2011 Semifinal Exam AAPT UNITEDSTATES PHYSICS TEAM AIP 2011 Semifinal Exam DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor This examination consists of two parts. Part A has four questions and is allowed 90 minutes. Part B has two questions and is allowed 90 minutes. The first page that follows is a cover sheet.Examinees may keep the cover sheet for both parts of the exam. The parts are then identified by the center header on each page.Examinees are only allowed to do one part at a time,and may not work on other parts,even if they have time remaining Allow 90 minutes to complete Part A.Do not let students look at Part B.Collect the answers to Part A before allowing the examinee to begin Part B.Examinees are allowed a 10 to 15 minutes break between parts A and B. Allow 90 minutes to complete Part B.Do not let students go back to Part A. Ideally the test supervisor will divide the question paper into 3 parts:the cover sheet (page 2),Part A (pages 3-4),and Part B(pages 6-7).Examinees should be provided parts A and B individually,although they may keep the cover sheet. The supervisor must collect all examination questions,including the cover sheet,at the end of the exam,as well as any scratch paper used by the examinees.Examinees may not take the exam questions.The examination questions may be returned to the students after March 31.2011. Examinees are allowed calculators,but they may not use symbolic math,programming,or graphic features of these calculators.Calculators may not be shared and their memory must be cleared of data and programs.Cell phones,PDA's or cameras may not be used during the exam or while the exam papers are present.Examinees may not use any tables,books, or collections of formulas. Please provide the examinees with graph paper for Part A. Copyright C2011 American Association of Physics Teachers
2011 Semifinal Exam 1 AAPT UNITED STATES PHYSICS TEAM AIP 2011 Semifinal Exam DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor • This examination consists of two parts. • Part A has four questions and is allowed 90 minutes. • Part B has two questions and is allowed 90 minutes. • The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam. • The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining. • Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B. • Allow 90 minutes to complete Part B. Do not let students go back to Part A. • Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-4), and Part B (pages 6-7). Examinees should be provided parts A and B individually, although they may keep the cover sheet. • The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after March 31, 2011. • Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas. • Please provide the examinees with graph paper for Part A. Copyright c 2011 American Association of Physics Teachers
2011 Semifinal Exam Cover Sheet 2 AAPT UNITEDSTATES PHYSICS TEAM AIP 2011 Semifinal Exam INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN Work Part A first.You have 90 minutes to complete all four problems.Each question is worth 25 points.Do not look at Part B during this time. After you have completed Part A you may take a break. Then work Part B.You have 90 minutes to complete both problems.Each question is worth 50 points.Do not look at Part A during this time. Show all your work.Partial credit will be given.Do not write on the back of any page.Do not write anything that you wish graded on the question sheets. Start each question on a new sheet of paper.Put your AAPT ID number,your name,the question number and the page number/total pages for this problem,in the upper right hand corner of each page.For example, AAPT ID# Doe,Jamie A1-1/3 A hand-held calculator may be used.Its memory must be cleared of data and programs.You may use only the basic functions found on a simple scientific calculator.Calculators may not be shared.Cell phones,PDA's or cameras may not be used during the exam or while the exam papers are present.You may not use any tables,books,or collections of formulas. Questions with the same point value