2What is meant by “quantized”? QuantitySpecific and discrete quantityPackets of definite size.Quantized energy can be thought of as existing in very small packets of specific size.Atoms absorb and emit quanta of energy.
3Let us consider light…Electromagnetic SpectrumVisible spectrum
4Three models are used to describe light What model is used in geometric optics, like with lenses and mirrors?RayWhat model is used in studying diffraction and interference?WaveWhat model is used to study interaction of light and atoms?Particle ( photon)
5But light is said to have a “dual nature” What is that supposed to mean?Wave particle dualityWaves have both a wave and a particle componentWe describe the path of light as a rayEquationE = hf for a single photonE = nhf for multiple photonsh = plank’s constant6.63x10-34 J ∙ s (SI version)4.14x10-15 eV ∙ s (convenient)f = frequencyn = number of photons
6Conceptual checkpoint Which has more energy in its photons, a very bright, powerful red laser or a small key-ring red laser?Neither! They both have the same energy per photon. The big one has more power.Which has more energy in its photons, a red laser or a green laser?The green one has shorter wavelength and higher frequency. It has more energy per photon.
7The “electron-volt” (eV) The electron-volt is the most useful unit on the atomic level.If a moving electron is stopped by 1 V of electric potential, we say it has 1 electron-volt (or 1 eV) of kinetic energy.1 eV = 1.602×10-19 J
8What is the frequency and wavelength of a photon whose energy is 4 What is the frequency and wavelength of a photon whose energy is 4.0 x 10-19J?E = 4.0x10-19 Jh= 6.625x10-34 J ∙ sE = hff = E / h= 4.0x10-19 J / 6.625x10-34 J ∙ s= 6.04x1014 Hzλ = c/fλ = c/f= 3x108 m/s / 6.04x / s= 4.97x10-7 m= 497 nm
9How many photons are emitted per second by a He-Ne laser that emits 3 How many photons are emitted per second by a He-Ne laser that emits 3.0 mW of power at a wavelength of nm?P = 3.0 mW = Wλ = nm = 632.8x10-9 mP = E /tE = P ∙ tE = W ∙ 1 s= Jf = c / λ= 3x108 m/s / 632.8x10-9 m= 4.74x1014 HzE =n(hf)n =E / (hf)= J / (6.625x10-34 J ∙ s ∙ 4.74x1014 Hz)= 9.55x1014
10What are atoms composed of? Atoms consist of nuclei (protons and neutrons and electrons.What happens when an atom encounters a photon?The atom usually ignores the photon, but sometimes the atom absorbs the photon.If the photon is absorbed by the atom, what happens next?The photon disappears and winds up giving all its energy to the atom’s electrons.
11This is a graph of energy levels for a hypothetical atom Atom loses an electronIonization level0.0eVThird excited state-1.0eVHighest energy levelSecond excited state-3.0eVHigher energy levels. (Atoms has to absorb energy to get from the ground state.First excited state-5.5 eVNormal “unexcited” stateGround state (lowest energy level-11.5 eV
12What do we mean when we say the atoms energy levels are “quantized”? Ionization level0.0eVOnly certain energies are allowed.These are represented by the horizontal lines.The atom cannot exist at energies not shown in this graph!Third excited state-1.0eVSecond excited state-3.0eVFirst excited state-5.5 eVGround state (lowest energy level-11.5 eV
13Absorption of photon by Atom When a photon of light is absorbed by an atom, it causes an increase in the energy of the atom.The photon disappears, and the energy of the atom increases by exactly the amount of energy contained in the photon.The photon can be absorbed ONLY if it can produce an “allowed” energy increase in the atom.
14Absorption of photon by atom 0.0eVExited stateΔE = hfE = hfGround state-10.0eV
15Absorption SpectrumWhen an atom absorbs photons, it removes the photons from the white light striking the atom, resulting in dark bands in the spectrum.Therefore, a spectrum with dark bands in it is called an absorption spectrum.
16Absorption SpectrumIonized0.0eVAbsorption spectra always involve atoms going up in energy level.Ground state-10.0eV
17Emission of photon by atom When a photon of light is emitted by an atom, it causes a decrease in the energy of the atom.A photon of light is created, and the energy of the atom decreases by exactly the amount of energy contained in the photon that is emitted.The photon can be emitted ONLY if it can produce an “allowed” energy decrease in an excited atom.
18Emission of photon by atom 0.0eVExited stateΔE = hfE = hfGround state-10.0eV
19Emission SpectrumWhen an atom emits photons, it glows! The photons cause bright lines of light in a spectrum.Therefore, a spectrum with bright bands in it is called an emission spectrum.
