Problem Set 3: Solutions

Transcription

1 University of Alabama Department of Pysics and Astronomy PH 253 / LeClair Spring 2010 Problem Set 3: Solutions 1. Te energy required to break one OO bond in ozone O 3, OOO) is about 500 kj/mol. Wat is te maximum wavelengt of te poton tat as enoug energy to poto-dissociate ozone by breaking one of te OO bonds? O 3 f O + O 2 Note Avagadro s number is N A tings/mol. If we are to break te double oxygen bond in ozone, we need to supply a poton wit an energy greater or equal to te bond energy. Two adjacent oxygens are ultimately bonded togeter because tey gain E 500 kj/mol wort of energy between tem to stay tat way - if we want to coax tem apart and break te bond, we need to supply tat muc energy wit an incident poton. If we can convert 500 kj/mol to an energy per bond in, say, electron volts, we can find out wat sort of poton as enoug energy to break te bond. To do tis, we must create an unoly alliance of cemistry and pysics units: E J mol J bond 5.18 ev bond 1 mol bonds 1 ev J Now we are getting somewere! It takes just over 5 electron volts per bond to break an oxygen double bond in an ozone molecule. An incident poton wit at least tis muc energy can be absorbed by one of te oxygen atoms, wic will ten ave enoug energy to leave its bound state and break te bond. Tus, to break a single bond: E poton c 5.18 ev λ λ c 5.18 ev ev m 5.18 ev m 239 nm

2 Here we used our andy relationsip from te last problem - c ev m. A poton of wavelengt 239 nm or lower will break up an ozone molecule, wic is well into te ultraviolet UV). Tis is one way te ozone layer protects us - it absorbs armful UV radiation and prevents it from reacing te eart s surface. 2. Park 1.2 Sow tat it is impossible for a poton striking a free electron to be absorbed and not scattered. All we really need to do is conserve energy and momentum for poton absorption by a stationary, free electron and sow tat someting impossible is implied. Before te collision, we ave a poton of energy f and momentum /λ and an electron wit rest energy mc 2. Afterward, we ave an electron of energy γ 1)+mc 2 p 2 c 2 + m 2 c 4 i.e., te afterward te electron as acquired kinetic energy, but retains its rest energy) and momentum p e γmv. Momentum conservation dictates tat te absorbed poton s entire momentum be transferred to te electron, wic means it must continue along te same line tat te incident poton traveled. Tis makes te problem one dimensional, wic is nice. Enforcing conservation of energy and momentum, we ave: initial) final) 1) f + mc 2 p 2 c 2 + m 2 c 4 energy conservation variant 1 2) f + mc 2 γ 1) mc 2 energy conservation variant 2 3) λ p e γmv momentum conservation 4) From tis point on, we can approac te problem in two ways, using eiter expression for te electron s energy. We ll do bot, just to give you te idea. First, we use conservation of momentum to put te electron momentum in terms of te poton frequency: λ p e c λ f p ec 5) Now substitute tat in te first energy conservation equation to eliminate p e, square bot sides, and collect terms: f + mc 2 ) 2 p 2 c 2 + m 2 c 4 ) 2 2 f 2 + m 2 c 4 ) 2 6) 2 f 2 + 2fmc 2 + m 2 c 4 2 f 2 + m 2 c 4 7) 2fmc 2 0 f 0 p e v 0 8) Tus, we conclude tat te only way a poton can be absorbed by te stationary electron is if its

