-7 kg / u Reduced mass µ = ( ) = 66 10 6 kg ( ) m m O O = m O m O + m O 66 10 6 kg = = 133 10 6 kg The fundamental vibrational frequency is ω = k, where k = 10 N/m µ ω = 10N / m 133 10 6 kg = 16 1014 Hz, E vibr = n +1/ ( )!ω, n = 0,1, ( )( 16 10 14 Hz) Ground state v = 0, E 0 =!ω = 1055 10 34 J s = 663 10 1 J 1 ev In ev, ΔE 0 = 663 10 1 J 16 10 19 J = 00415eV (05 point) First excited state v = 1, ( )( 16 10 13 Hz) E 1 = 3hν 0 = 3 1055 10 34 J s = 199 10 0 J (05 point) In ev, ΔE 1 = 199 10 1 1 ev J 16 10 19 J = 014eV (05 point) B) Find the vibration quantum number that approximately corresponds to its 15-eV dissociation energy Hint: see dissociation equation on the equation sheet E vibr = ( n +1/ )!ω U 0, where U0 is the dissociation energy The dissociation energy is found when E vibr = 0, which gives 0 = ( n +1/ )!ω U 0 n = U 0!ω 1 Using U 0 = 15eV, ω = 16 10 14 Hz, and h = 4136 10 5 ev s The vibrational quantum number that approximately corresponds to its 15-eV dissociation energy is v = U 0!ω 1 = 15eV 658 10 16 ev s After rounding off v = 18 ( )( 16 10 14 Hz) 1 = 176 C) Is O infrared active? Briefly explain your answer For a vibrational mode to be active, the vibration must change the dipole moment Since O is a homonulcear diatomic molecule, it possesses no dipole moment, and its one stretch vibration mode does not change the dipole Hence it is infrared inactive
Question 5 Microwave Spectroscopy: The rotational transition from the l = to the l = 1 state in CO is accompanied by the emission of a 955 10 4 ev photon A) Use this information to find the rotational inertia of the CO molecule Use E l = l( l +1) " I, and the transition For the l l 1transition, the change of energy is ΔE = E l E l 1 = " l I, which equals the photon energy is ΔE = 955 10 4 ev 16 10 19 J i ev 1 = 153 10 J ==! l I h = 666 10 34 J i s or! = h π = 1054 10 34 Jis, and l =, I =! l ΔE = 146 10 46 kg i m I =! l ΔE Use B) What is the bond length between the C and O atoms Data: Mass of carbon m C = 1u ; Mass of oxygen m O = 16u ; 1 u = 166 10 7 kg Mass of carbon m C = 1 u 166 10-7 kg / u Mass of oxygen m O = 1 6u 166 10-7 kg / u Reduced Mass µ = m C m O m C + m O = ( ) = 199 10 6 kg ( ) = 66 10 6 kg ( 199 10 6 kg) ( 66 10 6 kg) 199 10 6 kg + 66 10 6 kg = 114 10 6 kg I = µr R = I / µ = 146 10 46 kg m /114 10 6 kg = 113 10 10 m