Presentation on theme: "Early Quantum Theory and Models of the Atom"— Presentation transcript:
1 Early Quantum Theory and Models of the Atom Chapter 27Early Quantum Theory and Models of the Atom
2 27.2 Planck’s Quantum Hypothesis; Blackbody Radiation All objects emit radiation whose total intensity is proportional to the fourth power of their temperature. This is called thermal radiation; a blackbody is one that emits thermal radiation only.
3 27.2 Planck’s Quantum Hypothesis; Blackbody Radiation blackbody radiation curves for three different temperatures. Note that frequency increases to the left.The frequency of peak intensity increases linearly with temperature.
4 27.2 Planck’s Quantum Hypothesis; Blackbody Radiation This spectrum could not be reproduced using 19th-century physics.A solution was proposed by Max Planck in 1900:The energy of atomic oscillations within atoms cannot have an arbitrary value; it is related to the frequency:The constant h is now called Planck’s constant.
5 27.2 Planck’s Quantum Hypothesis; Blackbody Radiation Planck found the value of his constant by fitting blackbody curves:Planck’s proposal was that the energy of an oscillation had to be an integral multiple of hf. This is called the quantization of energy.
6 27.3 Photon Theory of Light and the Photoelectric Effect Einstein suggested that, given the success of Planck’s theory, light must be emitted in small energy packets:These tiny packets, or particles, are called photons.
7 27.3 Photon Theory of Light and the Photoelectric Effect If light strikes a metal, electrons are emitted.The effect does not occur if the frequency of the light is too lowthe kinetic energy of the electrons increases with frequencyW0 : work function
8 27.3 Photon Theory of Light and the Photoelectric Effect kinetic energy vs. frequency:f0 is the thresholdfrequency
9 Photocells Photocells are an application of the photoelectric effect When light of sufficiently high frequency falls on the cell, a current is produced
10 The energy of the photon is Example: Barium has a work function of 2.48 eV. What is the maximum kinetic energy of electrons if the metal is illuminated by UV light of wavelength 365 nm? What is their speed?The energy of the photon isJ.sThe maximum kinetic energy of the photoelectrons isWe find the speed from
11 27.4 Energy, Mass, and Momentum of a Photon Because a photon must travel at the speed of light its momentum is given by:Note mass of a photon is zero.
12 The momentum of the photon is Example: Calculate the momentum of a photon of yellow light of wavelength6.00x10-7 m.The momentum of the photon is
13 Pair ProductionThe equation E = m c2 implies that it is possible to convert mass into energy and vice versa.One example of the conversion of energy to mass is pair production.A high energy photon known as a gamma ray traveling near the nucleus of an atom may disappear and an electron and a positron may appear in its place.The electron and the positron have the same mass and carry the same magnitude of electric charge; however, the electron is negatively charged and the positron is positively charged.The minimum energy of a gamma ray required for the pair production of electron and positron is about 1.02 MeV (See EXAMPLE 27-9; p. 765).If the energy of the gamma ray is above this amount, then excess energy is shared equally between the particles in the form of kinetic energy.
14 Wave Particle Duality; the Principle of Complementarity Young's interference experiment and single slit diffraction indicate that light is a wave.The photoelectric effect and the Compton effect indicate that light is a particle.Light is a phenomena that exhibits both the properties of waves ad the properties of particles. This is known as wave-particle duality.Niels Bohr proposed the principle of complementarity which says that for any particular experiment involving light, we must either use the wave theory or the particle theory , but not both. The two aspects of light complement one another.
15 Wave Nature of MatterJust as light exhibits properties of both particles and waves, particles such as electrons, protons, and neutrons also exhibit wave propertiesIn 1923, Louis de Broglie suggested that the wavelength of a particle of mass m traveling at speed v is given by is the de Broglie wavelength of the particle
16 Example:Calculate the wavelength of a 0.21 kg ball traveling at 0.10 m/s.We find the wavelength from = h/p = h/mv = (6.63 x 10–34 J · s)/(0.21 kg)(0.10 m/s)= 3.2 x10–32 m.
17 Atomic SpectraEmission spectra are produced by a high voltage placed across the electrodes of a tube containing a gas under low pressure. The light produced can be separated into its component colors by a diffraction grating. Such analysis reveals a spectra of discrete lines and not a continuous spectrum.HydrogenMercury
18 27.11 Atomic Spectra: Key to the Structure of the Atom In 1885, J. Balmer developed a mathematical equation which could be used to predict the wavelengths of the four visible lines in the hydrogen spectrum. Balmer's formula statesThe wavelengths of electrons emitted from hydrogen have a regular pattern:(27-9)This is called the Balmer series. R is the Rydberg constant:n = 3 (red light)n = 4 (blue light)n = 5 (violet light)and n = 6 (violet light)HydrogenMercury
19 27.11 Atomic Spectra: Key to the Structure of the Atom Other series include the Lyman series for the UV-light:And the Paschen series for the infrared light:
20 Niels Bohr Physicist“The opposite of a correct statement is a false statement. But the opposite of a profound truth may well be another profound truth.” —Niels Bohr
21 Bohr Model1. The electron travels in circular orbits about the positively charged nucleus. However, only certain orbits are allowed.
