Presentation on theme: "LIGHT and Planck's Constant DO NOW: Using your textbooks answer the following 1. What is mass spectrometry and how is it used? 2. Define light 3. What."— Presentation transcript:
LIGHT and Planck's Constant DO NOW: Using your textbooks answer the following 1. What is mass spectrometry and how is it used? 2. Define light 3. What is the formula for Planck’s constant and what does each variable represent
Light l The study of light led to the development of the quantum mechanical model. l Light is a kind of electromagnetic radiation. (radiant energy) l Electromagnetic radiation includes many kinds of waves l All move at 3.00 x 10 8 m/s ( c)
Parts of a wave Wavelength Amplitude Orgin Crest Trough
Parts of Wave l Orgin - the base line of the energy. l Crest - high point on a wave l Trough - Low point on a wave l Amplitude - distance from origin to crest l Wavelength - distance from crest to crest Wavelength - is abbreviated Greek letter lambda.
Waves To understand the electronic structure of atoms, one must understand the nature of electromagnetic radiation. The distance between corresponding points on adjacent waves is the wavelength ().
Waves The number of waves passing a given point per unit of time is the frequency (). For waves traveling at the same velocity, the longer the wavelength, the smaller the frequency.
Electromagnetic Radiation All electromagnetic radiation travels at the same velocity: the speed of light (c), 3.00 10 8 m/s. Therefore,c =
Frequency l The number of waves that pass a given point per second. l Units are cycles/sec or hertz (hz) Abbreviated the Greek letter nu c =
Frequency and wavelength l Are inversely related l As one goes up the other goes down. l Different frequencies of light is different colors of light. l There is a wide variety of frequencies l The whole range is called a spectrum
Radiowave s Microwave s Infrared. Ultra- violet X-Rays GammaRays Low energy High energy Low Frequency High Frequency Long Wavelength Short Wavelength Visible Light
Wave model of light cannot explain several phenomena: ◦Emission of light from hot objects (black body radiation) ◦Emission of electrons from metal surfaces on which light shines (photoelectric effect) ◦Emission spectra of light from electronically excited gas atoms (emission spectra)
The Nature of Energy The wave nature of light does not explain how an object can glow when its temperature increases. Max Planck explained it by assuming that energy comes in packets called quanta.
Light is a Particle l Energy is quantized. l Light is energy l Light must be quantized l These smallest pieces of light are called photons. l Energy and frequency are directly related.
Energy and frequency E = h x n E is the energy of the photon n is the frequency h is Planck’s constant h = 6.6262 x 10 -34 Joules sec. joule is the metric unit of Energy
The Nature of Energy Einstein used this assumption to explain the photoelectric effect. He concluded that energy is proportional to frequency: E = h where h is Planck’s constant, 6.63 10 −34 J-s.
Photoelectric effect Light shining on a clean surface causes the surface to emit electrons A minimum frequency of light, different for different metals, is required for the emission of electrons Einstein deduced each photon must have an energy equal to Planck's constant times the frequency of the light
The Nature of Energy Therefore, if one knows the wavelength of light, one can calculate the energy in one photon, or packet, of that light: c = E = h
The Math l Only 2 equations c = λ ν E = h ν l Plug and chug.
Examples l What is the wavelength of blue light with a frequency of 8.3 x 10 15 hz? l What is the frequency of red light with a wavelength of 4.2 x 10 -5 m? l What is the energy of a photon of each of the above?
PES PHOTOELECTRON SPECTROSCOPY (PES) Interpret spectroscopic data and extract information on atomic structure Low resolution PES of atoms provides evidence of shell model
HOW DOES PES WORK? Light consists of photons, each of which has energy E = hv, where h is Planck’s constant and v is frequency of light
IN PHOTOELECTRIC EFFECT Incident light ejects electrons from material This requires the photon to have sufficient energy to eject the electron
PES determines energy needed to eject electrons
PES Measurement of these energies provides a method to deduce the shell structure of an atom
PES Intensity of photoelectron signal at a given energy is a measure of # of electrons in that energy level
PES Electron structure of atoms with multiple electrons can be inferred from evidence provided by PES
6.11, 6.12, 6.14, 6.17, 6.18 HW
Line Spectra and the Bohr model How color tells us about atoms
Atomic Spectrum How color tells us about atoms
Prism l White light is made up of all the colors of the visible spectrum. l Passing it through a prism separates it.