are not necessarily of the same difficulty. In order to maintain exam security,do not communicate any information about the questions (or their answers/solutions)on this contest until after April 1,2011. Possibly Useful Information.You may use this sheet for both parts of the exam. g=9.8 N/kg G=6.67×10-11Nm2/kg2 k=1/4re0=8.99×109N.m2/c2 km=4o/4r=10-7T.m/A c=3.00×108m/s k3=1.38×10-23J/K NA=6.02×1023(mol)-1 R=NA=8.31J/mol.K)） o=5.67×10-8J/(s·m2.K4) e=1.602×10-19C 1eV=1.602×10-19J h=6.63×10-34J.s=4.14×10-15eV.s me=9.109×10-31kg=0.511MeV/c2(1+x)m≈1+nx for<1 sin0≈0-03forl9l≤1 cos0≈1-02forl0l<1 Copyright C2011 American Association of Physics Teachers
2011 Semifinal Exam Cover Sheet 2 AAPT UNITED STATES PHYSICS TEAM AIP 2011 Semifinal Exam INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time. • After you have completed Part A you may take a break. • Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time. • Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets. • Start each question on a new sheet of paper. Put your AAPT ID number, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example, AAPT ID # Doe, Jamie A1 - 1/3 • A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. • Questions with the same point value are not necessarily of the same difficulty. • In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after April 1, 2011. Possibly Useful Information. You may use this sheet for both parts of the exam. g = 9.8 N/kg G = 6.67 × 10−11 N · m2/kg2 k = 1/4π0 = 8.99 × 109 N · m2/C 2 km = µ0/4π = 10−7 T · m/A c = 3.00 × 108 m/s kB = 1.38 × 10−23 J/K NA = 6.02 × 1023 (mol)−1 R = NAkB = 8.31 J/(mol · K) σ = 5.67 × 10−8 J/(s · m2 · K4 ) e = 1.602 × 10−19 C 1eV = 1.602 × 10−19 J h = 6.63 × 10−34 J · s = 4.14 × 10−15 eV · s me = 9.109 × 10−31 kg = 0.511 MeV/c 2 (1 + x) n ≈ 1 + nx for |x| 1 sin θ ≈ θ − 1 6 θ 3 for |θ| 1 cos θ ≈ 1 − 1 2 θ 2 for |θ| 1 Copyright c 2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 3 Part A Question Al Single bubble sonoluminescence occurs when sound waves cause a bubble suspended in a fluid to collapse so that the gas trapped inside increases in temperature enough to emit light.The bubble actually undergoes a series of expansions and collapses caused by the sound wave pressure variations. We now consider a simplified model of a bubble undergoing sonoluminescence.Assume the bub- ble is originally at atmospheric pressure Po =101 kPa.When the pressure in the fluid surrounding the bubble is decreased,the bubble expands isothermally to a radius of 36.0 um.When the pressure increases again,the bubble collapses to a radius of 4.50 um so quickly that no heat can escape. Between the collapse and subsequent expansion,the bubble undergoes isochoric(constant volume) cooling back to its original pressure and temperature.For a bubble containing a monatomic gas, suspended in water of T=293 K,find a.the number of moles of gas in the bubble, b.the pressure after the expansion, c.the pressure after collapse, d.the temperature after the collapse,and e.the total work done on the bubble during the whole process. You may find the following useful:the specific heat capacity at constant volume is Cy =3R/2 and the ratio of specific heat at constant pressure to constant volume is y=5/3 for a monatomic gas. Solution We consider the bubble to be filled with an ideal monatomic gas,so originally:PoVo =nRTo. The bubble undergoes 3 processes:1)isothermal expansion,2)adiabatic collapse (no heat escapes),and 3)isochoric (constant volume)cooling.The final process is isochoric,so we know that the bubble's collapsed volume is equal to its original volume,so 2=%, and PoVo=PoV2 nRTo. Rearranging, PoV2 n= RTo 乃π3 n= RTo 0量·新~(4.50×10-6m3 101,000 3.86×10-11 n= moles 8.31m0K293K 2430 Copyright C2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 3 Part A Question A1 Single bubble sonoluminescence occurs when sound waves cause a bubble suspended in a fluid to collapse so that the gas trapped inside increases in temperature enough to emit light. The bubble actually undergoes a series of expansions and collapses