Particles and Waves 34

PARTICLES AND WAVES 34

EXERCISES Section 34.2 Blackbody Radiation

15. INTERPRET This is a problem about blackbody radiation. We want to explore the connection between temperature and the radiated power. DEVELOP From the Stefan-Boltzmann law (Equation 34.1), PAT= σ 4, we see that the total radiated power, or luminosity, of a blackbody is proportional to T4. 4 EVALUATE Doubling the absolute temperature increases the luminosity by a factor of 2 = 16. ASSESS A blackbody is a perfect absorber of electromagnetic radiation. As the temperature of the blackbody increases, its radiated power also goes up.

16. INTERPRET This problem explores blackbody characteristics of a star. We are to find the power radiated, the peak emission wavelength, and the median emission wavelength of the star Rigel. DEVELOP To a good approximation, the surface of Rigel radiates like a blackbody, so the power radiated per unit area may be found from the Stefan-Boltzmann law (Equation 34.1), dividing both sides by the area A: P = σT 4 A The peak and median wavelengths may be found from Equations 34.2a and 34.2b. EVALUATE (a) The power radiated per unit area is P 4 ==σT 4824482⎡⎤5.67 × 10− W/ m ⋅× K 1.0 10 K= 5.7× 10 W/m . A ⎣⎦()() (b) From Equation 34.2a, the peak wavelength is 2.898 mm⋅ K λ ==290 nm peak 1.00× 104 K (c) From Equation 34.2b, the median wavelength is 4.11mm⋅ K λ ==411nm median 1.00× 104 K ASSESS The median wavelength is longer than the peak wavelength because of the long-wavelength tail of the temperature distribution (see Figure 34.2).

17. INTERPRET We are given the temperature of a blackbody (i.e., the Earth) and asked to find the wavelengths that correspond to peak radiance and median radiance. DEVELOP The wavelength at which a blackbody at a given temperature radiates the maximum power is given by Wien’s displacement law (Equation 34.2a):

λpeakT =2.898 mm⋅ K Similarly, the median wavelength, below and above which half the power is radiated, is given by Equation 34.2b:

λmedianT =4.11 mm⋅ K

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EVALUATE Using the above formulas, we obtain 2.898 mm⋅ K λ ==10.1 μm peak 288 K

4.11 mm⋅ K λ ==14.3 μm median 288 K

ASSESS The wavelengths are in the infrared. Note that λλmedian> peak.

18. INTERPRET This problem involves blackbody radiation. We are given the wavelength at which the peak power is emitted, and are asked to find the temperature. DEVELOP Apply Equation 34.2a

λpeakT =2.898 mm⋅ K

with λpeak = 40 µm. EVALUATE The asteroid’s temperature is 2.898 mm⋅ K T ==72 K 40 μm

ASSESS This is a cold asteroid. According to NASA, the temperature of a typical asteroid in the asteroid belt is around 200 K.

19. INTERPRET We are to find the wavelength for the peak radiance of solar blackbody radiation, and the median wavelength. In both cases, we’ll use the per-unit-wavelength basis; Equations 34.2a and 34.2b.

DEVELOP Wien’s law (Equation 34.2a) gives us the peak wavelength: λpeakT =2.898 mm⋅ K. The median

wavelength is given by Equation 34.2b: λmedianT =4.11 mm⋅ K. The temperature of the Sun is T = 5800 K, so we can use these equations to solve for the respective wavelengths. EVALUATE Inserting the temperature gives 2.898 mm⋅ K (a) λ ==5.000× 10−4 mm= 500.0 nm peak 5800 K 4.11 mm⋅ K (b) λ ==7.086× 10−4 mm= 708.6 nm median 5800 K ASSESS The peak wavelength is near the center of the visible spectrum (green) and the median wavelength is just beyond the visible in the near-infrared region. Section 34.3 Photons

20. INTERPRET This problem explores the connection between frequency and energy. We are given the frequency of photons and are asked to find the energy in electron volts (eV). DEVELOP Apply Equation 34.6, E = hf. EVALUATE (a) For f = 1.0 MHz, Ehf==()()4.136 ×⋅× 10−−−15 eV s 1.0 10 6 Hz= 4.1× 10 9 eV (b) For f = 5.0 × 1014 Hz, Ehf==()()4.136 ×⋅× 10−15 eV s 5.0 10 14 Hz= 2.1eV (c) For f = 3.0 × 1018 Hz, Ehf==()()4.136 ×⋅× 10−15 eV s 3.0 10 18 Hz= 12 keV ASSESS The energy for the photon of part (a) corresponds roughly to radio frequencies, that for part (b) is visible light, and that for part (c) is X-ray radiation.