If the light is not white l By heating a gas with electricity we can get it to give off colors. l Passing this light through a prism does something different.
Atomic Spectrum l Each element gives off its own characteristic colors. l Can be used to identify the atom. l How we know what stars are made of.
These are called discontinuous spectra Or line spectra unique to each element. These are emission spectra The light is emitted given off.
The Nature of Energy Another mystery involved the emission spectra observed from energy emitted by atoms and molecules.
The Nature of Energy One does not observe a continuous spectrum, as one gets from a white light source. Only a line spectrum of discrete wavelengths is observed.
The Nature of Energy Niels Bohr adopted Planck’s assumption and explained these phenomena in this way: 1.Electrons in an atom can only occupy certain orbits (corresponding to certain energies).
The Nature of Energy Niels Bohr adopted Planck’s assumption and explained these phenomena in this way: 2.Electrons in permitted orbits have specific, “allowed” energies; these energies will not be radiated from the atom.
The Nature of Energy Niels Bohr adopted Planck’s assumption and explained these phenomena in this way: 3.Energy is only absorbed or emitted in such a way as to move an electron from one “allowed” energy state to another; the energy is defined by E = h
The Nature of Energy The energy absorbed or emitted from the process of electron promotion or demotion can be calculated by the equation: E = −R H ( ) 1nf21nf2 1ni21ni2 - where R H is the Rydberg constant, 2.18 10 −18 J, and n i and n f are the initial and final energy levels of the electron.
Limitations of the Bohr Model Electrons exist only in certain discrete energy levels, which are described by quantum numbers Energy is involved in the transition of an electron from one level to another
The Wave Nature of Matter Louis de Broglie suggested that if light can have material properties, matter should exhibit wave properties. He demonstrated that the relationship between mass and wavelength was h = Planck’s constant and v = velocity Quantity mv is called its momentum = h mv
The Uncertainty Principle Heisenberg showed that the more precisely the momentum of a particle is known, the less precisely is its position known: In many cases, our uncertainty of the whereabouts of an electron is greater than the size of the atom itself! (x) (mv) h4h4
HOMEWORK 6.23, 6.24, 6.30, 6.33, 6.35a,
Quantum Theory 1. What did Schrödingers equation lead to? 2. What is the principal reason we must consider the uncertainty principle when discussing electrons and other subatomic particles but not when discussing our macroscopic world?
Quantum Mechanics Erwin Schrödinger developed a mathematical treatment into which both the wave and particle nature of matter could be incorporated. It is known as quantum mechanics.
Quantum Mechanics The wave equation is designated with a lower case Greek psi (). The square of the wave equation, 2, gives a probability density map of where an electron has a certain statistical likelihood of being at any given instant in time.
Quantum Mechanics Electron density refers to the probability of finding an electron in a specified location High dot density, high Ψ 2 value, high probability of finding an electron in this region Low dot density, low Ψ 2 value, low probability of finding an electron in this region
Quantum Numbers Solving the wave equation gives a set of wave functions, or orbitals, and their corresponding energies. Each orbital describes a spatial distribution of electron density. An orbital is described by a set of three quantum numbers.
Principal Quantum Number, n The principal quantum number, n, describes the energy level on which the orbital resides. The values of n are integers ≥ 0. Increase in n means the electron has a higher energy and is therefore less tightly bound to the nucleus For Hydrogen E n = -2.178x10 -18 / n 2 (Bohr calculated the energies corresponding to the allowed orbits for the electron)
Azimuthal Quantum Number, l This quantum number defines the shape of the orbital. Allowed values of l are integers ranging from 0 to n − 1. We use letter designations to communicate the different values of l and, therefore, the shapes and types of orbitals.
