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* AP PHYSICS B Atomic and Wave/Particle Physics Teacher Packet AP* is a trademark of the College Entrance Examination Board. The College Entrance Examination Board was not involved in the production of this material. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic and Wave/Particle Physics Objective To review the student on the concepts, processes and problem solving strategies necessary to successfully answer questions on atomic and nuclear physics. Standards Atomic and nuclear physics is addressed in the topic outline of the College Board AP* Physics Course Description Guide as described below. V. Atomic and Nuclear Physics A. Atomic physics and quantum effects 1. Photons, the photoelectric effect, Compton scattering, x-rays 2. Atomic energy levels 3. Wave-particle duality B. Nuclear physics 1. Nuclear reactions (including conservation of mass number and charge) 2. Mass-energy equivalence AP Physics Exam Connections Topics relating to atomic physics, quantum effects, and nuclear physics are tested every year on the multiple choice and virtually every year on the free response portion of the exam. The list below identifies free response questions that have been previously asked over atomic and nuclear physics. These questions are available from the College Board and can be downloaded free of charge from AP Central. http://apcentral.collegeboard.com. 2008 2007 2006 2005 2004 2003 2002 2000 Free Response Questions Question 7 2008 Form B Question 7 2007 Form B Question 6 2006 Form B Question 7 2005 Form B Question 6 2004 Form B Question 7 2003 Form B Question 7 2002 Form B Question 5 ® Question 7 Question 7 Question 6 Question 7 Question 6 Question 7 Question 7 Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics What I Absolutely Have to Know to Survive the AP* Exam • • • • • Determine the frequency, wavelength, momentum or energy of a photon (the smallest particle of light). Apply the photoelectric effect, in which the absorption of photons of a certain frequency causes electrons to be emitted from a metal surface. Understand the Compton effect which verifies the photon nature of light by showing that momentum is conserved in a collision between a photon and an electron. Determine the de Broglie wavelength for particles, such as electrons, that exhibit wave properties. Determine the binding energy in the nucleus as a result of some of the mass of the particles (the mass defect) in the nucleus being converted into energy by using ΔE = ( Δm ) c 2 . Key Formulas and Relationships E = hf = p= hc λ E hf h = = c c λ K max = hf − φ Energy of a photon, where h is Planck's constant (6.63 ×10 −34 J is or 4.14 ×10 −15ev is), f is the frequency of the light, c is the speed of light, and λ is the wavelength. (Note that the frequency of the light is independent of the medium.) Momentum of a photon (Note that photons have no rest mass, but do possess momentum). Einstein's photoelectric equation relates the energy of the incident photon to the maximum kinetic energy of the ejected electron and the work function of the photoemissive surface. φ = hf 0 Work function of a photoemissive surface where f 0 is the threshold frequency (minimum frequency of light to initiate photo electron emission from the surface). λ= f = h p de Broglie wavelength where λ is the wavelength associated with a particle of momentum, p. E0 − E f h Photon frequency of emitted light when an electron makes a transition to an orbit with a lower energy level, from one of higher energy, E0 to E f . ΔE = ( Δm ) c 2 Change in the energy of a system, ΔE is related to the change in mass, Δm of that system (mass energy equivalence). ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics Important Concepts • Electrons In 1897, JJ Thomson investigated cathode rays and determined they were beams of negatively charged particles. He measured the mass to charge ratio of the cathode rays by measuring how much they were deflected by a magnetic field and how much energy they carried. He called the particles that made up the cathode ray “corpuscles,” but the name did not stick and today we call these negatively charged particles, electrons. Later, Robert Millikan used an atomizer to create tiny drops of oil that were given a charge and allowed to fall between two charged plates. The mass of the drop was calculated by the rate at which it fell when the electric field was turned off. The electric field was then switched on and by adjusting the voltage, the drop could be made to rise, fall, or stay steady. The voltage needed to keep the drop steady is directly related to the amount of charge on the drop (the downward gravitational force is balanced by the upward electric force). Millikan found that the amount of charge was always a whole number multiple of 1.6 × 10 −19 C , which he reasoned was the fundamental electric charge. Coupling the charge of the electron with Thomson’s mass to charge ratio, the mass of the electron could now be determined as 9.11× 10−31 kg . • Photons One of the conundrums facing physicists in the early 1900’s was the spectrum of blackbody radiation. A blackbody is an object that radiates light, but does not reflect it. As a black body was heated, light at different frequencies was generated. It was noted that at each temperature there was a specific wavelength that was most intense. As the temperature increased, the curve became more peaked with a shorter width as seen in the diagram below. In addition, the maximum value of the peak moved to a shorter wavelength as the temperature increased. Classical wave theory could not explain the observed energy distribution. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics Planck suggested that energy is radiated in discrete packets or bundles called quanta rather than a smooth continuum of values. This assumption worked and indicated that light was emitted from the hot blackbodies in discrete amounts like particles instead of waves. The energy of a thermal oscillator is not continuous, but is a discrete quantity. E = nhf where n=1,2,3,... Einstein extended this idea by adding that all emitted radiation is quantized. He averred that light is composed of discrete quanta called photons, rather than waves. Thus each photon has an energy given by E = hf where h represents Plank’s constant (6.63 × 10−34 J is or 4.14 × 10−15evis) and f the frequency. According to Einstein, photons travel at the speed of light in vacuum and have no rest mass, but they do acquire momentum. Hence for photons, their h momentum is found by using p = and not p = mv . The energy of a photon is given by the equations below. λ E = hf c= fλ hc E= = pc λ o Photoelectric Effect In the figure above, two metal plates (electrodes) are placed in a vacuum tube and connected to a voltage source. Electrons flow from the negative cathode to the positive anode. If photons of the right frequency strike the metal plate, then electrons are ejected from the cathode and pulled across to the anode. The current may be measured with an ammeter. This is called the photoelectric effect. The voltage may be reversed to stop the flow of electrons. This applied voltage is called the stopping voltage and is related to the KE of the electrons, such that KEmax = qVstop . ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics The kinetic energy, K, of an ejected electron displaced by a photon of energy hf is given by: K max = hf − φ where the work function φ is the minimum energy needed to free an electron from a photoemissive surface like a metal. No photoemission occurs if the frequency of the incident light is below a certain threshold frequency given by: φ f0 = h Einstein’s theory explained several aspects of the photoelectric effect that could not be explained by the classical theory. o The kinetic energy of the photon is dependent on the light’s frequency. o No photoemission occurs for light below a certain threshold frequency. o Current flows immediately when the light’s frequency is greater than the threshold frequency. • If a graph of kinetic energy of the ejected electrons versus the frequency of the photons is plotted, then it can be seen that o o o o o o o The slope of the graph is Planck’s constant. All metals (photoemissive surfaces) have the same slope. The threshold frequency is the x intercept, below which no electrons are emitted. Different metals have different threshold frequencies. The work function is the negative y intercept. Different metals have different work functions. The threshold frequency for any metal is the work function divided by Planck’s constant. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics • Example I Light is shined on a photoemissive surface of work function φ = 2.0 eV and electrons are released with a kinetic energy KEmax = 4.0 eV. A. What voltage, called the stopping voltage Vstop, would be necessary to stop the emission of electrons? Stopping the emission of electrons requires work equal to the maximum kinetic energy of the electrons; thus, it would take 4.0 V to stop electrons with a kinetic energy of 4.0 eV. KEmax = W = qeVstop Vstop 1 point For the correct equation for finding the stopping voltage. 1 point For correct substitution of the maximum kinetic energy. −19 KEmax ( 4.0 eV ) (1.6 x10 J/eV ) = = qe 1.6 x10−19 C 1 point For the correct answer with units. Vstop = 4.0 Volts B. Determine the energy of each of the incoming photons in electron volts (eV) and Joules (J). E photons = KEmax + φ 1 point For a correct equation for the photoelectric effect. E photons = 4.0 eV + 2.0 eV 1 point For the correct substitution for the work function. E photons = 6.0 eV 1 point For the correct answer with units for electron volts. 6.0 eV (1.6 x10−19 J/eV ) = 9.6 x10−19 J 1 point For the correct answer with units for Joules. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics C. Determine the frequency of the incoming photons in Hertz (Hz). f = 1 point For the correct equation for the energy of a photon. E photon h 9.6 x10−19 J f = 6.6 x10−34 J /Hz 1 point For the correct energy in Joules or consistent with part B. f = 1.5 x1015 Hz 1 point For the correct answer with units. • Compton effect: Photons impart momentum to subatomic particles in collisions that follow the laws of conservation of momentum and energy. Compton aimed x-rays of a certain frequency at electrons, and when they collided and scattered, the x-rays were measured to have a lower frequency or higher wavelength indicating less energy and momentum. The scattered photon did not appear to change speed during the collision, but rather changed its frequency. The scattering of x-ray photons from an electron with a loss in energy of the x-ray photon is called the Compton effect. Thus photons may act as both particles and as waves depending upon the situation. λ′ − λ = h (1 − cos θ ) me c where λ ′ is the wavelength of a photon after being scattered by a collision with an electron, λ is the wavelength of the photon before being scattered, h is Planck's constant, me is the mass of the electron, c is the speed of light and θ is the angle by which the photon's heading changes (between 0° and 180°). ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics • Wave-Particle Duality Louis de Broglie suggested that if light can behave as a particle, then possibly a particle, like an electron could behave like a wave. The de Broglie wavelength of a particle is related to Planck’s constant and the momentum of the particle. λ= h h = mv p Note that objects with very large masses, will have very small wavelengths and thus behave like particles. As masses approach smaller values, however, such as the mass of an electron, the wavelengths can become very noticeable. This is true because Planck’s constant is so small. In 1927, Davisson and Germer experimentally verified de Broglie’s hypothesis. They studied the diffraction of electrons by a nickel crystal and showed that the wavelength of a diffracted beam of electrons corresponded to the de Broglie wavelength. • Energy Levels Niels Bohr proposed that the electrons in the atom existed in certain discrete energy levels. The quantized energy level model explained the stability of atoms and the absorption and emission spectra from atoms. The major features of quantized energy levels include: o Electron orbits are only allowed at certain distances from the nucleus, not in between. While in these allowed orbits or levels, the electrons will emit no light. o Electrons only emit light if they make a transition from a higher level to a lower level. The photon energy of the emitted light will account for the energy difference between the two levels. The position of the electron becomes more stable. o Electrons can absorb photon energies but only if the energy is the exact amount needed to raise the electron from a lower level to a higher level. The electron is said to be in an excited state when it absorbs energy and the stability of its position decreases. o The ground state or first level, n=1, is the level closest to the nucleus. An electron in this state is firmly attached to the atom. o The last level or n = ∞ , the electron is now free and no longer bound to the atom. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics • Example II The electron energy levels for a mercury atom are shown below. E=0 E4 = - 4.95 eV E3 = -5.52 eV E2 = -5.74 eV E1 = - 10.38 eV A. An electron in the ground state absorbs a photon which causes the electron to be raised to the second energy level. Determine the following for the absorbed photon. i. wavelength ii. frequency i. 1 point For a correct equation for energy absorbed by photon E = E2 − E1 = −5.74 eV − ( −10.38 eV ) = 4.64 eV 1 point For the correct value for E λ= hc 1240 eV nm = = 267.24 nm = 2.67 x10−7 m 4.64 eV E 1 point For a correct equation for calculating the wavelength of the photon 1 point For the correct answer with units 1 point For a correct equation for the frequency of the photon ii. f = c λ = 3.00 x108 m/s = 1.12 x1015 Hz 2.67 x10−7 m 1 point For the correct answer with units ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics B. State whether or not the absorbed photon is in the visible range. Explain your answer. This photon is not in the visible range of wavelengths, which is 400 nm to 700 nm. 1 point For the correct answer, based on answer in A(i). 1 point For stating that the visible range is approximately 400 nm to 700 nm. C. The electron absorbs a second photon of energy 10.00 eV while it is in the second energy level of the atom. Find the kinetic energy for the electron after it absorbs the 10.00 eV photon. 