21. INTERPRET We’re asked to express the range of human eye sensitivity in terms of photon energies. DEVELOP The photon energy is given by Equation 34.6: E = hf , or in terms of wavelength: Ehc= /.λ EVALUATE The limits of human eye sensitivity are hc 1240 eV⋅ nm E == =1.8 eV min λ 700 nm max hc 1240 eV⋅ nm Emax == =3.1 eV λmin 400 nm ASSESS We’ve used the common shorthand of hc =1240 eV⋅ nm.

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22. INTERPRET We are to find the energy of the photons produced by a microwave oven and the rate at which they are produced for a 900-W oven. DEVELOP Apply Equation 34.6, E = hf to find the photon energy E. To find the photon production rate, divide the oven power by the photon energy (in joules). EVALUATE (a) For f = 2.4 GHz, Ehf==()()4.136 ×⋅× 10−−15 eV s 2.4 10 9 Hz= 9.9× 10 6 eV . (b) The photon production rate is =1 625 J/s⎛⎞ 1eV PE 3.9 1026 s− 1 ==−−619⎜⎟× 9.9×× 10 eV/photon⎝⎠ 1.6 10 J ASSESS The production rate is quite large because the photon energy is quite small.

23. INTERPRET The problem asks for a comparison of the power output by a red laser and a blue laser. The lasers emit photons at the same rate, but the photon energy of each laser is different. DEVELOP Using λ = c/f and Equation 34.6, E = hf, the ratio of the photon energies is Efλ blue== blue red Efred redλ blue EVALUATE Using the above equation, the ratio of the energies is E λ 650 nm blue== red =1.44 Eredλ blue 450 nm Since the lasers emit photons at the same rate, this is also the ratio of their power outputs. Thus the power of the blue laser is 1.44 times that of the red laser. ASSESS Blue lasers, with shorter wavelength (higher frequency), are more energetic than red lasers.

24. INTERPRET We are to find the minimum work function that would allow photons to be ejected by 900-nm light.

DEVELOP Equation 34.7 Kmax = hf − φ gives the maximum kinetic energy for electrons ejected by light at a frequency f from a material with a work function φ. Use c = λf to convert this to a function of wavelength. The result is hc φ =−K λ max

The minimum work function occurs for Kmax = 0. EVALUATE For Kmax = 0, the work function is φλ==hc ()()1240 eV ⋅ nm 945 nm= 1.31eV . ASSESS The work function for many materials is greater than this, so 945-nm light would not be able to eject electrons from these materials. Section 34.4 Atomic Spectra and the Bohr Atom

25. INTERPRET This problem is about the energy levels of a hydrogen atom using the Bohr model. We are interested in the wavelengths of the first three lines in the Lyman series. DEVELOP The wavelength can be calculated using Equation 34.9: 111⎛⎞ =RH ⎜⎟22− λ ⎝⎠nn21 71− where RH =1.097× 10 m is the Rydberg constant and n2 =1 for the Lyman series.

EVALUATE The first three lines correspond to n1 = 2, 3, and 4, and the wavelengths are, respectively,

22 2 11nn12 n 1 λ ===22 2 122 nm, 103 nm, and 97.2 nm RnH1−− n 2 Rn H11 −1 −1 Note that RH ==()0.01097 nm 91.2 nm, which is the Lyman series limit. ASSESS The wavelengths are less than 400 nm. Therefore, the Lyman spectral lines are in the ultraviolet regime.