caused by the sound wave pressure variations. We now consider a simplified model of a bubble undergoing sonoluminescence. Assume the bubble is originally at atmospheric pressure P0 = 101 kPa. When the pressure in the fluid surrounding the bubble is decreased, the bubble expands isothermally to a radius of 36.0 µm. When the pressure increases again, the bubble collapses to a radius of 4.50 µm so quickly that no heat can escape. Between the collapse and subsequent expansion, the bubble undergoes isochoric (constant volume) cooling back to its original pressure and temperature. For a bubble containing a monatomic gas, suspended in water of T = 293 K, find a. the number of moles of gas in the bubble, b. the pressure after the expansion, c. the pressure after collapse, d. the temperature after the collapse, and e. the total work done on the bubble during the whole process. You may find the following useful: the specific heat capacity at constant volume is CV = 3R/2 and the ratio of specific heat at constant pressure to constant volume is γ = 5/3 for a monatomic gas. Solution We consider the bubble to be filled with an ideal monatomic gas, so originally: P0V0 = nRT0. The bubble undergoes 3 processes: 1) isothermal expansion, 2) adiabatic collapse (no heat escapes), and 3) isochoric (constant volume) cooling. The final process is isochoric, so we know that the bubble’s collapsed volume is equal to its original volume, so V2 = V0, and P0V0 = P0V2 = nRT0. Rearranging, n = P0V2 RT0 n = P0 4 3 πr3 2 RT0 n = 101, 000 N m2 · 4 3 π · (4.50 × 10−6 m)3 8.31 J mol·K · 293 K = 3.86 × 10−11 2430 moles Copyright c 2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 4 a n=1.58×10-14 noles. Process 1:Isothermal expansion This process is isothermal,so Ti=To and 乃M=nRT1=nRTo B=nm=1.58×10-14 moles831nJ moR·293K 巧 等r·(3.60×10-5m3 b) A=1970=197Pa. The work done by the bubble is: W1 nRTo ln M 6 W=1.58×10-14 moles8.31-J .293K,1n (3.60×10-5)3 mol.K (4.50×10-6)3 W1=2.40×10-10J So,the work done on the bubble during the expansion is: W1=-2.40×10-10J. Process 2:Adiabatic collapse For an adiabatic process D?=P② B AY For a monatomic gas y=5/3 so, P2= 197品·3.60×10-5m5 (4.50×10-6m)5 c) P2=6.46×106Pa. And 及 nR T2= 6.46×106Pa专π·(4.50×10-6m)3 158×10-4 moles-8,31mdK d) T2 =18800 K.Lord,have mercy!That's hot! The work done by the bubble during an adiabatic process is W2=-△Einternal=-nC,△T Copyright C2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 4 a) n = 1.58 × 10−14 moles. Process 1: Isothermal expansion This process is isothermal, so T1 = T0 and P1V1 = nRT1 = nRT0 P1 = nRT0 V1 = 1.58 × 10−14 moles · 8.31 J mol·K · 293 K 4 3 π · (3.60 × 10−5 m)3 b) P1 = 197 N m2 = 197 Pa. The work done by the bubble is: W1 = nRT0 ln V1 V0 W1 = 1.58 × 10−14 moles · 8.31 J mol·K · 293 K · ln (3.60 × 10−5 ) 3 (4.50 × 10−6) 3 W1 = 2.40 × 10−10 J So, the work done on the bubble during the expansion is: W1 = −2.40 × 10−10 J. Process 2: Adiabatic collapse For an adiabatic process P1V γ 1 = P2V γ 2 P2 = P1V γ 1 V γ 2 For a monatomic gas γ = 5/3 so, P2 = 197 N m2 · (3.60 × 10−5 m)5 (4.50 × 10−6 m)5 c) P2 = 6.46 × 106 Pa. And T2 = P2V2 nR T2 = 6.46 × 106 Pa · 4 3 π · (4.50 × 10−6 m)3 1.58 × 10−14 moles · 8.31 J mol·K d) T2 = 18800 K. Lord, have mercy! That’s hot! The work done by the bubble during an adiabatic process is W2 = −∆Einternal = −nCv∆T Copyright c 2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 5 where C=3R/2. W2=-1.58×10-14 moles 2 831J mol.K ·(18800-293)K W3=-3.64×10-9J The work done on the bubble is then W2=3.64×10-9J Process 3:Isochoric cooling The work done on the bubble during an isochoric process is zero,so W3=0J. The total work is then the sum of the work on the bubble Wtotal W1 W2 W3 Witotal=-2.40×10-10J+3.64×10-9J+0J e) Wtotal=3.4×109J. Question A2 A thin,uniform rod of length L and mass M=0.258 kg is suspended from a point a distance R away from its center of mass.When the end of the rod is displaced slightly and released it executes simple harmonic oscillation.The period,T,of the oscillation is timed using an electronic timer. The