3 frequency is zero, i.e., if tere is no poton to begin wit! Clearly, tis is silly. We can also use te second variant of te conservation of energy equation along wit momentum conservation to come to an equally ridiculous conclusion: f c λ γ 1) mc2 energy conservation variant 2 9) λ γmv or c γmvc momentum conservation 10) λ γmvc γ 1) mc 2 11) γ 1) c γv γ 1 v 1 γ c 1 γ 2 definition of γ) 13) ) γ γ γ 2 14) γ 2 2γ + 1 γ ) 15) γ 1 v 0 16) Again, we find an electron recoil velocity of zero, implying zero incident poton frequency, wic means tere is no poton in te first place! Conclusion: stationary electrons cannot absorb potons, but tey can Compton scatter tem. 3. Park 1.3 Wat is te expected recoil velocity of a sodium atom wic at rest emits a quantum of its λ589.0 nm radiation? We need only conservation of momentum. Initially, te sodium atom of mass m is at rest. After te poton emission, te poton carries away momentum p /λ, and conservation of momentum dictates tat te sodium atom ave equal and opposite momentum pγmv: 0 γmv λ γv v 1 v 2 /c 2 mλ 2 m 2 λ 2 v 2 1 v 2 /c 2 19) ) ) v m 2 λ 2 v2 c 2 20) ) ) 1/2 v ± mλ m 2 c 2 λ 2 21) 17) 18) Te atomic mass of sodium is m u kg, leading to v 0.03 m/s. Wit tis

4 small velocity, we really did not require relativity. Using p mv for te sodium atom s momentum, we find v mλ 22) wic is consistent wit a Taylor expansion of our relativistic result for v c. Incidentally, anoter way to determine if relativity is really required is to compare te rest energy of te sodium atom and poton. If te latter is relatively small, relativity is not required. f mc 2 1 λ p poton mc 23) If te poton as a negligible fraction of te atom s rest energy, or equivalently its momentum is small compared to mc, te relativistic correction is negligible. Anoter ting to tink about: if te poton carries away energy, te sodium atom as also effectively lost a mass m f/c 2 owing to mass-energy equivalence. Tis mass is negligibly small in most cases, but we will come back to tis point wen we consider nuclear reactions. 4. Oanian Suppose tat a poton is Compton scattered from a proton instead of an electron. Wat is te maximum wavelengt sift in tis case? Te only difference from normal Compton scattering is tat te proton is eavier. We simply replace te electron mass in te Compton wavelengt sift equation wit te proton mass, and note tat te maximum sift is at θ π: λ max m p c m 2.64 fm 24) Fantastically small. Tis is rougly te size attributed to a small atomic nucleus, since te Compton wavelengt sets te scale above wic te nucleus can be localized in a particle-like sense. 5. Te Compton sift in wavelengt λ is independent of te incident poton energy E i f i. However, te Compton sift in energy, E E f E i is strongly dependent on E i. Find te expression for E. Compute te fractional sift in energy for a 10 kev poton and a 10 MeV poton, assuming a scattering angle of 90. Te energy sift is easily found from te Compton formula wit te substitution λc/e:

5 λ f λ i c c 1 cos θ) 25) E f E i mc ce i ce f E i E f 1 cos θ mc E E f E i E E i Ef mc 2 26) ) Ei E f mc 2 1 cos θ) ) 27) 1 cos θ) 28) Tus, te fractional energy sift is governed by te poton energy relative to te electron s rest mass, as we migt expect. In principle, tis is enoug: one can plug in te numbers given for E i and θ, solve for E f, and ten calculate E/E i as requested. Tis is, owever, inelegant. One sould really solve for te fractional energy cange symbolically, being bot more elegant and enligtening in te end. Start by dividing bot sides of te equation above by E i to isolate E f : E f + 1 E f 1 cos θ) 29) E i mc2 [ 1 1 E f + 1 ] 1 cos θ) 30) E i mc2 E f 1 1/E i + 1 cos θ) /mc 2 mc 2 E i mc 2 + E i 1 cos θ) 31) Now plug tat in to te expression for E we arrived at earlier: ) E 1 E i mc 2 E E i mc 2 E i mc 2 + E i 1 cos θ) E i 1 cos θ) mc 2 + E i 1 cos θ) ) 1 cos θ) 32) E i 1 cos θ) mc2 1 + E 33) i mc 2 1 cos θ) Tis is even more clear opefully): Compton scattering is strongly energy-dependent, and te relevant energy scale is set by te ratio of te incident poton energy to te rest energy of te electron, E i /mc 2. If tis ratio is large, te fractional sift in energy is large, and if tis ratio is small, te fractional sift in energy becomes negligible. Only wen te incident poton energy is an appreciable fraction of te electron s rest energy is Compton scattering significant. Te numerical values required can be found most easily by noting tat te electron s rest energy is mc kev, wic means we don t need to convert te poton energy to joules. One sould find:

6 E 0.02 E i 10 kev incident poton, θ 90 34) E 0.95 E i 10 MeV incident poton, θ 90 35) Consistent wit our symbolic solution, for te 10 kev poton te energy sift is negligible, wile for te 10 MeV poton it is extremely large. Conversely, tis means tat te electron acquires a muc more significant kinetic energy after scattering from a 10 MeV poton compared to a 10 kev poton. 6. Sow tat te relation between te directions of motion of te scattered poton and te recoiling electron in Compton scattering is 1 tan θ/2) 1 + f ) i m e c 2 tan ϕ 36) Let te electron s recoil angle be ϕ and te scattered exiting) poton s angle be θ. Conservation of momentum gets us started. Te initial poton momentum is /λ i, te final poton momentum is /λ f, and te electron s momentum we will simply denote p e. p e sin ϕ p f sin θ 37) p e cos ϕ + p f cos θ p i 38) We can rearrange te second equation to isolate p e cos ϕ: p e cos ϕ p i p f cos θ 39) Now we can divide Eq. 37 by Eq. 39 to come up wit an expression for tan ϕ: tan ϕ p f sin θ p i p f cos θ sin θ p i /p f cos θ 40) We now need a substitution for p i /p f to eliminate p f. For tis, we can use te Compton equation, wic we can rearrange to yield λ f /λ i p i /p f in terms of λ i alone, noting p/λ. λ f λ i 1 cos θ) mc 41) λ f p i cos θ) 1 + f i 1 cos θ) λ i p f mcλ i mc2 42)

7 For te last line, we used te relationsip λf c. Substituting tis in Eq. 40, we eliminate p i and p f in favor of f i alone, wic we need in our final expression. tan ϕ sin θ p i /p f cos θ sin θ 1 + f i mc 2 1 cos θ) cos θ 1 + f i mc 2 sin θ ) 1 cos θ) 43) Wit te aid of a rater obscure trigonometric identity, we can obtain te desired result. Specifically: 1 cos θ sin θ tan ) θ 2 44) Using tis in Eq. 43, 1 + f ) i mc 2 tan ϕ 1 tan θ/2) 45) If we define a dimensionless energy/momentum α i f i as is te Compton equation: mc 2 mcλ i p i mc te result is somewat simpler, α i ) tan ϕ tan θ/2) 46) α i 1 + α i 1 cos θ) α f Compton) 47) Tis simplification as utility, because it will allow us to derive te electron energy in a more compact fasion for te last question see below). 7. Frenc & Taylor 1.8 A radio station broadcasts at a frequency of 1 MHz wit a total radiated power of 5 kw. a) Wat is te wavelengt of tis radiation? b) Wat is te energy in electron volts) of te individual quanta tat compose te radiation? How many potons are emitted per second? Per cycle of oscillation? c) A certain radio receiver must ave 2 µw of radiation power incident on its antenna in order to provide an intelligible reception. How many 1 MHz potons does tis require per second? Per cycle of oscillation? d) Do your answers for parts b) and c) indicate tat te granularity of electromagnetic radiation can be neglected in tese circumstances? a) Radio waves are just ligt, so knowledge of te frequency gives us te wavelengt: λ c 300 m 48) f