26. INTERPRET This problem involves finding the spectral line in the Paschen series that corresponds to the wavelength 1282 nm.

DEVELOP The wavelengths in the Paschen series for hydrogen are given by Equation 34.9, with n2 = 3 and

n1 = 4, 5, 6,… , which gives 9n2 λ = 1 Rn−129 H1()− −1 EVALUATE With λ =1282 nm and RH = 0.01097 nm , one finds 912820.010979nn22 11=()×() − or 91.41× n ==5 1 14.1− 9 which corresponds to the second line in this series. ASSESS The wavelength 1282 nm is in the infrared portion of the spectrum.

27. INTERPRET This problem is about the ionization energy of a hydrogen atom in its ground state. We want to find the wavelength that corresponds to a photon carrying this much energy.

DEVELOP The energy of the ground state of hydrogen is given by Equation 34.12b (with n = 1): E1 =−13.6 eV.

Therefore, the ionization energy is EEI1==13.6 eV (the subscript “I” is for ionization). For a photon whose wavelength is λ, the energy it carries is (Equation 34.6) E ==hf hc λ .

EVALUATE A photon with energy EI =13.6 eV has wavelength hc 1240 eV⋅ nm λ == =91.2 nm EI 13.6 eV −1 ASSESS This is the same as the Lyman series limit (Equation 34.9 with n2 =1 and n1 =∞) RhcH = 13.6 eV, and lies in the ultraviolet.

28. INTERPRET We are to find the energy level of a Bohr hydrogen atom that has a diameter of 5.18 nm. DEVELOP The diameter of a hydrogen atom in the Bohr model is (Equation 34.13) 5.18 nm so 22,sornan==22 = 49.0nm n 0 20.0529() nm or n = 7. EVALUATE This is the sixth excited state. ASSESS The radius of this state is two orders of magnitude larger than that of the ground state. Section 34.5 Matter Waves

29. INTERPRET In this problem, we are asked to find the de Broglie wavelength of the Earth orbiting the Sun and an electron moving at the given speed. DEVELOP For nonrelativistic momentum, Equation 34.14 becomes hh λ == p mv

EVALUATE (a) Using the orbital speed given and the Earth’s mass from Appendix E gives h 6.626×⋅ 10−34 J s λ == =3.7× 10−63 m mv ()5.97× 1024 kg() 30 km/s (b) For the given electron, h 6.626×⋅ 10−34 J s λ == =73nm mv ()()9.11×× 10−31 kg 10 10 3 m/s ASSESS The Earth’s de Broglie wavelength is much smaller than the smallest physically meaningful distance.

30. INTERPRET This problem involves finding the momentum of an electron with the given de Broglie wavelength.

DEVELOP For a nonrelativistic electron (),vc<< we can use the classical expression p = mv for momentum. The de Broglie wavelength is then λ = h/(mv). EVALUATE Solving for velocity gives h 6.63×⋅ 10−34 J s v == =73cm/s mλ ()()9.11×× 10−−31 kg 1.0 10 3 m ASSESS This velocity is far less than the speed of light, so we were justified in using the classical expression for momentum.

31. INTERPRET The problem asks what relative speed must an electron have in order to have the same de Broglie wavelength as a proton. DEVELOP Since λ = hp/ (Equation 34.14), the same de Broglie wavelength means the same momentum.

EVALUATE At non-relativistic speeds, the equal momenta implies mvp pee= mv, or −27 v mp 1.672× 10 kg e 1836 ==−31 = vmpe9.109× 10 kg This says that the electron will need to be moving 1836 times faster than the proton. ASSESS The problem would be more complicated if the momenta were relativistic: p = γ mv (Equation 33.7).