Azimuthal Quantum Number, l Value of l0123 Type of orbitalspdf
Magnetic Quantum Number, m l Describes the three-dimensional orientation of the orbital. Values are integers ranging from -l to l: −l ≤ m l ≤ l. Therefore, on any given energy level, there can be up to 1 s orbital, 3 p orbitals, 5 d orbitals, 7 f orbitals, etc.
Magnetic Quantum Number, m l Orbitals with the same value of n form a shell. Different orbital types within a shell are subshells.
Questions: 1. Without referring to Table 6.2 predict the number of subshells in the fourth shell, that is for n = 4? 2. Give the label for each of these subshells? 3. How many orbitals are in each of these subshells?
Questions: 1. What is the designation for the subshell with n=5 and l=1? 2. How many orbitals are in this subshell? 3. Indicate the values of m 1 for each of these orbitals
HOMEWORK: 1. 6.49, 6.51, 6.52
AIM: QUANTUM THEORY DO NOW: Textbook #6.53, 6.54, 6.55
s Orbitals Value of l = 0. Spherical in shape. Radius of sphere increases with increasing value of n.
s Orbitals Observing a graph of probabilities of finding an electron versus distance from the nucleus, we see that s orbitals possess n−1 nodes, or regions where there is 0 probability of finding an electron.
p Orbitals Value of l = 1. Have two lobes with a node between them.
d Orbitals Value of l is 2. Four of the five orbitals have 4 lobes; the other resembles a p orbital with a doughnut around the center.
Summary s p d f # of shapesMax electronsStarts at energy level 121 362 5103 7144
Spin Quantum Number, m s In the 1920s, it was discovered that two electrons in the same orbital do not have exactly the same energy. The “spin” of an electron describes its magnetic field, which affects its energy.
Spin Quantum Number, m s This led to a fourth quantum number, the spin quantum number, m s. The spin quantum number has only 2 allowed values: +1/2 and −1/2.
Pauli Exclusion Principle No two electrons in the same atom can have exactly the same energy. For example, no two electrons in the same atom can have identical sets of quantum numbers.
Pauli Exclusion Principle To put more than one electron in an orbital and satisfy the Pauli Exclusion principle, need to assign different ms value (only 2 possible values for ms) Orbital can have a max of 2 electrons and they must have opposite spins
Energies of Orbitals For a one- electron hydrogen atom, orbitals on the same energy level have the same energy. That is, they are degenerate.
Energies of Orbitals The presence of more than one electron greatly changes the energies of the orbitals
Energies of Orbitals As the number of electrons increases, though, so does the repulsion between them. Therefore, in many- electron atoms, orbitals on the same energy level are no longer degenerate.
By Energy Level l First Energy Level l only s orbital l only 2 electrons l 1s 2 l Second Energy Level l s and p orbitals are available l 2 in s, 6 in p l 2s 2 2p 6 l 8 total electrons
By Energy Level l Third energy level l s, p, and d orbitals l 2 in s, 6 in p, and 10 in d l 3s 2 3p 6 3d 10 l 18 total electrons l Fourth energy level l s,p,d, and f orbitals l 2 in s, 6 in p, 10 in d, and 14 in f l 4s 2 4p 6 4d 10 4f 14 l 32 total electrons
By Energy Level l Any more than the fourth and not all the orbitals will fill up. l You simply run out of electrons l The orbitals do not fill up in a neat order. l The energy levels overlap l Lowest energy fill first.
Aim: Electron Configurations DO NOW: answer the following questions: 1. What information does an electron configuration give us? 2. What are valence electrons and where are they located in the electron configuration? 3. What do you think core electrons are?
Electron Configurations Distribution of all electrons in an atom Consist of ◦Number denoting the energy level
Electron Configurations Distribution of all electrons in an atom Consist of ◦Number denoting the energy level ◦Letter denoting the type of orbital
Electron Configurations Distribution of all electrons in an atom. Consist of ◦Number denoting the energy level. ◦Letter denoting the type of orbital. ◦Superscript denoting the number of electrons in those orbitals.
Orbital Diagrams Each box represents one orbital. Half-arrows represent the electrons. The direction of the arrow represents the spin of the electron.
Hund’s Rule “For degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximized.”