following data is recorded for the period as a function of R.What is the local value of g?Do not assume it is the canonical value of 9.8 m/s2.What is the length,L,of the rod?No estimation of error in either value is required.The moment of inertia of a rod about its center of mass is (1/12)ML2. R T R (m) (s) (m) (s) 0.050 3.842 0.211 2.074 0.075 3.164 0.302 1.905 0.102 2.747 0.387 1.855 0.156 2.301 0.451 1.853 0.198 2.115 0.588 1.900 You must show your work to obtain full credit.If you use graphical techniques then you must plot the graph;if you use linear regression techniques then you must show all of the formulae and associated workings used to obtain your result. Solution The period of a physical pendulum is given by I T=2r1 mgR= 2π bL2+必 9R Copyright C2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 5 where Cv = 3R/2. W2 = −1.58 × 10−14 moles · 3 2 8.31 J mol·K · (18800 − 293) K W2 = −3.64 × 10−9 J The work done on the bubble is then W2 = 3.64 × 10−9 J Process 3: Isochoric cooling The work done on the bubble during an isochoric process is zero, so W3 = 0 J. The total work is then the sum of the work on the bubble Wtotal = W1 + W2 + W3 Wtotal = −2.40 × 10−10 J + 3.64 × 10−9 J + 0 J e) Wtotal = 3.4 × 10−9 J. Question A2 A thin, uniform rod of length L and mass M = 0.258 kg is suspended from a point a distance R away from its center of mass. When the end of the rod is displaced slightly and released it executes simple harmonic oscillation. The period, T, of the oscillation is timed using an electronic timer. The following data is recorded for the period as a function of R. What is the local value of g? Do not assume it is the canonical value of 9.8 m/s2 . What is the length, L, of the rod? No estimation of error in either value is required. The moment of inertia of a rod about its center of mass is (1/12)ML2 . R T (m) (s) 0.050 3.842 0.075 3.164 0.102 2.747 0.156 2.301 0.198 2.115 R T (m) (s) 0.211 2.074 0.302 1.905 0.387 1.855 0.451 1.853 0.588 1.900 You must show your work to obtain full credit. If you use graphical techniques then you must plot the graph; if you use linear regression techniques then you must show all of the formulae and associated workings used to obtain your result. Solution The period of a physical pendulum is given by T = 2π s I mgR = 2π s 1 12L2 + R2 gR Copyright c 2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 6 A little math,and T2R 12=R2 94r2- 12 This is of the form mx +b=y if we let y=R2 and T2R T= 4r2 R T T2R/4R2 R 0.050 3.842 0.0187 0.0025 0.075 3.164 0.0190 0.0056 0.102 2.747 0.0195 0.0104 0.156 2.301 0.0209 0.0243 Filling out a table of data,we get 0.198 2.115 0.0224 0.0392 The corresponding graph of 0.211 2.074 0.0230 0.0445 0.302 1.905 0.0278 0.0912 0.387 1.855 0.0337 0.1498 0.451 1.853 0.0392 0.2034 0.588 1.900 0.0538 0.3457 T2R42 versus R2ought yield a straight line such that the slope isg and the intercept is 04000 0.3500 0300 f0)=9.786x-0.180 R2=1.000 02500 02000 01500 0.100 00500 001500030002500s0000g5000400.045000000.a5000G00 9=9.7923m/s2 and L=1.470m Copyright C2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 6 A little math, and g T 2R 4π 2 − 1 12 L 2 = R 2 . This is of the form mx + b = y if we let y = R 2 and x = T 2R 4π 2 Filling out a table of data, we get R T T 2R/4π 2 R2 0.050 3.842 0.0187 0.0025 0.075 3.164 0.0190 0.0056 0.102 2.747 0.0195 0.0104 0.156 2.301 0.0209 0.0243 0.198 2.115 0.0224 0.0392 0.211 2.074 0.0230 0.0445 0.302 1.905 0.0278 0.0912 0.387 1.855 0.0337 0.1498 0.451 1.853 0.0392 0.2034 0.588 1.900 0.0538 0.3457 The corresponding graph of T 2/R/4π 2 versus R2 ought yield a straight line such that the slope is g and the intercept is − 1 12L 2 . g = 9.7923 m/s 2 and L = 1.470 m Copyright c 2011 American Association of Physics Teachers