20Emission of photon by atom Ionized0.0eVEmission spectra always involve atoms going down inenergy level.Ground state-10.0eV
22What is the frequency and wavelength of the light that will cause the atom shown to transition from the ground state to the first excited state? Draw the transition.Ionization levelh = 4.14x10-15 eV ∙ s0.0eVΔE = hff = Δ E / hThird excited state-1.0eVf = (11.5 – 3.0) / 4.14x10-15Second excited state-3.0eVf = 1.45x1015 HzFirst excited state-5.5 eVλ= c / fλ= 3x108 / 1.45x1015Ground state(lowest energy levelλ= 2.07x10-7 mλ= 207 nm-11.5 eV
23What is the longest wavelength of light that when absorbed will cause the atom shown to ionize from the ground state? Draw the transition.Ionization levelh = 4.14x10-15 eV ∙ s0.0eVΔE = hff = Δ E / hThird excited state-1.0eVf = (11.5 ) / 4.14x10-15Second excited state-3.0eVf = 2.78x1015 HzFirst excited state-5.5 eVλ= c / fλ= 3x108 / 2.78x1015Ground state(lowest energy levelλ= 1.08x10-7 mλ= 1-8 nm-11.5 eV
24The atom shown is in the second excited state The atom shown is in the second excited state . What frequencies of light are seen in its emission spectrum? Draw the transition.Ionization levelh = 4.14x10-15 eV ∙ s0.0eVThird excited state1-1.0eVΔE = hff = Δ E / hSecond excited state-3.0eVf = (11.5 – 3.0 ) / 4.14x10-1512f = 2.053x1015 HzFirst excited state-5.5 eV3Ground state(lowest energy level-11.5 eV
25The atom shown is in the second excited state The atom shown is in the second excited state . What frequencies of light are seen in its emission spectrum? Draw the transition.Ionization levelh = 4.14x10-15 eV ∙ s0.0eVThird excited state2-1.0eVΔE = hff = Δ E / hSecond excited state-3.0eVf = (5.5-3 ) / 4.14x10-1512f = 6.09x1014 HzFirst excited state-5.5 eV3Ground state(lowest energy level-11.5 eV
26The atom shown is in the second excited state The atom shown is in the second excited state . What frequencies of light are seen in its emission spectrum? Draw the transition.Ionization levelh = 4.14x10-15 eV ∙ s0.0eVThird excited state3-1.0eVΔE = hff = Δ E / hSecond excited state-3.0eVf = ( ) / 4.14x10-1512f = 1.45x1015 HzFirst excited state-5.5 eV3Ground state(lowest energy level-11.5 eV
27The atom shown is in the second excited state The atom shown is in the second excited state . What frequencies of light are seen in its emission spectrum? Draw the transition.1f = 2.053x1015 HzIonization level0.0eVThird excited state-1.0eV2f = 6.09x1014 HzSecond excited state-3.0eV12First excited state3f =1.45x1015 Hz-5.5 eV3Ground state(lowest energy level-11.5 eV
28Atoms absorbing photons increase in energy We’ve seen that if you shine light on atoms, they can absorb photons and increase in energy.The transition shown is the absorption of an 8.0 eV photon by this atom.You can use Planck’s equation to calculate the frequency and wavelength of this photon.Ionization level0.0eVPhoton eV with largest allowed energy-4.0eVGround state(lowest energy level-12 eV
29QuestionNow, suppose a photon with TOO MUCH ENERGY encounters an atom?If the atom is “photo-active”, a very interesting and useful phenomenon can occur…This phenomenon is called thePhotoelectric Effect.
30Photoelectric Effect E = W0 + KE e- Eph Kinetic energy Some “photoactive” metals can absorb photons that not only ionize the metal, but give the electron enough kinetic energy to escape from the atom and travel away from it.The electrons that escape are often called “photoelectrons”.The binding energy or “work function” is the energy necessary to promote the electron to the ionization level.The kinetic energy of the electron is the extra energy provided by the photon.E = W0 + KEe-EphKinetic energy0.0eVIonization level-4.0eVW0 = Work functionPhoton energyGround state(lowest energy level-12 eV
31Photon Energy = Work Function + Kinetic Energy hf = Φ + Kmax Photoelectric EffectPhoton Energy = Work Function + Kinetic Energyhf = Φ + KmaxKmax = hf – ΦK:max Kinetic energy of “photoelectrons”hf: energy of the photonΦ: binding energy or “work function” of the metal.E = W0 + KEe-EphKinetic energy0.0eVIonization level-4.0eVW0 = Work functionPhoton energyGround state(lowest energy level-12 eV
32Suppose the maximum wavelength a photon can have and still eject an electron from a metal is 340 nm. What is the work function of the metal surface?The longest wavelength is the lowest energy, and will provide no “extra” kinetic energy for the electron.Kmax = hf – Φ0 J = hf – Φf = v / λΦ = hv / λΦ = hfΦ = (4.14x10-15eV ∙ 3x108 m/s) / 340x10-9 mΦ = 3.65 eV
33QuestionSuppose you collect Kmax and frequency data for a metal at several different frequencies. You then graph Kmax for photoelectrons on y-axis and frequency on x-axis. What information can you get from the slope and intercept of your data?Slope: Planck’s ConstantIntercept: Φ - binding energy or “work function”
34The Photoelectric Effect experiment The Photoelectric Effect experiment is one of the most famous experiments in modern physics.The experiment is based on measuring the frequencies of light shining on a metal (which is controlled by the scientist), and measuring the resulting energy of the photoelectrons produced by seeing how much voltage is needed to stop them.Albert Einstein won the Nobel Prize by explaining the results.