32. INTERPRET We are to find the de Broglie wavelength of electrons with various kinetic energies. DEVELOP We shall use Equation 34.14 λ = h/p for the de Broglie wavelength. Because the largest kinetic energy, 10 keV, is small compared to the electron’s rest energy mc2 = 511 keV, we can use the nonrelativistic expressions K = p2/(2m), so the expression for the de Broglie wavelength becomes hhc λ == . 2mK 2mc2 K EVALUATE (a) For K = 10 eV, 8 ()()4.136×⋅× 10−15 eV s 3.00 10 m/s λ ==12 pm 2511MeV10eV()() (b) For K = 1.0 eV, 8 ()()4.136×⋅× 10−15 eV s 3.00 10 m/s λ ==39 pm 2511MeV1.0eV()() (c) For K = 10 keV, 8 ()()4.136×⋅× 10−15 eV s 3.00 10 m/s λ ==0.39 pm 2511MeV10keV()() ASSESS For part (c), the relativistic relation p =+KmcKc222 leads to a result that differs by about 0.0005%. Section 34.6 The Uncertainty Principle

33. INTERPRET We want to find the minimum uncertainty in the velocity of a proton, given the uncertainty in its position. DEVELOP To find ∆v, use the uncertainty principle, ∆∆xp ≥ (Equation 34.15) with ∆p =mv∆ and ∆x = 1 fm. EVALUATE The above equation gives

∆p ()197.3 MeV⋅ fm/c 7 ∆vc=≥= ==0.21 6× 10 m/s mmx∆ ()()938 MeV 1 fm ASSESS The quantity ∆p = ∆xc=197.3 MeV/ is barely small enough compared to mc = 938 MeV/c to justify using the nonrelativistic relation p = mv, but this is good enough for the purpose of approximation.

34. INTERPRET This problem involves the uncertainty principle, which we shall use to determine the precision with which we can measure an electrons velocity and position.

DEVELOP The uncertainty principle (Equation 34.15) is ∆∆xp ≥ . For the speeds considered in this problem, we can use the nonrelativistic expression p = mv for momentum, which leads to ∆∆≥xm() v 6.63×⋅ 10−34 J s m ≥= =310kg×−29 ∆∆xv ()2210m2m/sπ ()× −6 () where we have used ∆x = 2 µm and ∆v = 2 m/s. Insert the mass of the electron and proton to see if the inequality is satisfied. −31 EVALUATE For an electron, m = 9.11 × 10 kg, so the inequality is not satisfied. The mass of a proton is m = 1.67 × 10−27 kg, so the inequality is satisfied. ASSESS Thus, we cannot determine both the velocity and position to the desired precision for an electron, but we can do so for a proton.

35. INTERPRET In this problem, we want to find the uncertainty in the position of a proton given the uncertainty in its velocity. DEVELOP To find ∆x, we use the uncertainty principle, ∆∆xp ≥ (Equation 34.15), where ∆p =mv∆. We take the uncertainty in velocity to be the full range of variation given; that is, ∆v =0.25 m/s−−() 0.25 m/s = 0.50 m/s . EVALUATE The position uncertainty of the proton is 1.055×⋅ 10−34 J s ∆≥x = =130 nm mv ∆ ()1.67× 10−27 kg() 0.50 m/s ASSESS The smaller the uncertainty ∆v in velocity, the greater the uncertainty ∆x in position.

36. INTERPRET We are given the uncertainty in the speed of an electron and are asked to find the uncertainty in its position. DEVELOP The electron is moving at (50× 106 m/s)c vc==8 0.17 3.00× 10 m/s so we can use the classical approximation for the momentum, p = mv. Thus, the uncertainty in momentum is ∆p =mv∆ where ∆v = (0.20)(50 × 106 m/s). Insert this into the expression for the uncertainty principle (Equation 34.15) to find the minimum uncertainty in position. EVALUATE This minimum uncertainty in position ∆x is 6.63×⋅ 10−34 J s ∆≥x = = =12 pm ∆∆pmv()29.1110kg0.205010m/sπ ()××−31()() 6 ASSESS AS a rule of thumb, relativistic formulas should be used for vc≥ 0.3 .