2011 Semifinal Exam Part A Question A3 A light bulb has a solid cylindrical filament of length L and radius a,and consumes power P. You are to design a new light bulb,using a cylindrical filament of the same material,operating at the same voltage,and emitting the same spectrum of light,which will consume power nP.What are the length and radius of the new filament?Assume that the temperature of the filament is approximately uniform across its cross-section;the filament doesn't emit light from the ends;and energy loss due to convection is minimal. Solution Since the new bulb emits the same spectrum of light,the emitted power is simply proportional to the area: P 2TaL P aL If the resistivity of the filament is p,the resistance is Ta2 and therefore the power is also given by p=y、V2ma2 五= pL a2 Combining our conditions, a x p2/3 L Pl/3 So the new filament must have length n2/3a and length n/3L. Question A4 In this problem we consider a simplified model of the electromagnetic radiation inside a cubical box of side length L.In this model,the electric field has spatial dependence E(x,y,z)=Eo sin(kzx)sin(kuy)sin(kzz) where one corner of the box lies at the origin and the box is aligned with the x,y,and z axes.Let h be Planck's constant,kB be Boltzmann's constant,and c be the speed of light. a.The electric field must be zero everywhere at the sides of the box.What condition does this impose on kz,ky,and k:?(Assume that any of these may be negative,and include cases where one or more of the ki is zero,even though this causes E to be zero.) b.In the model,each permitted value of the triple (ky,k)corresponds to a quantum state. These states can be visualized in a state space,which is a notional three-dimensional space Copyright C2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 7 Question A3 A light bulb has a solid cylindrical filament of length L and radius a, and consumes power P. You are to design a new light bulb, using a cylindrical filament of the same material, operating at the same voltage, and emitting the same spectrum of light, which will consume power nP. What are the length and radius of the new filament? Assume that the temperature of the filament is approximately uniform across its cross-section; the filament doesn’t emit light from the ends; and energy loss due to convection is minimal. Solution Since the new bulb emits the same spectrum of light, the emitted power is simply proportional to the area: P ∝ 2πaL P ∝ aL If the resistivity of the filament is ρ, the resistance is R = ρL A = ρL πa2 and therefore the power is also given by P = V 2 R = V 2πa2 ρL P ∝ a 2 L Combining our conditions, a ∝ P 2/3 L ∝ P 1/3 So the new filament must have length n 2/3a and length n 1/3L. Question A4 In this problem we consider a simplified model of the electromagnetic radiation inside a cubical box of side length L. In this model, the electric field has spatial dependence E(x, y, z) = E0 sin(kxx) sin(kyy) sin(kzz) where one corner of the box lies at the origin and the box is aligned with the x, y, and z axes. Let h be Planck’s constant, kB be Boltzmann’s constant, and c be the speed of light. a. The electric field must be zero everywhere at the sides of the box. What condition does this impose on kx, ky, and kz? (Assume that any of these may be negative, and include cases where one or more of the ki is zero, even though this causes E to be zero.) b. In the model, each permitted value of the triple (kx, ky, kz) corresponds to a quantum state. These states can be visualized in a state space, which is a notional three-dimensional space Copyright c 2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 8 with axes corresponding to k,ky;and k:.How many states occupy a volume s of state space, if s is large enough that the discreteness of the states can be ignored? c.Each quantum state,in turn,may be occupied by photons with frequency ==ckl, where lkl=Vka2+ky2+2 In the model,if the temperature inside the box is T,no photon may have energy greater than kBT.What is the shape of the region in state space corresponding to occupied states? d.As a final approximation,assume that each occupied state contains exactly one photon.What is the total energy of the photons in the box,in terms of h,kB,c,T,and the volume of the box V?Again,assume that the temperature is high enough that there are a very large number of occupied states.