版权说明:本文档由用户提供并上传,收益归属内容提供方,若内容存在侵权,请进行举报或认领
文档简介
1、Quantum Theory and the Electronic Structure of AtomsChapters 7 & 87.1 Light and Classic Quantum Theory 1. Classic theory of Light Different waves have different color, and different wavelength. Particles (Newton, 1680): an array of particles, emitted from light source, that move in space in one dime
2、nsional direction. forcespeedAcceleration speedOne can determine the exact position and speed at the same time. Wave (Huygens, 1690): an elastic vibrator, emitted from light source, that move in space in three dimensional direction. Kinetic energyWavelength (l l)7.1 Light and Classic Quantum Theory
3、Maxwell (1893) wrote an equation for light as a waveLight is an electromagnetic wave that spread in space.For a light in vacuum, u = c = 3.00 x 108 m/sFrequency n is the number of waves that pass in 1 s.u = l = l n nabout 1.28 s from moon to Earth Wavelength l is the distance between two successive
4、waves. Speed u is the distance of waves that pass in 1 s.Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves.Wavelength (l l)a0 amplitude At a certain position (x), the wave change with t.At a certain time (t), the wave change with x.The energy i
5、ntensityEnergy changescontinuously !Energy passed through a unit area per unit timePlancks Quantum Hypothesis (1901): For a single quantum, the smallest quantity of energy : E = h vPlanck constant h = 6.63 x 10-34 JsEnergy change is only by hv, 2hv, 3 hv, 4 hv . but never by 1.5 hv or 3.06 hvThen in
6、 1918, he won Nobel Prize in physics.2. Classic quantum theory of Light (Plank, 1900)Classical theory explains it well at low v, but very bad at high v. Energy is emitted or absorbed in discrete units (quantum). Black body radiation: curve E vs vExplains all results well.3. The Particle Nature of Li
7、ght: Photoelectric Effect: hnvoltage currentV0Experimental evidence:(1) minimum frequency of light (v0)(2) v0 is dependent of metal. (3) The current light intensity.h v = p cParticlenatureWavenatureEinstein Photon Theory (1905)hn = KE + BEKE = hn - BELight are both wave and particle:1. Wave nature (
8、l): E = hv2. Particle nature(m): momentum p = mc3. Energy E = mc2 It explains well the evidence:(1) minimum frequency of light (v0)(2) v0 is dependent of metal. (3) The current light intensity.Here BE is the binding energy of the metal lattice electron7.2 Dual Nature of Electron 1. Hydrogen Emission
9、 line spectrum: Bohrs Planetary model1. e- only has a specific energy values (quantized) 2. light is emitted as e- moves from one energy level to a lower energy levelBohrs Model of H Atom (1913)En = -RH( )1n2n (principal quantum number) = 1,2,3,RH (Rydberg constant) = 2.18 x 10-18J = 13.6 eVEphoton
10、= DE = Ef - Ei3. ifDE = RH( )1n21n2nf = 1ni = 2nf = 1ni = 3nf = 2ni = 3Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)Ephoton = DE = 1.55 x 10-19 Jl = 6.63 x 10-34 (Js) x 3.00 x 108 (m/s)/1.55 x 10-19Jl = 1280 nmCalculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops
11、from the n = 5 state to the n = 3 state.Ephoton = h c/ ll = h c / EphotonifDE = RH( )1n21n2Ephoton =De Broglie (1924) reasoned that e- is both particle and wave.2pr = nl l = h/muu = velocity of e- m = mass of e-2. Dual nature of the Electron: Matter as wavesWhat is the de Broglie wavelength (in nm)
12、associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s?l = 6.63 x 10-34 / (2.5 x 10-3 x 15.6)l = 1.7 x 10-32 m = 1.7 x 10-23 nmNobel Prize in 1929& 7. 3 Quantum Mechanical Description of Electrons in Atoms Schrdinger Wave EquationBasis:1)Bohrs theory, good to explain the H emission spectrum, b
13、ut failed for the spectrum of other elements.2)Dual nature of electron (m and l l).In 1926, Schrodinger wrote an equation for electron in the atom. = E E = V + KETotal energy of the systemPotential energykinetic energy (psi): wave function, or atomic orbital that describes the movement of electron i
14、n the atom in three dimensional space.: Hamiltonian Operator2: Laplacian Operator2Uncertainty principle (1927, W K Heisenberg): DX Dp h/4p It is impossible to determine precisely both the position and momentum of a particle at the same time.1. Application of Schrdinger to Hydrogen Atom zxy(x,y,z)r =
15、 E (x,y,z) = E (x,y,z) hereAfter transferred into sphere coordinateGeneral solution for Hydrogen AtomSame as Bohrs result, ie. Only dependent of n;Angular partSpacial partn=1 l =0 ml=0E1s=-13.6 eV(Z is atomic number, and for H, Z=1)n=2 l =0 ml=0n=2 l =1 ml=0n=2 l =1 ml=-1n=2 l =1 ml=+1E* = -3.4 eV r
16、 distance from the nucleus 2ElectrondensityBut the probability to find e in space (DV) is P= Y2 D DV, highest at r = 0.529 90 % of electron density foundl = 0n=1n=2n=3ml = -1ml = 1ml = 0ml = -2ml = -1ml = 0ml = 1ml = 27.6n: Principal quantum numbern = 1, 2, 3, 4, .distance of e- from the nucleus, an
17、d determination of energyHere n, l, m are called as the quantum numberl: angular momentum quantum numberl = 0, 1, 2, 3, n-1n = 1, l = 0n = 2, l = 0 , 1n = 3, l = 0, 1, 2Shape of the “volume” of space that the e- occupies l = 0 1 2 3Orbital s p d fml : magnetic quantum numberml = -l, ., 0, . +lif l =
18、 1 (p orbital), ml = -1, 0, 1if l = 2 (d orbital), ml = -2, -1, 0, 1, 2orientation of the orbital in spaceEnergy of orbitals in a single electron atomEnergy only depends on principal quantum number nEn = -RH( )1n2n=1n=2n=3Ground stateexcited state2. Application of Schrdinger to Many-electron atomszx
19、y(xi,yi,zi)ri = E rjrij(xj,yj,zj)For the ith electron, the average net potential is spherically approximated as:shielding constanti = Ei i Effective charge of the nucleusGeneral solutionEnergy of orbitals in a multi-electron atomEnergy depends on n and ln=1 l = 0n=2 l = 0n=2 l = 1n=3 l = 0n=3 l = 1n
20、=3 l = 2Order of orbitals (filling) in multi-electron atom1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s3. The Building-Up Principle of electrons in the orbitals:1. Electrons occupy first the lowest energy orbitals available.2. Each orbital accommodates only two electrons, and these two electrons must have opp