37. INTERPRET The neutron is confined in the uranium nucleus with ∆x equal to the diameter of the nucleus. We are to find the minimum energy of the neutron using the uncertainty principle. DEVELOP Using the same reasoning as given in Example 34.6, for a neutron (940mc2 = MeV)confined to a uranium nucleus (15∆≈x fm), the uncertainty principle requires that

2 p2 1 ⎛⎞ K =≥⎜⎟ 222mm⎝⎠∆ x EVALUATE From the above equation, we find the minimum kinetic energy to be

22 2 11⎛⎞ ⎛⎞c 1197.3⎛⎞ MeVfm⋅ K == =⎜⎟ =23 keV min ⎜⎟2 ⎜⎟ ⎜⎟ 22mx⎝⎠∆∆ 2 mcx ⎝⎠ 2 2940() MeV215⎝⎠() fm 2 ASSESS This is smaller than the 5 MeV estimated for the nucleon in Example 34.6 by a factor of 15 = 225 because ∆x is 15 times larger. Most estimates of nuclear energies for single-particle states, based on the uncertainty principle, give values of the order of 1 MeV, consistent with experimental measurements.

PROBLEMS

38. INTERPRET This problem involves blackbody radiation, which we can use to find the power emitted per unit area by the lamp within the given wavelength range. DEVELOP The power emitted is dP= R()λλ, T d (Equation 34.3) where 2π hc2 RT()λ, = λ 5/()ehcλ kT −1 Since dλ = 2 nm is such a small interval around 500 nm, integration is not necessary. 2162− −2 EVALUATE With 23.7410π hc =× Wm ⋅ and hc kB =1.44× 10 K ⋅ m, and with λ = 500 nm and T = 3000 K, we find

−−51 dP=()3.74×⋅ 10−−16 W m 2() 2 nm()() 5.0 × 10 7 m e 9.60 − 1= 2 kW/m 2 to a single significant figure. ASSESS If we assume an average of half this value is emitted over the entire visible spectrum, then the surface area of a 60-W incandescent light bulb should be approximately 60 W s ≈=7.5 cm2 0.8 kW/m2 which seems like a reasonable order-of-magnitude estimate.

39. INTERPRET We are given the temperature of the Sun, which we shall treat as a blackbody, and asked to compare its radiance at two different wavelengths. DEVELOP The radiance of a blackbody is given by Equation 34.4: 2π hc2 RT()λ, = λ 5/()ehcλ kT −1 This equation allows us to compare the radiance at two different wavelengths. EVALUATE From the above equation (also see Example 34.1), the ratio of the blackbody radiances for the two given wavelengths is

5 5 hc/λ1 kT RT()λ 2, ⎛⎞λ1 ⎛⎞e −15⎛⎞⎛ 146.9 ⎞ −2 ===5.4× 10 ⎜⎟⎜⎟hc/λ2 kT ⎜⎟⎜5 ⎟ RT()λλ12,122.6610⎝⎠⎝⎠ e −×⎝⎠⎝ ⎠ −2 where λλ12===500 nm, 200 nm,Thck 5800 K, and = 1.449×⋅ 10 m K. ASSESS The characteristic radiance as a function of wavelength is shown in Figure 34.2. For a given wavelength, the radiance increases with temperature.

40. INTERPRET We are to compare the Rayleigh-Jeans law to the Planck formula for blackbody radiation at the three wavelengths given. DEVELOP The ratio of the radiances for the Rayleigh-Jeans and Planck laws is

21πλλckT45 ehc /λ kT − hc/λ kT ReR-J ()()−1 ==2 RP 2πλhc hc() kT Accurate values of hcλ kT, for T ==2000 K and λ 1 mm,10 μm, and 1 μm are 7.245× 10−3 , 0.7245, and 7.245, respectively. EVALUATE The percent difference is ⎛⎞R ⎜⎟R-J −×1100% ⎝⎠RP which equals (a) 0.36%, (b) 47%, and (c) (1.9 × 104)% for the three given wavelengths. ASSESS The error becomes increasingly large as the wavelengths approach the visible part of the spectrum.