1 point For subtracting 5.74 eV from 10.00 eV. KE = 10.00 eV − 5.74 eV = 4.26eV KE = 6.82 x10−19 J • 1 point For a correct answer with units, either eV or J. Nuclear Physics A nucleus is composed of two types of nucleons called protons and neutrons. The atomic number, denoted by Z is the number of protons and defines what kind of element it is. In a neutral atom, the number of electrons is equal to the number of protons. If electrons are gained or lost, then the resulting particle is an ion of that element. The atomic mass number, denoted by A is the number of protons and neutrons. N is the number of neutrons, such that A=Z+N. Atoms that have the same number of protons, but different number of neutrons are known as isotopes. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics • Mass Defect The nitrogen atom 147 N is composed of 7 protons and 7 neutrons, which gives a total of 14 atomic mass units (u). But if these particles are combined into a 147 N nitrogen nucleus, the resulting mass of the nitrogen nucleus is 14.003074 u. The sum of the masses of the protons and neutrons is 7(1.007 825 u) = 7.054775 u 7(1.008 665 u) = 7.060655 u 14.115430 u Notice that when we add the mass of the nucleons and compare it to the mass of the nucleus, we find that some of the mass is missing. This is called the mass defect, Δm . Δm = mass of protons + mass of neutrons - mass of nucleus Δm = 14.115430 u − 14.003074 u = 0.112356 u Einstein discovered that mass could be converted into energy and energy into mass. The missing mass is a form of energy that may be released if the nucleus is pulled apart. The missing energy is called binding energy and represents the amount of energy that must be supplied in order to break the nucleus into its individual protons and neutrons. Conversely it is the amount of energy released when the nucleus is formed from individual protons and neutrons. Thus the binding energy of the nitrogen nucleus is 2 ⎛ 1.6605 × 10−27 kg ⎞ 8 ΔE = ( Δm ) c 2 = ⎜ 0.112356 u ⎟ ( 2.997 ×10 m / s ) 1u ⎝ ⎠ −11 ΔE = 1.68 × 10 J This does not seem like a lot of energy, but remember it is for just one atom. The process of making a mole of nitrogen atoms releases a great deal of energy. ⎛ 1.68 ×10−11 J ⎞⎛ 6.022 ×1023 atoms ⎞ 13 E =⎜ ⎟⎜ ⎟ = 1.01×10 J atom mole ⎝ ⎠⎝ ⎠ • Nuclear Reactions In any nuclear reaction the following are true: o Conservation of charge: the total atomic number before the reaction must be the same as the total atomic number after the reaction. o Conservation of nucleon number: the total atomic mass before the reaction must be the same as the total atomic mass after the reaction. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics • Radioactive Decay There are four types of particles that can be emitted when an element undergoes radioactive decay. A 4 o Alpha decay 238 92 U → z X + 2 He An alpha particle is a helium nucleus, consisting of 2 protons and 2 neutrons. When an element emits an alpha particle, its nucleus loses 2 atomic numbers and 4 mass numbers, and thus changes into another element, called the daughter element. o Beta decay 146 C → ZA X + −10 e Beta decay is really just a neutron emitting an electron and becoming a proton. Thus, the daughter element resulting from beta decay is one atomic number higher than the parent nucleus, but the mass number essentially does not change. o Gamma decay ZA X * → ZA X + photon The excited nucleus emits a gamma photon when it relaxes, so only the energy of the nucleus changes and there is no change for A nor Z. o Positron decay ZA X → Z −A1Y + +10 e + 00ν A positron is exactly like an electron except for the fact that it is positively charged. Positron decay equations are typically not included on the AP Physics B exam. • Nuclear Fission Nuclear fission is when an atomic nuclei splits into two or more pieces. Fission is usually caused artificially by shooting a slow neutron at a large atom such as uranium which absorbs the neutron and splits into two smaller atoms, along with the release of more neutrons and some energy. 94 1 U + 01n → 140 54 Xe + 38 Sr + 2 0 n + 160 MeV 235 92 Since the reaction produces two new neutrons, these can be used to split new Uranium nuclei (4, 8, 16, 32, etc.) in an uncontrolled chain reaction. A controlled chain reaction can be obtained if the number of neutrons can cause a new fission with a multiplication factor of one. • Nuclear Fusion Fusion occurs when small nuclei combine into larger nuclei and energy is released. 13 H + 12 H → 24 He + 01n Note that the sum of the atomic numbers on the left must equal the sum of the atomic numbers on the right. The same is true of the mass numbers. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics Free Response Question 1 (15 pts) A. Electrons in the ground state of a gas are excited to the n = 3 energy level as shown in the Energy Level diagram above. What are the energies of all the possible photons that could be emitted by this gas? (4 points max) 1 point for a correct equation for the energy emitted by a photon E = E0 − E f E = −1eV − (−10eV ) = 9 eV E = −4eV − (−10eV ) = 6 eV E = −1eV − (−4eV ) = 3 eV 1 point for the correct answer with units 9 eV 1 point for the correct answer with units 6 eV 1 point for the correct answer with units 3 eV ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics B. Determine the possible wavelengths for the emitted photons. 