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Get Full Access to An Introduction To Thermal Physics - 1 Edition - Chapter 5 - Problem 20p
Get Full Access to An Introduction To Thermal Physics - 1 Edition - Chapter 5 - Problem 20p

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# The first excited energy level of a hydrogen atom has an

ISBN: 9780201380279 40

## Solution for problem 20P Chapter 5

An Introduction to Thermal Physics | 1st Edition

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Problem 20P Problem 20P

The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of S = k ln 4, since for this given value of the energy, the multiplicity is 4. Question: For what temperatures is the Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what temperatures is it negative? (Comment: When F for the level is negative, the atom will spontaneously go from the ground state into that level, since F = 0 for the ground state and F always tends to decrease. However, for a system this small, the conclusion is only a probabilistic statement; random fluctuations will be very significant.)

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4/4/16 EXAMAPRIL14 th Ø 4/12NickTreichishostingareviewsession@415pminFulmer125 Ø 4/12Finneganishostingareviewsessioninthepit@7 Ø 4/11@515theChemclubishostingareviewsession AXE • • n=thenumberofbondinggroupsonthecentralatom • m=thenumberoflonepairsonthecentralatom • ANYtypeofbond(single,double,triple)isonebondinggroup • n+m=thenumberofelectrongroupsonthecentralatom ElectronGeometry • Thenumberofelectrongroupsdeterminesthis • Thereareonly5,seetable10.1 • Determinestheidealbondangles-elementstryandarrangethemselvesasfar awayfromeachotheraspossiblebecausetheelectronsrepulsethemselves MolecularGeometry(shape)-electrongeometryandthenumberoflonepairs determinethemoleculargeometry • Themolecularshape,thebonpolarities,andtheformalchargedistribution determinethemolecularpolarity 4/6/16 BondAngles • ElectronGeometrydeterminestheidealbondangles(theyareapproximate) Linear:180degreeangles Triganolplanar:120degreeangles Tetrahedral:109.5degreeangles Trigonalbipyramidal:90,120degreeangles Octahedral:90degreeangles • Lonepairstakeupmorespacethanbondingpairs • Doublebondstakeupmorespacethanasinglebonds • ImportantNote:Whendealingwithresonancestructures,theamountof spacethe“rotating”doublen=bondtakesupisnegligiblebecauseitis technicallyalldoublebonds ▯ • ElectronGeometry:TriganolBipyramidal • ℎ:Linearduetothe180degreeangle HowdowedothebondanglesforthismoleculeByusingtheideathatdouble bondstakeupmorespacethanasingle,andlonepairstakemorespacethanbonds • OctahedralElectronGeometry: ▯ ▯ • Molecularshape:squareplanar,duetothe90degreeangle • BondanglesforF-Xe-Fis90degrees • ElectronGeometry:Tetrahedral • MolecularShape:Triganolpyramidal • BondAnglesforH-N-Hislessthan109.5becausethelonepairwillmakethe anglesmallerthanthatofanormaltetrahedralmolecule NoticehowitisTetrahedralwiththelonepair,fortheshapethelonepairisnot drawn PolarandNon-PolarMolecules • Somethingispolarifithasanegativesideandanonpolarside • Inmoleculesthisisdeterminedbysymmetryaroundthecentralatom • If a molecule is symmetric (by polar bonds)around the central atom, the charges cancel eachother out • If the molecule is not semetric(by polar bonds) than the molecule is polar • A numeric measure of polarity is the Dipole moment(μ) • The more polar a molecule the larger the dipole moment • Is ▯ polar or non-polar Non-polar • Is ▯polar or non polar Non polar • Is ▯olar or non polar Polar • Look at the bonds and the way the bonds are arranged to see if they are polar or not! ValenceBondTheory • Acovalentbondisproducedbytheoverlapoforbitalsintheregion betweenthetwoatoms • Thegreatertheoverlap,thestrongerthebond. • But,theatomicorbitalsdonotalwayspointintherightdirectionsto producetheshapesthatthemoleculesaresupposedtohave Toexplainthis,itisassumedthattheatomicorbitalsaremixedtoproduce‘hybrid’ orbitalsthatpointinthecorrectdirection. • Hybridorbitalsarenamedfortheorbitalsthatcontributetothem.For Example isacombinationofonesorbitalandtwoporbitals • ImportantNote:Allorbitalsthataremixedmustcomefromthesame shell.ForExample3pand3scanbehybrids,but2pand4scannotbe combinedtogether. • The▯totalnumberoforbitalsisunchanged(Therewillbe2orbitals,three orbitalsandsoon) EachhybridhasitsownGeometryassociatedwithit,asshownbelow NoticehowforthegroundstateofCarbonthe2sorbitalisfull,butthe2porbitalis not▯evenhalffull,thismakescarbonunstablesothetwoorbitalscombinetoform ,makingcarbonstable • Singlebondsareformedbyoverlappingorbitalsinbetweenthe atoms.Thesebondsarecalledsigma(σ)bonds. • Doubleandtriplebondsalsocontainaσ-bondbutthesecond(andthird bond)cannotformdirectlybetweentheatoms(becausethereisalreadya pairofelectronsthere) • Theadditionalbondsareformedbyunhybridizedp-orbitalsthat overlapintheareastoeithersideofthesigmabond.Thesebondsare calledpi(π)bonds. • Adoublebondconsistsofaσ-bondandaπ-bond. • Atriplebondconsistsofaσ-bondandtwoπ-bonds. ThisDiagramshouldhelpyouvisualizetheprincipalofsigmaandpibonds

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##### ISBN: 9780201380279

An Introduction to Thermal Physics was written by and is associated to the ISBN: 9780201380279. The full step-by-step solution to problem: 20P from chapter: 5 was answered by , our top Physics solution expert on 07/05/17, 04:29AM. This full solution covers the following key subjects: Energy, level, excited, state, atom. This expansive textbook survival guide covers 10 chapters, and 454 solutions. The answer to “The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of S = k ln 4, since for this given value of the energy, the multiplicity is 4. Question: For what temperatures is the Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what temperatures is it negative? (Comment: When F for the level is negative, the atom will spontaneously go from the ground state into that level, since F = 0 for the ground state and F always tends to decrease. However, for a system this small, the conclusion is only a probabilistic statement; random fluctuations will be very significant.)” is broken down into a number of easy to follow steps, and 143 words. This textbook survival guide was created for the textbook: An Introduction to Thermal Physics , edition: 1. Since the solution to 20P from 5 chapter was answered, more than 305 students have viewed the full step-by-step answer.

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