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Σ=Zr/A0 A0=0.053Nm Θ Is the Polar Angle Down from Z

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Σ=Zr/A0 A0=0.053Nm Θ Is the Polar Angle Down from Z s=Zr/a0 a0=0.053nm q is the polar angle down from z FIGURE 12.23: Orbital Energy Levels for H Atom The energies are given by this expression– and only depend on n, not on the other quantum numbers. 2 4 2 $ Z me ' $ Z ' 18 E 2.18*10 − J = −& 2 2 2 ) = −& 2 ) % n 8ε0 h ( % n ( € l: Angular Momentum ml: The Direction of the Angular Momentum l ranges from 0 to n-1; ml ranges from –l to l It is possible to prove that there are n2 orbitals for any value of n PollEverywhere Question 51 Statements from Nice Chem websites about nodes. Which is true? 1. For s orbitals, n is the number of angular nodes. 2. Radial nodes are always spherical and angular nodes are planar. 3. Thus, n tells how many radial nodes an orbital will have. A. T T F B. T F T C. F T T D. T F F E. F F F Nodes in Wavefunctions • Node: wave function is zero on a plane or a sphere or a cone or another geometric surface, wave function changes sign on crossing this boundary • Total number of nodes for yn is n-1 Angular vs. Radial Nodes The shapes of the probability distribution, angular vs. radial nodes, relate to the value of quantized angular momentum for the orbital “l”. Angular nodes are functions of angles: e.g. planes or cones There are l angular nodes (0 for s, 1 for p, 2 for d etc.) Radial nodes: function of radius e.g. spheres. There are (n-l-1) radial nodes. From Wikipedia Orbitals are often depicted in terms of the 90% or 95% or sometimes 67% probability surface Angular vs. Radial Nodes 4p vs. 2p at right. Which is which? http://winter.group.shef.ac.uk/orbitron/AOs/2p/index.html Node for 3dz2 How many nodes does this wave function have? Where does this wavefunction have a node? Radial? Angular? A 0, radial B 1, r= 2a0. radial C 1, q=0, angular D 2, q=arccos(+/-sqrt(1/3)), angular E 2, r= 1.9, 7.1, angular Node for 3dz2 How many nodes does this wave function have? Where does this wavefunction have a node? Radial? Angular? A 0, radial B 1, r= 2a0. radial C 1, q=0, angular D 2, q=arccos(+/-sqrt(1/3)), angular E 2, r= 1.9, 7.1, angular 3 cos2 q - 1 = 0 q= Arc.cos (+/- sqrt (1/3)) q= 0.9553 rad or 2.1862 rad q = 54.73 deg or 125.248 deg Two cones, at “magic angle” and negative magic angle Node for 3dz2 How many nodes does this wave function have? Where does this wavefunction have a node? ANS: Its found at a specific angle or angles, so they are angular nodes. This is a cone above, and one below. arccos 1 .995, 2.19rad A 0, θ = (± 3 ) = B 1, r= 2a 0 = 54.735,125.478degrees C 1, q=0 D 2, cos2q=1/3 E 2, r= 1.9, 7.1 STATEMENTS FROM NICE CHEM WEBSITES: WHICH TRUE, WHICH IS FALSE? 1. For s orbitals, n is the number of angular nodes 2. Radial nodes are spherical and angular nodes are always planar 3. Thus, n tells how many radial nodes an orbital will have A. T T F B. T F T C. F T T D. T F F E. F F F CAVEAT EMPTOR… STATEMENTS FROM “NICE LOOKING” CHEM WEBSITES, CORRECTED 1. For s orbitals, n-1 is the number of angular nodes 2. Radial nodes are spherical and angular nodes are planar, conical sections or other angular dependent surfaces…. 3. Thus, n-1- l tells how many radial nodes an orbital will have PollEverywhere Question 52 Ionization Energies The (1st) ionization is the minimum energy needed to release an electron from the atom. The 2nd ionization energy is the minimum energy needed to remove a second electron. Which of the following statements is correct? + • For the transitions n1 ® n2, the frequency is larger for H than for He . • The ionization energy of the H atom is smaller than the second ionization energy of the He atom. • The 1s orbital in He+ is larger (in the sense that the probability density is