41. INTERPRET This problem is about blackbody radiation. We are given the wavelengths that correspond to peak radiance and asked to find the temperature of the blackbody. We also want to compare the radiance at two different wavelengths. DEVELOP The wavelength at which a blackbody at a given temperature radiates the maximum power is given by

Wien’s displacement law (Equation 34.2a): λpeakT =2.898 mm⋅ K. For part (b), to compare the radiance at two different wavelengths, we use Equation 34.4: 2π hc2 RT()λ, = λ 5/()ehcλ kT −1 EVALUATE (a) Equation 34.2a gives 2.898 mm⋅×⋅ K 2.898 106 nm K T == =5.19× 103 K λpeak 558 nm (b) Using the result obtained in (a), we have hc() kT = 2.768 μm, and the ratio of the radiances is 5 R()λ = 382 nm ⎛⎞694⎛⎞e2.768/0.694 − 1 ==⎜⎟⎜⎟2.768/0.382 0.748 Re()λ =694 nm⎝⎠ 382⎝⎠− 1 ASSESS The wavelengths considered are in the visible spectrum. Note that the characteristic radiance as a function of wavelength is shown in Figure 34.2. For a given wavelength, the radiance increases with temperature.

42. INTERPRET This problem examines the Compton wavelength of a proton and the energy of a gamma ray with an equivalent wavelength. 2 DEVELOP The constant in Equation 34.8 gives the Compton wavelength ()λC ==hmcpp hcmc to find the

Compton wavelength. Use Equation 34.6, Ehfhc== λC , to find the energy of a photon of the same wavelength. EVALUATE (a) The proton’s Compton wavelength (the constant in Equation 34.8) is hhc1240 MeV⋅ fm λC ==2 = =1.32 fm mcpp mc 938 MeV (b) The energy of a photon with this wavelength (Equation 34.6) is

hc 2 Ehf== = mcp =938 MeV λC ASSESS The energy of the gamma ray is the same as the rest energy of a proton.

43. INTERPRET We are given the power output at various frequencies and asked to find the rate of photon emission. DEVELOP The rate dN/dt of photon emission is the electromagnetic power output divided by the photon energy: dN P P Pλ == = dt E hf hc where we have used Equation 34.6 E = hf. EVALUATE (a) For the antenna, the rate is dN P 1.0 kW == =1.7× 1028 s− 1 dt hf ()6.626×⋅ 10−34 J s() 89.5 MHz (b) For the laser, we have

dN Pλ ()()1.0 mW 633 nm 15− 1 ==8 =3.2× 10 s dt hc ()()6.626×⋅ 10−34 J s 3.00 × 10 m/s (c) Similarly, for the X-ray machine, the rate is

dN Pλ ()()2.5 kW 0.10 nm 18− 1 ==8 =1.3× 10 s dt hc ()()6.626×⋅ 10−34 J s 3.00 × 10 m/s

ASSESS For a general device at a given power output, the rate of photon production decreases with the energy of the photon; the more energetic the photons, the smaller the rate of production because each photon carries more energy.

44. INTERPRET This problem involves the photoelectric effect. Given the maximum kinetic energy with which electrons emerge from an aluminum surface, we are to find the wavelength of the illuminating radiation. DEVELOP Einstein’s equation for the photoelectric effect (Equation 34.7) gives hc K =−φ max λ

hc λ = Kmax +φ where φ = 4.28 eV (see Table 34.1) is the work function of Al.

EVALUATE For Kmax = 1.3 eV, the wavelength is

−15 8 hc ()()4.136×⋅× 10 eV s 3.00 10 m/s λ == =220 nm Kmax ++φ 1.3 eV 4.28 eV to two significant figures. ASSESS This wavelength is in the ultraviolet portion of the electromagnetic spectrum.

45. INTERPRET This problem is about the photoelectric effect. We want to find the cutoff frequency and the maximum energy of electrons ejected by shining light with the given frequency on copper.

DEVELOP At the cutoff frequency, Kmax = 0 and the photon energy equals the work function, φ = hfcutoff (see Equation 34.7), which we can find in Table 34.1. For part (b), apply Equation 34.7 to find the maximum kinetic energy possible for the given frequency f

Khfmax =−φ

EVALUATE (a) The work function of copper is φCu = 4.65 eV. Therefore, the cutoff frequency is

φ 4.65 eV 15 f ==Cu =1.12 × 10 Hz cutoff h 4.136×⋅ 10−15 eV s (b) The maximum kinetic energy of the ejected electrons is Khf 4.136 10−15 eV s 1.8 10 15 Hz 4.65 eV 2.79 eV max =−φ =()()×⋅× −= ASSESS Upon illuminating copper with photons at 7.44 eV, it takes 4.65 eV to overcome the work function of copper, leaving the electrons with 2.79 eV of kinetic energy.