36Strange results in the Photoelectric Effect experiment Voltage necessary to stop electrons is independent of intensity (brightness) of light. It depends only on the light’s frequency (or color).Photoelectrons are not released below a certain frequency, regardless of intensity of light.The release of photoelectrons is instantaneous, even in very feeble light, provided the frequency is above the cutoff.
37Voltage versus current for different intensities of light.Number of electrons (current) increases with brightness, but energy of electrons doesn’t!“Stopping Voltage”Vs, the voltage needed to stop the electrons, doesn’t change with light intensity. That means the kinetic energy of the electrons is ndependent of how bright the light is.
38Voltage versus current for different frequencies of light. Energy ofelectronsincreases asthe energy ofthe lightincreases.“Stopping Voltage”Vs changes with light frequency. That means the kinetic energy of the photoelectrons is dependent on light color.
39Experimental determination of the Kinetic Energy of a photoelectronThe kinetic energy of photoelectrons can be determined from the voltage (stopping potential) necessary to stop the electron.If it takes 6.5 Volts to stop the electron, it has 6.5 eV of kinetic energy.
41Question: Does a photon have mass? A photon has a fixed amount of energy (E = hf).We can calculate how much mass would have to be destroyed to create a photon (E=mc2).
42Calculate the mass that must be destroyed to create a photon of 340 nm light. λ = 340x10-9 m h = 6.63x10-34 J∙sh = 6.63x10-34 (kg ∙m2 / s2)∙sE = hff = c / λE = mc2hf = mc2hc / λ = mc2h / λ = mcm = h / (λc)m = 6.63x / (340x10-9 ∙3x108 )=
43Yes Momentum of a Photon Does a photon have momentum? A photon’s momentum is calculated byp = E / c = hf / c = h / λ
44We have experimental proof of the momentum of photons Compton scatteringProof that photons have momentum.High-energy photons collided with electrons exhibit conservation of momentum.Work Compton problems just like other conservation of momentum problems – except the momentum of a photon uses a different equation.
45What is the frequency of a photon that has the same momentum as an electron with speed 1200 m/s? melectron = 9.11x10-31 Kgh = 6.63x10-34 J∙sh = 6.63x10-34 (kg ∙m2 / s2)∙sp = mvp = 9.11x10-31 Kg ∙ 12 m / s=p = hf / cf = P c / hf = ( kg ∙ m/s) ∙ (3x108 m/s) / (6.63x10-34 (kg ∙m2 / s2)∙s)f =
46Wave-Particle Duality Waves act like particles sometimes and particles act like waves sometimes.This is most easily observed for very energetic photons (gamma or x-Ray) or very tiny particles (elections or nucleons)
47Particles and Photons both have Energy A moving particle has kinetic energyE = K = ½ mv2A particle has most of its energy locked up in its mass.E = mc2A photon’s energy is calculated using its frequencyE = hf
48Particles and Photons both have Momentum For a particle that is movingp = mvkg ∙ m/sFor a photonp = h/λ(kg ∙m2 / s2)∙s / m = kg ∙ m / sCheck out the units! They are those of momentum.
