Chem 1061 1st Edition Lecture 19 Outline of Last Lecture 1. Calculating Frequency2. Quantum Theory3. Rydberg Equation4. Atomic Model5. Bohr’s Model6. Bohr’s Model of the Hydrogen AtomOutline of Current Lecture 7. Energy Levels in H atom8. Ionization Energy9. Conversions10. Energy Levels in H Atom11. Absorbance12. Wave Particle DualityCurrent LectureEnergy Levels in H atomsE(n) = -2.18 x 10^-18 (z^2/n^2) z= number of protonsE(n) = quantum number for stationary stateOnly for one electron species Energy Difference between Two Levels∆E = E(final) = E(initial)∆E = -2.18 x 10^-18 J [z^2/n(final)^2 –z^2/n(initial)^2] N(final) < N(initial) Atom EMITS energy ∆E is NEGATIVEN(final) > n(initial) Atom ABSORBS energy ∆E is POSITIVE]These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Example: Calculate ∆E for transition (movement of e- between the states) from n=3 to n=2 in a H atom.-2.18x10^-18J[(1^2/2^2) – (1^2/3^2)] = ∆E-2.18 x 10^-18J(1/4-1/9)=-3.0278 x 10^-19 J/atom∆E(Atom) = E(photon)Ionization EnergyIonization: remove e- completelyH(g) H+(g) + e-Transition from e- in n=1 to n= infinity∆E = E(final) – E(initial)H atom∆E = -2.18 x 10^-18 J [(1^2 / infinity^2) – (1^2/1^2)∆E = -2.18 x 10^-18 J (0-1)∆E= 2.18 x 10^-18Convert J/atom KJ/mol2.18 x 10^-18/1atom x 6.022 x 10^23 atoms/1mol x 1 KJ/1000J = 1.31 x 10^3 KJ/molEnergy Levels in H atomE(n) = -2.18 x 10^-18 J (z^2/n^2)E(1)=-2.18 x 10^-18E(2)=-5.4 x 10^-19E(3)=-2.4 x 10^-19E(10)=-2.18 x 10^-20Absorbance (pg 300-301)The energy absorbedEnergy absorbed is proportional to concentration ( # of molecules)A(1)/c(1) = A(2)/C(2)A=absorbanceC= concentration Wave Particle Duality of Matter and energyE=hc/wavelengthRelates energy of a photonE=mc^2 Mass- energy equivalenceMc^2= hc/wavelengthRearrange to solve for wavelengthWavelength = h/muc u because it’s the speed of a particle nowmass needs to be in
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