1 point for a correct equation for calculating the wavelength of the photon (5 points max) hc E 1240eVnm λ= = 138 nm 9eV 1240eVnm λ= = 207 nm 6eV 1240eVnm λ= = 413 nm 3eV λ= 1 point for correct values of E or answers consistent with part A 1 point for correct answer with units 138 nm 1 point for correct answer with units 207 nm 1 point for correct answer with units 413 nm C. State whether or not any of the emitted photons are in the visible range. Justify your answer. 1 point for the correct answer based on part B (2 points max) The emitted photon of wavelength 413 nm is in the visible range, since the visible range is approximately 400 nm to 700 nm. ® 1 point for stating that the visible range is approximately 400 nm to 700 nm Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics D. Determine the frequency of the absorbed photons that caused the electrons to be excited to the third energy level. (4 points max) E = E0 − E f = −1eV − ( −10eV ) = 9 eV E = hf f = E = h ⎛ 1.6 ×10−19 J ⎞ ⎟ eV ⎝ ⎠ = 2.17 ×1015 Hz −34 ( 6.63 ×10 J is ) ( 9eV ) ⎜ 1 point for a correct equation for the energy absorbed by the photon 1 point for the correct value of E 1 point for the correct equation for the frequency of a photon 1 point for the correct answer with units Alternate solution hc 1240eV inm = 138 nm λ= = E 9eV c 3.0 × 108 m / s = 2.17 × 1015 Hz f = = λ 1.38 ×10−7 m ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics Question 2 (15 pts) Kinetic Energy versus frequency Light is shined on a photoemissive surface. The graph above shows the maximum kinetic energy of each electron ( ×10−20 J ) versus the frequency of the incoming light ( ×1014 Hz ). ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics A. On the graph above, draw the line that is your estimate of the best straight-line fit to the data points. 1 point for at least one point above and one point below the best fit line (2 points max) 1 point for a straight line that intersects the x axis between 4 × 1014 Hz and 5 × 1014 Hz B. Using your graph, find a value for Planck’s constant, and briefly explain how you found the value. 1 point for recognition that Planck’s constant is the slope of the best fit line (3 points max) Planck’s constant is equal to the slope of the graph. Choose two points on the best fit line to find the slope. ( 5 ×1014 Hz, 2 ×10−20 J ) (8 ×10 14 Hz , 21× 10 −20 J) −20 −20 ΔKE ( 21× 10 J − 2 × 10 J ) h = slope = = Δf (8 ×1014 Hz − 5 ×1014 Hz ) 1 point for picking two points off the best fit line 1 point for a reasonable value for Planck’s constant with correct units and a reasonable margin of error h = 6.33 ×10−34 J is ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics C. From the graph, estimate the threshold frequency of the photoemissive surface. (2 points max) The threshold frequency is the x intercept, where the graph crosses the frequency axis. From the graph, f0 = 4.6 x 1014 Hz. 1 point for recognition that the threshold frequency is the x intercept 1 point for correct answer within a reasonable margin of error D. Determine the work function of the photoemissive surface. (2 points max) φ = hf 0 φ = ( 6.63 ×10−34 J i s )( 4.6 ×1014 Hz ) φ = 3.0 ×10−19 J or 1.9 eV The work function can be found by extending the graph and finding the negative y intercept. 30 × 10 −20 J ® 1 point for using the threshold frequency for calculating the work function or for recognition that the work function is the negative y intercept 1 point for correct answer with units Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics E. Determine the de Broglie wavelength of the ejected electrons that are ejected by a photon of frequency 8 × 1014 Hz . (4 points max) From the graph, the KE associated with the frequency is 2.1x10-19 J. 1 KE = mv 2 2 2 ( 2.1× 10−19 J ) 2 KE = v= 9.11×10−31 kg m v = 6.79 × 105 m/s h λ= mv 6.63 × 10−34 J / Hz λ= ( 9.11×10−31 kg )( 6.79 ×105 m / s ) 1 point for finding the KE associated with the frequency 1 point for finding the speed of the electron using KE 1 point for the de Broglie equation 1 point for the correct answer including units λ = 1.07 ×10−9 m F. What voltage, called the stopping voltage, would be necessary to stop the emission of electrons that are ejected by a photon of frequency 8 × 1014 Hz . (2 points max) From the graph, the KE associated with a frequency of 8 × 1014 Hz is 2.1× 10−19 J Stopping the emission of electrons requires work equal to the maximum KE of the electrons; thus, it would take 1.3 V. 1 point for the correct equation for finding the stopping voltage 1 point for correct answer with units KEmax = W = qeVstop Vstop = KEmax 2.1×10−19 J = = 1.3 V qe 1.6 ×10−19 C ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics Multiple Choice 1. Light of various frequencies is shined on a metal surface and in each case the stopping potential is measured. The work function for the metal surface is given by φ , the charge of the electron by e , and h is Planck’s constant. In terms of the given