27 27.12 The Bohr AtomBohr proposed that values energy states were quantized. Then the spectrum could be explained as transitions from one level to another.If an electron falls from one orbit, also known as energy level, to another, it loses energy in the form of a photon of light. The energy of the photon equals the difference between the energy of the orbits.
28 A hydrogen atom can absorb only those photons of light which will cause the electron to jump froma lower level to a higher level. Thus the energy ofthe photon must equal the difference in the energybetween the two levels.
30 Binding energy or ionization energy: minimum energy required to remove and electron from the ground state.The ionization energy for hydrogen is 13.6 eV.
31 Example: How much energy is needed to ionize a hydrogen atom in the n = 2 sate ? 3.4 eVExample: Calculate the ionization energy of doubly ionized lithium, Li2+ , which has Z = 3.Doubly ionized lithium is like hydrogen, except that there are three positive charges (Z = 3) in the nucleus. The square of the product of the positive and negative charges appears in the energy term for the energy levels. We can use the results for hydrogen, if we replace e2 by Ze2:
33 27.3 Photon Theory of Light and the Photoelectric Effect If light strikes a metal, electrons are emitted.The effect does not occur if the frequency of the light is too lowthe kinetic energy of the electrons increases with frequency
34 27.3 Photon Theory of Light and the Photoelectric Effect If light is a wave, theory predicts:1. Number of electrons and their energy should increase with intensity2. Frequency would not matter
35 27.3 Photon Theory of Light and the Photoelectric Effect If light is particles, theory predicts:Increasing intensity increases number of electrons but not energyAbove a minimum energy required to breakatomic bond, kinetic energy will increaselinearly with frequencyThere is a cutoff frequency below which noelectrons will be emitted, regardless of intensity
36 Example: A 60 W light bulb operates at about 2. 1% efficiency Example: A 60 W light bulb operates at about 2.1% efficiency. Assuming that all the light is green light (λ =555 nm) determine the number of photons per second given off by the bulb. Light energy emitted per second:0.02160 J/s=1.3 J/sThe energy of a single photon is:E =hf=hc/λE =(6.6310-34 Js)(3108 m/s)/(55510-9 m)E =3.5810-19 J/photonNumber of emitted photons per second:(1.3 J/s)/(3.5810-19 J/photon)=3.61018 photons/s
37 The Compton Effect Dl=l-l0=lC(1-cosq) l>l0 The experiment was performed by Arthur H. Compton (American Scientist, ). An x-ray photon collides with a stationary electron. The scattered photon and the recoil electron depart the collision in different directions.Dl=l-l0=lC(1-cosq)l0 is the incoming wavelength and l is the emitted wavelengthlC=h/(mec)=2.4310-12 m is the Compton wavelength
38 Dl=lC(1-cosq) Dl=4.8610-12 m (b) Dl=2.4310-12 m (1-cos30) Example: Determine the change in the photon’s wavelength that occurs when an electron scatters an x-ray photon (a) at =180 and (b) =30.Dl=lC(1-cosq) (a) Dl=2.4310-12 m (1-cos180)Dl=4.8610-12 m(b) Dl=2.4310-12 m (1-cos30)Dl=(0.134)(2.4310-12 m)Dl =3.2610-13 m
39 Bohr’s Model of the Hydrogen Atom 1. The electron travels in circular orbits about the positively charged nucleus.However, only certain orbits are allowed.2. The allowed orbits have radii (rn) wherern = (0.53 nm) n2 and n = 1, 2, 3, etc.3. The orbits have angular momentum (L) given by L = mv rn = n h/2 where n = 1,2,3,4. If an electron falls from one orbit, also known as energy level, to another, it losesenergy in the form of a photon of light. The energy of the photon equals the differencebetween the energy of the orbits.5. The energy level of a particular orbit is given by E = -13.6eV/n2 where n = 1,2,3,If n = 1, the electron is in its lowest energy level and it would take 13.6 eV to removeit from the atom (ionization energy).6. A hydrogen atom can absorb only those photons of light which will cause theelectron to jump from a lower level to a higher level. Thus the energy of the photonmust equal the difference in the energy between the two levels.
40 27.12 The Bohr AtomAn electron is held in orbit by the Coulomb force:
41 27.12 The Bohr AtomThe lowest energy level is called the ground state; the others are excited states.
42 27.11 Atomic Spectra: Key to the Structure of the Atom An atomic spectrum is a line spectrum – only certain frequencies appear. If white light passes through such a gas, it absorbs at those same frequencies.Atomic hydrogen emissionHelium emissionSolar absorption
43 Balmer seriesIn 1885, J. Balmer developed a mathematical equation which could be used to predict the wavelengths of the four visible lines in the hydrogen spectrum. Balmer's formula states1/ = R(1/22 – 1/n2), n=3, 4, 5, …n = 3 (red light)n = 4 (blue light)n = 5 (violet light)and n = 6 (violet light)
44 For the spectral lines in the ultaraviolet (UV) region , the so-called Lyman series is used and is given by1/ = R(1/12 – 1/n2), n=2,3, 4, …Lyman seriesAnd the wavelengths in the infrared region are given by the so-called Paschen series is used and is given by1/ = R(1/32 – 1/n2), n= 4, 5,..Paschen series