shifted outward) than the 1s orbital in H. 2 4 2 $ Z me ' $ Z ' 18 E 2.18*10 − J = −& 2 2 2 ) = −& 2 ) % n 8ε0 h ( % n ( € • b) The ionization energy of the H atom is smaller than the second ionization energy of the He atom. • Section 12.4, The Bohr Model • In both cases, only one electron is present. For helium, Z = 2; for hydrogen, Z = 1. Ionization energy is related to Z2, so the second ionization energy for helium is 4 times as great as the ionization energy for the hydrogen atom. PollEverywhere Question 53 Transition Energies and Light Frequency A Li2+ atom undergoes the n=3 to n=1 transition. What would the energy of the emitted light be? -17 2 4 2 A) 3.3 x 10 J $ Z me ' $ Z ' −18 E = − 2 = −& )2.18 *10 J B) 1.7 x 10 17 J & 2 2 ) 2 % n 8ε0 h ( % n ( C) 9.0 x 10-17 J D) 1.7 x 10-17 J E) 1.6 x 10-18 J € Can we use the provided equation for Li2+? (Is Li2+ a one electron atom?) 2 4 2 $ Z me ' $ Z ' 18 E 2.18*10 − J = −& 2 2 2 ) = −& 2 ) % n 8ε0 h ( % n ( € Emission. Final energy is lower, photon is given off. DE= Ei - Ef is positive when f >i, higher energy minus lower energy. 2 2 2 2 2 2 2 Ep = ( [-(Z /ni )R ] - [-(Z /nf )R]) = Z (1/nf - 1/ni )R 2 2 2 -18 Ep= Z (1/nf - 1/ni )2.18*10 J • Photon energy is positive and is equal to energy lost in molecule. Absorption. Final energy is higher. Photon is absorbed. 2 2 2 -18 Ep= Z (1/ni - 1/nf )2.18*10 J. • Photon energy is positive and is equal to energy gain in molecule. A Li2+ atom undergoes the n=3 to n=1 transition. 2 2 2 -18 Ep = Z (1/nf - 1/ni )2.18*10 J 2 2 2 -18 -17 Ep = 3 (1/1 - 1/3 )R = 8 x 2.18 x 10 J = 1.7 x 10 J = Ep. PollEverywhere Question 53 Transition Energies and Light Frequency A Li2+ atom undergoes the n=3 to n=1 transition. What would the energy of the emitted light be? 2 2 2 -18 -17 Ep= Z (1/nf - 1/ni )2.18*10 J = 1.7 x 10 J A) 3.3 x 10-17 J B) 1.7 x 10 17 J C) 9.0 x 10-17 J D) 1.7 x 10-17 J E) 1.6 x 10-18 J PollEverywhere Question 54 Transition Energies and Light Frequency A Li2+ atom undergoes the n=3 to n=1 transition. What would the -17 wavelength of the emitted light be? Recall Ep= DE = 1.7 x 10 J 8 -34 2 -1 -34 Ep=hn; ln= c; c = 3x10 M/s; h = 6.6310 kg M s = .6310 Js A) 114 nm or 1140 Angstroms is in the UV range B) 11.4 nm or 114 Angstroms is in the soft X-ray range C) 1.1 nm or 11.4 A is in the X-ray range D) 1.1 A or 114 Pm is in the X-ray range E) 0.11A or 11.4 Pm l=hc/DE=(6.63 x 10-34Js)(3x108 M/s)/17.4 x 10-18 J =1.14 x 10-8 M −34 2 8 ⎡6.6x10 Kgm / s•3x10 m / s⎤ −8 λ = hc / E = ⎢ ⎥ =1.1x10 m ⎣ 1.7x10−17 Kgm2s−2 ⎦ 11.4 nM or 114 Angstroms is in the soft X-ray range PollEverywhere Question 54 Transition Energies and Light Frequency A Li2+ atom undergoes the n=3 to n=1 transition. What would the -17 wavelength of the emitted light be? Recall 1.7 x 10 J = DE=Ep; 8 -34 2 -1 -34 Ep=hn; ln= c; c = 3x10 M/s; h = 6.6310 kg M s = .6310 Js A) 114 nm or 1140 Angstroms is in the UV range B) 11.4 nm or 114 Angstroms is in the soft X-ray range C) 1.1 nm or 11.4 A is in the X-ray range D) 1.1 A or 114 Pm is in the X-ray range E) 0.11A or 11.4 Pm PollEverywhere Question 55 Transition Energies and Light Frequency For which of the following electron transitions in a hydrogen atom does the emitted light have the longest wavelength? 2 2 2 -18 Ep= Z (1/nf - 1/ni )2.18*10 J a) n = 4 to n = 3 b) n = 4 to n = 2 c) n = 4 to n = 1 d) n = 3 to n = 2 e) n = 2 to n = 1 ⎛ 1 1 ⎞ E Z 2 2.18*10−18 J Δ = ⎜ 2 − 2 ⎟ ⎝ ni n f ⎠ • a) n = 4 to n = 3 • Section 12.4, The Bohr Model • Energy is inversely related to wavelength: the smaller the energy change, the longer the wavelength. The transition n = 4 to n = 3 has the smallest energy change of all of the options. Question 56 The Hydrogen Lamp A strong line is predicted for H n=1 to n=2 at 121.57 nm and is observed experimentally.
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