49Particles and Photons both have a Wavelength For a photonl = c/fFor a particle, which has an actual mass, this equation still worksλ= h/p where p = mvThis is referred to as the deBroglie wavelength
50We have experimental proof that particles have a wavelength Davisson-Germer ExperimentVerified that electrons have wave properties by proving that they diffract.Electrons were “shone” on a nickel surface and acted like light by diffraction and interference.We’ll study diffraction in the next unit, and return to this experiment then…
51What is the momentum of photons that have a wavelength of 620 nm? λ = 620x10-9 mh = 6.63x10-34 J∙sp = h / λ= 6.63x10-34 J∙s / 620x10-9 m=
52What is the wavelength of a 2,200 kg elephant running at 1.2 m/s? m = 2200 kgv = 1.2 m / sp = mv= 2200 kg ∙1.2 m/s=p = h / λλ = h / p= 6.63x10-34 J∙s / 12 m / s=
54What are isotopes?What characteristics do isotopes of the same element share?What characteristics do isotopes of the same element share?Atomic numberWhat characteristics do isotopes of the same element not share?Atomic massRadioactivity
55IsotopesIsotopes have the same atomic number and different atomic mass.Isotopes have similar or identical chemistry.Isotopes have different nuclear behavior.Examples:Naming a NucleusCCC121314666
56p n e e Elementary Particles mass charge mass charge mass charge 1 Proton1masschargen1NeutronmasschargeeeElectron-1+1Negative charePositive chare
57Nuclear reactionsNuclear Decay: a spontaneous process in which an unstable nucleus ejects a particle and changes to another nucleus.Alpha decayBeta decayBeta MinusPositronFission: a nucleus splits into two fragments of roughly equal size.Fusion: Two nuclei combine to form another nucleus.
58A nucleus ejects an alpha particle, which is just a helium nucleus. Decay ReactionsAlpha decayA nucleus ejects an alpha particle, which is just a helium nucleus.Beta decayA nucleus ejects a negative electron.Positron decayA nucleus ejects a positive electron.Simulations
59Alpha (α) DecayAlpha particle (helium nucleus) is released. Alpha decay only occurs with very heavy elements.239235+4PuUHe94922
60Beta (β-)DecayA beta particle (negative electron) is released. Beta decay occurs when a nucleus has too many neutrons for the protons present. A neutron converts to a proton. An antineutrino is also released.1414++γCNe67-1
61+ + γ He H e Beta (β+)Decay Positron (positive electron) is released. Positron decay occurs when a nucleus has too many protons for the neutrons present. A proton converts to a neutron. A neutrino is also released.22++γHeHe211
62Neutrino and Anti-Neutrino Proposed to make beta and positron decay obey conservation of energy.These particles possess energy and spin, but do not possess mass or charge.They do not react easily with matter, and are extremely hard to detect.
63Gamma Radiation, γGamma radiation is electromagnetic in nature.Gamma photons are released by atoms which have just undergone a nuclear reaction when the excited new nucleus drops to its ground state.The high energy in a gamma photon is calculated by E = hf.
64+ + γ Th Pa e Complete the reaction, identify the type of decay. 234 e9190-1
65+ Th Ra He Complete the reaction for the alpha decay of Thorium-232 228+4ThRaHe90882
67FissionFission occurs when an unstable heavy nucleus splits apart into two lighter nuclei, forming two new elements.Fission can be induced by free neutrons.Mass is destroyed and energy produced according to E = mc2.••
68Neutron-induced fission Neutron-induced fission produces a “chain reaction.” What does that mean?Nuclear power plants operate by harnessing the energy released in fission in by controlling the chain reaction.Nuclear weapons depend upon the initiation of an uncontrolled fission reaction.
69Critical MassThe neutrons released from an atom that has undergone fission cannot immediately be absorbed by other nearby fissionable nuclei until they slow down to “thermal” levels.How can this concept be used to explain why a chain reaction in nuclear fission will not occur unless a “critical mass” of the fissionable element is present at the same location?
70Nuclear ReactorsNuclear reactors produce electrical energy through fission.Advantages are that a large amount of energy is produced without burning fossil fuels or creating greenhouse gases.A disadvantage is the production of highly radioactive waste.Another simulation appears at
71Nuclear WeaponsNuclear weapons have been used only twice, although they have been tested thousands of times.Weapons based on nuclear fission involve slamming together enough material to produce an uncontrolled fission chain reaction.Little Boy was dropped on Hiroshima and contained U-235 produced in Oak Ridge, TN.
72FissionFission occurs only with very heavy elements, since fissionable nuclei are too large to be stable.A charge/mass calculation is performed to balance the nuclear equation.Mass is destroyed and energy produced according to E = mc2.
73FusionFusion occurs when two light nuclei come together to form a new nucleus of a new element.Fusion is the most energetic of all nuclear reactions.Energy is produced by fusion in the sun.Fusion of light elements can result in non-radioactive waste.
74FusionFusion is the reaction that powers the sun, but it has not been reliably sustained on earth in a controlled reaction.Advantages to developing controlled fusion would be the tremendous energy output and the lack of radioactive waste products.Disadvantages are – we don’t know if we’ll be technically able to do it on earth!
75Mass defectThis strange term is used to indicate how much mass is destroyed when a nucleus is created from its component parts.The mass defect is generally much, much less than the mass of a proton or neutron, but is significant nonetheless.The loss of mass results in creation of energy, according to E = mc2.