Electron Configurations l The way electrons are arranged in atoms. l Aufbau principle- electrons enter the lowest energy first. l This causes difficulties because of the overlap of orbitals of different energies. l Pauli Exclusion Principle- at most 2 electrons per orbital - different spins
PARAMAGNETISM l Diamagnetism - All of the electrons are spin-paired in diamagnetic elements so their subshells are completed, causing them to be unaffected by magnetic fields. l Paramagnetism - Paramagnetic elements are strongly affected by magnetic fields because their subshells are not completely filled with electrons.
PARAMAGNETISM PARAMAGNETISM Which of the following elements would be expected to be paramagnetic? Diamagnetic?He, Be, Li, N He: 1s 2 subshell is filled Be: 1s 2 2s 2 subshell is filled Li: 1s 2 2s 1 subshell is not filled N: 1s 2 2s 2 2p 3 subshell is not filled Answer Li and N are paramagnetic. He and Be are
Electron Configuration Hund’s Rule- When electrons occupy orbitals of equal energy they don’t pair up until they have to. Let’s determine the electron configuration for Phosporus Need to account for 15 electrons
l The first to electrons go into the 1s orbital l Notice the opposite spins l only 13 more Increasing energy 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 7p 6d 4f 5f
l The next electrons go into the 2s orbital l only 11 more Increasing energy 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 7p 6d 4f 5f
The next electrons go into the 2p orbital only 5 more Increasing energy 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 7p 6d 4f 5f
The next electrons go into the 3s orbital only 3 more Increasing energy 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 7p 6d 4f 5f
Increasing energy 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 7p 6d 4f 5f The last three electrons go into the 3p orbitals. They each go into seperate shapes 3 upaired electrons 1s 2 2s 2 2p 6 3s 2 3p 3
The easy way to remember 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 6f 7s 7p 7d 7f 1s 2 2 electrons
Fill from the bottom up following the arrows 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 6f 7s 7p 7d 7f 1s 2 2s 2 4 electrons
Fill from the bottom up following the arrows 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 6f 7s 7p 7d 7f 1s 2 2s 2 2p 6 3s 2 12 electrons
Fill from the bottom up following the arrows 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 6f 7s 7p 7d 7f 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 20 electrons
Fill from the bottom up following the arrows 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 6f 7s 7p 7d 7f 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 38 electrons
Fill from the bottom up following the arrows 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 6f 7s 7p 7d 7f 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 56 electrons
Fill from the bottom up following the arrows 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 6f 7s 7p 7d 7f 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 6 7s 2 88 electrons
Fill from the bottom up following the arrows 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 6f 7s 7p 7d 7f 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 6 7s 2 5f 14 6d 10 7p 6 108 electrons
Exceptions to Electron Configuration
Orbitals fill in order l Lowest energy to higher energy. l Adding electrons can change the energy of the orbital. l Half filled orbitals have a lower energy. l Makes them more stable. l Changes the filling order
Write these electron configurations l Titanium - 22 electrons l 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 2 l Vanadium - 23 electrons 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 3 l Chromium - 24 electrons l 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 4 is expected l But this is wrong!!
Chromium is actually l 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5 l Why? l This gives us two half filled orbitals. l Slightly lower in energy. l The same principal applies to copper.
Copper’s electron configuration l Copper has 29 electrons so we expect l 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 9 l But the actual configuration is l 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 l This gives one filled orbital and one half filled orbital. l Remember these exceptions
Periodic Table We fill orbitals in increasing order of energy. Different blocks on the periodic table, then correspond to different types of orbitals.
Some Anomalies Some irregularities occur when there are enough electrons to half-fill s and d orbitals on a given row.
Some Anomalies For instance, the electron configuration for copper is [Ar] 4s 1 3d 5 rather than the expected [Ar] 4s 2 3d 4.
Some Anomalies This occurs because the 4s and 3d orbitals are very close in energy. These anomalies occur in f-block atoms, as well.
AP CHEM PRACTICE – finish for HW Complete the circled problems in the packet Text #6.68, 6.69, 6.70, 6.73 STUDY TEST ON WED!