quantities and fundamental constants, the slope of this straight line is A) B) C) D) E) h e h φ h φ h φ e K max = hf − φ and K max = eVstop eVstop = hf − φ A φ ⎛h⎞ Vstop = ⎜ ⎟ f − e ⎝e⎠ y = mx + b The slope of the line is ® h e Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics 2. An X ray photon collides with a stationary electron. After the collision, the electron departs with a velocity v and the X ray is scattered at an angle as shown above. Which statements about this collision are correct? I. Energy is conserved. II. Momentum is conserved. III. The frequency of the scattered X ray decreases. A) B) C) D) E) I only II only III only I and II only I, II, and III Energy and momentum are conserved and λ = E h h ⇒ p= . p λ Thus, a decrease in the photon’s momentum corresponds to an increase in the photon’s wavelength or a decrease in its frequency, since momentum and wavelength are inversely proportional to each other. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics 3. A photoemissive surface is irradiated with light and no electrons are emitted. All of the following will result in no electrons being emitted EXCEPT A) B) C) D) E) decreasing the wavelength of the light decreasing the frequency of the light decreasing the intensity of the light increasing the intensity of the light increasing the wavelength of the light A Not all frequencies of light cause electrons to be ejected from a photoemissive surface. Photoemissive materials have various threshold frequencies necessary to eject electrons. Decreasing the wavelength of the light, means increasing the frequency and this will cause the ejection of electrons from the photoemissive surface when the frequency equals or exceeds the threshold frequency. 4. Seven protons and seven neutrons are brought together to form a nitrogen nucleus, but the mass of the nitrogen nucleus is less than the sum of the masses of the individual particles that make up the nucleus. This missing mass, called the mass defect, has been A) B) C) D) E) converted into the binding energy of the nucleus given off in a radioactive decay process converted into electrons converted into energy to hold the electrons in orbit emitted as light A Each particle that makes up the nucleus gives up a little mass to be converted into energy by E = mc2 to bind the nucleus together. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics Questions 5-6: Consider the electron energy level diagram for hydrogen shown below. 5. Electrons from a quantity of hydrogen gas drop from the state of n = 3 to n = 1 in one or more jumps. The maximum number of spectral lines of different frequencies that can be produced from these transitions is: A) B) C) D) E) 1 2 3 4 5 C Possible transitions are from n=3 to n=2, then from n=2 to n=1, and from n=3 to n=1. 6. In which of the following transitions, will the emitted photon have the greatest wavelength? A) B) C) D) E) n=5 to n=4 n=4 to n=3 n=3 to n=2 n=2 to n=2 All transitions will emit a photon with the same wavelength. A hc so the transition with the smallest energy which is from n=5 to E n=4 will emit a photon with the greatest wavelength. λ= ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics 7. Carbon, 146C , undergoes beta decay. The resulting product X has an atomic configuration of A) 10 4 X B) 14 5 X C) 14 6 X D) 14 7 X E) 16 8 X C → ZA X + −10 e 14 6 D C → 147 X + −10 e 14 6 8. Uranium decays into an isotope of Thorium as shown in the nuclear equation below. U→ 238 92 A) B) C) D) E) Th + ?+ kinetic energy What other particle is formed in this reaction? 234 90 alpha particle beta particle gamma ray neutron positron A 4 2 He alpha particle ® U→ 238 92 Th + 24 He + kinetic energy 234 90 Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Atomic & Wave/Particle Physics 9. Light exhibits a wave-particle duality. An apple does not exhibit wave like properties because A) B) C) D) E) apples have a large mass apples are made of particles apples do not travel at speeds close to the speed of light on Earth, gravity acts on particles like apples, but not waves the laws of quantum mechanics can only be applied to subatomic particles h We typically don’t notice the wave properties of objects mv moving around us because Planck’s constant is incredibly small compared to their mass. Nevertheless, the wavelength of any moving mass is inversely proportional to its momentum. λ= A 10. An electron and a proton have the same wavelength. Which of the following statements are correct? I. They have the same momentum. II. They have the same speed. III. The kinetic energy of the proton is less than that of the electron. A) B) C) D) E) I only II only III only I and III only I, II, and III h . p The mass of the proton is greater than that of the electron and hence, the kinetic energy of the proton is less than that of the electron since The momentum of the electron and proton are the same since λ = D mv 2 ( mv ) p2 . K= = = 2 2m 2m 2 ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org