46. INTERPRET This problem involves the photoelectric effect. Given the electric potential energy difference needed to stop electrons emitted from a surface by the given radiation, we are to determine the work function of the material. DEVELOP The electron’s maximum kinetic energy is expended in crossing the stopping potential (see text), so

eVsmax= K

where Vs = 1.8 V. Apply Equation 34.7 to find the work function; hc Khf=−φφ=− max λ EVALUATE (a) for λ = 365 nm, the work function is hc φ =−K λ max hc =−eV λ s 8 ()()6.626×⋅ 10−34 J s 3.00 × 10 m/s =−1.8 eV= 1.6 eV 365 nm

(b) At the new wavelength, hc 1240 eV⋅ nm eV =−φ =−1.60 eV= 2.79 eV s λ 280 nm or

Vs = 2.8 V to two significant figures. ASSESS The stopping potential increases because it takes a bigger “hill” to stop the electrons emitted by the 280-nm radiation, which contains photons of higher energy than radiation at 365 nm.

47. INTERPRET We are asked to explain why plants are green using the absorption peaks in the chlorophyll molecule. DEVELOP We can convert the wavelength peaks into energy peaks using Ehc= /,λ and the shorthand hc =1240 eV⋅ nm. EVALUATE (a) The energy peaks in chlorophyll’s absorption spectrum are at 1240 eV⋅ nm E ==2.9 eV 1 430 nm

1240 eV⋅ nm E ==1.9 eV 2 662 nm (b) These absorption peaks correspond to blue and red wavelengths, near the limits of the human visible range. The light that is not absorbed is reflected, and this is what we humans observe. Since the reflected light is primarily in the green region of the visible spectrum between blue and red, we perceive plants to be green. ASSESS Plants also reflect a lot of infrared light with wavelengths longer than 700 nm.

48. INTERPRET This problem involves Compton scattering, which we can use to find the initial wavelength of photons that scatter at the given angle from electrons. DEVELOP Apply Equation 34.8, which describes Compton scattering (i.e., the scattering of photons off electrons). For a photon that loses half its initial energy, we have 1 f = f 2 0 so the wavelength shift ∆λ is cc2 ccc ∆λλλ=−0 =−=−= = λ ffo f000 f f EVALUATE Using this result for ∆λ in Equation 34.8 gives hh λθ=()1cos−== 2.43pm cc ASSESS The energy of these photons is about 0.5 MeV.

49. INTERPRET This problem is about the photoelectric effect. We are given the maximum speed of electrons ejected from potassium and asked to find the wavelength of the light that ejected the electrons. DEVELOP The maximum speed of the ejected electrons is related to the wavelength of the light by Einstein’s photoelectric effect equation (Equation 34.7): 1 hc Kmvhf==2 −φφ=− max2 max λ We shall use this equation to find λ. EVALUATE The maximum kinetic energy of the electron is

2 11⎛⎞v 1⎛⎞ 4.210m/s× 5 22max Kmvmcmax== max ()⎜⎟ =()511keV⎜⎟8 = 0.501eV 22⎝⎠c 2⎝⎠ 3.0010m/s×

From Equation 34.7 and using Table 34.1 to find the work function φ, we find the wavelength to be hc 1240 eV⋅ nm λ == =440 nm Kmax ++φ 0.501 eV 2.30 eV to two significant figures. 1 ASSESS Strictly speaking, the result should be reported as 44 ×10 nm to make the significant figures more obvious. The cutoff wavelength of potassium is

λβcutoff ==hc ()()1240 eV ⋅ nm 2.30 eV= 539 nm

For the photoelectric effect to take place, we require λλ≤ cutoff.