21、osing spins.Pauli exclusion principlespin quantum number ms= + or - To descript one electron completely, Four quantum numbers of n, l, ml and ms are needed.3. In the degenerate orbitals (same n and l), the electrons occupy the orbitals as many as possible with same spins. N (Z=7)1s2s2p(Z=7) N 1s2 2s
22、2 2p3Electron configurationHunds ruleWhat is the electron configuration of Mg?Mg 12 electrons1s 2s 2p 3s 3p 4s 1s2 2s22p6 3s22 + 2 + 6 + 2 = 12 electronsAbbreviated as Ne3s2Ne 1s22s22p6What are the possible quantum numbers for the last (outermost) electron in Cl?Cl 17 electrons1s 2s 2p 3s 3p 4s 1s2
23、2s22p6 3s23p52 + 2 + 6 + 2 + 5 = 17 electronsLast electron added to 3p orbitaln = 3l = 1ml = -1, 0, or +1ms = or -Noble gas coreFe 26 electrons1s 2s 2p 3s 3p 4s 3d 1s2 2s22p6 3s23p6 4s23d6Abbreviated as Ar4s23d6Ar 1s22s22p63s23p6What is the electron configuration of Fe?What is the electron configura
24、tion of Mo?Mo 42 electrons1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p1s2 2s22p6 3s23p6 4s23d104p6 5s24d4Abbreviated as Kr5s24d4Kr5s14d5Half-filled and fully-filled subshell are more stable. (p3, d5, f7)(p6, d10, f14)One e- choice7.8Outermost subshell being filled with electrons& 7. 4 Periodic Relationships Amo
25、ng the Elementsns1ns2ns2np1ns2np2ns2np3ns2np4ns2np5ns2np6d1d5d104f5fGround State Electron Configurations of the ElementsIonization energy is the minimum energy (kJ/mol) required to remove an electron from a gaseous atom in its ground state.I1 + X (g) X+(g) + e-I2 + X+ (g) X2+(g) + e-I3 + X2+ (g) X3+
26、(g) + e-I1 first ionization energyI2 second ionization energyI3 third ionization energyI1 I2 I3General Trend in First Ionization Energies8.4Increasing First Ionization EnergyIncreasing First Ionization EnergyElectron Configurations of Cations and AnionsNa Ne3s1Na+ NeCa Ar4s2Ca2+ ArAl Ne3s23p1Al3+ Ne
27、Atoms lose electrons so that cation has a noble-gas outer electron configuration.H 1s1H- 1s2 or HeF 1s22s22p5F- 1s22s22p6 or NeO 1s22s22p4O2- 1s22s22p6 or NeN 1s22s22p3N3- 1s22s22p6 or NeAtoms gain electrons so that anion has a noble-gas outer electron configuration.Electron Configurations of Cations of Transition MetalsWhen a cation is formed from an
温馨提示
- 1. 本站所有资源如无特殊说明,都需要本地电脑安装OFFICE2007和PDF阅读器。图纸软件为CAD,CAXA,PROE,UG,SolidWorks等.压缩文件请下载最新的WinRAR软件解压。
- 2. 本站的文档不包含任何第三方提供的附件图纸等,如果需要附件,请联系上传者。文件的所有权益归上传用户所有。
- 3. 本站RAR压缩包中若带图纸,网页内容里面会有图纸预览,若没有图纸预览就没有图纸。
- 4. 未经权益所有人同意不得将文件中的内容挪作商业或盈利用途。
- 5. 人人文库网仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对用户上传分享的文档内容本身不做任何修改或编辑,并不能对任何下载内容负责。
- 6. 下载文件中如有侵权或不适当内容,请与我们联系,我们立即纠正。
- 7. 本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。
最新文档
- 医疗设备采购合同履行承诺书(5篇)
- 我想象的未来作文300字(9篇)
- 智慧城市构建技术支撑保证承诺书8篇
- 企业团队建设活动策划方案创意及执行工具
- 网络安全保障长期承诺书8篇
- 企业审核与持续改进框架
- 产品开发流程与质量管理工具
- 企业形象识别系统(CIS)标准化模板
- 跨部门协作流程管理标准化工具
- 自我提升方案实施责任书(3篇)
- 图文广告服务投标方案(技术方案)
- 体彩笔试试题及答案
- 《城乡规划管理与法规系列讲座课件-建设项目规划与审批》
- 支气管哮喘患者护理查房
- DBJ∕T15-234-2021 广东省绿色建筑检测标准
- 2022秋季教科版2017版六年级 上册《科学》全册期末复习 知识总结 背诵归纳
- 统编版《复活》教学课件(共33张)
- 保安队排班表
- 超滤膜技术介绍及应用课件(PPT 36页)
- 矫正教育学习题集(DOC)
- 眼镜项目规划设计方案
评论
0/150
提交评论