8 b) Te energy of an individual poton is just f ev J. Te station s power P ) is te energy E) per unit time t) emitted, and must just be te energy per poton times te number of potons per unit time. If we call te number of potons per unit time N/ t, P E t f N t N t P f potons/s 49) Tere are 10 6 periods of oscillation per second, so tat means tat tere are approximately potons/period being emitted. c) Tis is precisely te same as te previous question, except te relevant power is 2 µw instead of 5000 W. N t P f potons/s 50) Again, tere are 10 6 periods of oscillation per second, so tere are approximately potons/period being emitted. Tis is certainly enoug potons tat te granularity of electromagnetic radiation is utterly negligible for everyday power levels suc as tese. Wat would te power level ave to be for 1 MHz potons to ave a noticeable granularity? Rougly speaking, te sampling teorem says tat if a function xt) contains no frequencies iger tan B, it is completely determined by sampling at a rate of 1/2B. i. We could say ten tat te granularity in a signal would be noticeable in tis case if te potons were coming at less tan 2 per cycle of oscillation. Tat means N t P f 2 potons/period potons/sec 51) Wit te given poton frequency of 1 MHz, we find P W, a negligible amount of power. For potons of visible ligt, in te Hz range, te power is W, wic is close to te limit of uman vision. Wit dark-adapted scotopic vision, we detect about W/m 2 of green ligt 550 nm), wic means down to around potons/s for an average-sized eye. Just about enoug to notice te granularity, but not quite. ii i ttp://en.wikipedia.org/wiki/nyquist-sannon_sampling_teorem ii Actually, it is more complicated tan tis. Te sensors in te eye are capable of detecting single potons, but our neural ardware filters te incoming signals to smoot out tis granularity. If it didn t, we would be too distracted by te granularity in low ligt. See ttp://mat.ucr.edu/ome/baez/pysics/quantum/see_a_poton.tml for a nice discussion.

9 8. Frenc & Taylor 1.11 Te clean surface of sodium metal in vacuum) is illuminated wit monocromatic ligt of various wavelengts and te retarding potentials required to stop te most energetic potoelectrons are observed as follows: Wavelengt nm) Stopping potential V) Plot tese data in suc a way as to sow tat tey lie approximately) on a straigt line as predicted by te potoelectric equation, and obtain a value for and te work function of sodium in electron volts. Te plot we require is one of stopping potential versus frequency. Te slope ten yields /e, and te y intercept te work function. Stopping potential V) Stopping potential vs. frequency, Sodium V f 2.27 R e 4.13 ± 0.02) ev s ϕ 2.27 ± 0.02) ev frequency 10 Hz) Figure 1: Stopping potential versus incident poton frequency for sodium metal. Linear regression gives /e4.13 ± 0.02) ev s and ϕ2.27 ± 0.02 ev wit a correlation coefficient R Oanian Wat is te maximum energy tat a free electron initially stationary) can acquire in a collision wit a poton of energy 4 kev? We can exploit our results from problem 6 to come up wit a relatively simple expression for te electron energy. Te Compton equation, expressed in terms of te dimensionless energies α i f i /mc 2 and α f f f /mc 2, becomes: α i α f 1 + α i 1 cos θ) 52)

10 Conservation of energy dictates tat te electron energy E e must simply be te difference between incident and exiting poton energies: E e E i E f c c α i mc 2 α f mc 2 α i mc 2 1 α ) f λ i λ f α i ) [ ] E e α i mc α i mc 2 αi 1 cos θ) α i 1 cos θ) 1 + α i 1 cos θ) [ ] [ ] α E e mc 2 2 i 1 cos θ) αi 1 cos θ) f i 1 + α i 1 cos θ) 1 + α i 1 cos θ) 53) 54) 55) Wit sufficient interest, one can go on to sow two oter interesting relationsips: [ E e mc 2 2αi 2 ] 1 + 2α i α i ) 2 tan 2 ϕ 2 cos θ α i ) 2 tan 2 ϕ ) 57) However, we ave no need of tese relationsips at te moment... all we really need to do is maximize E e wit respect to θ. One could simply assert te maximum is clearly wen cos θ 1, i.e., θ π, but tis is unsatisfying and peraps a touc arrogant. We can set de/dθ 0 to be sure: [ ] de dθ α2 i mc 2 α i sin θ 1 + α i 1 cos θ)) 2 + sin θ 1 + α i 1 cos θ) + α i sin θ cos θ 1 + α i 1 cos θ)) ) 0 sin θ [ α i α i 1 cos θ) + α i cos θ] 59) 0 sin θ θ {0, π} 60) 61) Te solution θ 0 can be discarded, since tis corresponds to te poton going rigt troug te electron, an unpysical result. One sould also perform te second derivative test to ensure we ave found a maximum, but it is tedious and can be verified by a quick plot of Eθ). At θ π, te maximum energy of te electron tus takes a nicely simple form: ) 2αi E max f 62 ev 62) 1 + 2α i For te numerical answer, we noted tat α i f i /mc 2 4 kev) / 511 kev)