(Hint:divide state space into thin regions corresponding to photons of the same energy.) Note that while many details of this model are extremely inaccurate,the final result is correct except for a numerical factor. Solution a.We require that sin(L)=0,so that krL=nxπ for any integer n,and similarly for ky and k. b.The occupied states are equally spaced a distance f apart.Each can therefore be thought of as taking up volume and the number of states in the volumesis L3 c.A photon's energy is E=w=hck],wheren=Thus the occupied states obey ickl≤kBT 因s梁 This corresponds to a ball of radius in state space. d.As we have seen,the energy of a photon is proportional to its distance k from the origin in state space.Thus consider the spherical shell in state space between radius k and radius k+dk.The volume of this region is ds 4nk2dk Each photon in the region has energy fick,and from above there areds photons in the region.Therefore the photons in the region have total energy L dE huck. 3·4rk2dk Copyright C2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 8 with axes corresponding to kx, ky, and kz. How many states occupy a volume s of state space, if s is large enough that the discreteness of the states can be ignored? c. Each quantum state, in turn, may be occupied by photons with frequency ω = f 2π = c|k|, where |k| = q kx 2 + ky 2 + kz 2 In the model, if the temperature inside the box is T, no photon may have energy greater than kBT. What is the shape of the region in state space corresponding to occupied states? d. As a final approximation, assume that each occupied state contains exactly one photon. What is the total energy of the photons in the box, in terms of h, kB, c, T, and the volume of the box V ? Again, assume that the temperature is high enough that there are a very large number of occupied states. (Hint: divide state space into thin regions corresponding to photons of the same energy.) Note that while many details of this model are extremely inaccurate, the final result is correct except for a numerical factor. Solution a. We require that sin(kxL) = 0, so that kxL = nxπ for any integer nx, and similarly for ky and kz. b. The occupied states are equally spaced a distance π L apart. Each can therefore be thought of as taking up volume π 3 L3 , and the number of states in the volume s is L 3 π 3 s c. A photon’s energy is E = ¯hω = ¯hc|k|, where ¯h = h 2π . Thus the occupied states obey ¯hc|k| ≤ kBT |k| ≤ kBT ¯hc This corresponds to a ball of radius kBT ¯hc in state space. d. As we have seen, the energy of a photon is proportional to its distance |k| from the origin in state space. Thus consider the spherical shell in state space between radius k and radius k + dk. The volume of this region is ds = 4πk2 dk Each photon in the region has energy ¯hck, and from above there are L3 π3 ds photons in the region. Therefore the photons in the region have total energy dE = ¯hck · L 3 π 3 · 4πk2 dk Copyright c 2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 9 dE From above,.kranges from zero to=第是，so the total energy is B=u() Since the volume of the box is V=L3,and h=2mh,this cleans up to TknTV E二 h3c3 Copyright C2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 9 dE = 4 π 2 ¯hcL3 k 3 dk From above, k ranges from zero to kmax = kBT ¯hc , so the total energy is E = Z kmax 0 4 π 2 ¯hcL3 k 3 dk E = 4 π 2 ¯hcL3 · 1 4 kBT ¯hc 4 Since the volume of the box is V = L 3 , and h = 2π¯h, this cleans up to E = 8πkB 4 h 3c 3 T 4V Copyright c 2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 10 STOP:Do Not Continue to Part B If there is still time remaining for Part A,you should review your work for Part A,but do not continue to Part B until instructed by your exam supervisor. Copyright C2011 American Association of Physics Teachers
2011 Semifinal Exam Part A 10 STOP: Do Not Continue to Part B If there is still time remaining for Part A, you should review your work for Part A, but do not continue to Part B until instructed by your exam supervisor. Copyright c 2011 American Association of Physics Teachers