50. INTERPRET This problem involves the photoelectric effect. Given the maximum electron energy for two different illuminating wavelengths we are to find the work function of the material. DEVELOP The photoelectric effect equations (Equation 34.7) for the two experimental runs are hc K =−φ =2.8 eV max λ and hc hc K′ =−φφ=−=1.1eV max κλ′ 1.5 which we can solve for the work function φ and the initial wavelength λ. EVALUATE (a) Subtracting the two equations gives

−15 8 hc ()()4.136×⋅× 10 eV s 3.00 10 m/s λ == =243 nm 31.7eV() 31.7eV() Adding the two equations gives

−15 8 53.9eVhc 54.13610()()×⋅× eVs 3.0010m/s φ =−=−1.85 eV= 2.3 eV 62λ 6243nm() (b) From the calculation above, we find λ = 240 nm (to two significant figures). ASSESS This material is probably potassium, which has a work function of φ = 2.3 eV.

51. INTERPRET This problem is about Compton scattering of a photon with an electron. We are interested in the wavelength of the scattered photon and the kinetic energy of the electron. DEVELOP The Compton shift of wavelength is given by Equation 34.8: h ∆λθλθ=()()1cos−=C 1cos− mc

where λC ==hmc()2.43 pm is the Compton wavelength of the electron. By conservation of energy, the kinetic energy of the scattered electron is equal to the energy lost by the photon. EVALUATE (a) From Equation 34.8, the wavelength of the scattered photon is

λλ′ =+∆ λ=+150 pm 2.43 pm ⎣⎦⎡⎤ 1 − cos() 135°= 150 pm + 4.15 pm = 154 pm (b) The kinetic energy of the scattered electron is hc hc hc∆λ ()()1240 eV⋅ nm 4.15 pm KEE=−′ =−= = =222 eV λλ′′ λλ ()()150 pm 154 pm ASSESS For X rays, the wavelength is in the range 0.01–10 nm, so the detection of the Compton shift in X rays is difficult.

o 52. INTERPRET This problem involves Compton scattering of an X-ray at 90 from a stationary electron. We are to find the kinetic energy of the electron after the scattering event. DEVELOP In Compton scattering, the kinetic energy of the recoil electron equals the energy lost by the photon:

hc hc hc()λλ′ −− E λC ()1cos θ KEE=−′ =−= = λλ′′ λλλλ+C ()1cos− θ where we used Equation 34.8 for ∆λλ=′ −λ .

EVALUATE For the given data (90,100θλ=°= pm, Ehc = λ = 12.4 keV, and=2.43pm), λC we find Eλ 2.43 K ==C ()12.4 keV = 290 eV λλ+ C 102.43 to two significant figures. ASSESS This (nonrelativistic) energy corresponds to a speed of

2 22KKc 8 2294eV()7 v == =3.00× 10 m/s= 10 m/s mmc2 ()511keV

53. INTERPRET This problem involves the photoelectric effect. We are given enough information to find the work function of the material, and are asked if this material will emit electrons when illuminated with radiation at a longer wavelength and, if so, what will be the maximum electron energy. DEVELOP To determine whether or not the photoelectric effect can occur, we shall first find the work function from Einstein’s photoelectric effect equation (Equation 34.7). Using the data for the blue light, this gives the work function of the photocathode material as hc 1240 eV⋅ nm φ =EK−=− K =−0.85 eV= 2.03 eV maxλ max 430 nm EVALUATE The energy of a photon of the red light is only hc 1240 eV⋅ nm ==1.96 eV λ 633 nm and therefore is insufficient to eject photoelectrons.

ASSESS A wavelength of 633 nm is greater than the cutoff wavelength of λφcutoff ==hc 610 nm for the photocathode material.

54. INTERPRET We investigate the relation between a cosmic-ray particle’s interaction time and uncertainty in its energy measurements. DEVELOP From Equation 34.16, the uncertainties in the energy and time are constrained by ∆∆≥Et . Therefore, the uncertainty in its measured energy will be inversely proportional to the interaction time of the particle: ∆≥E /. ∆t EVALUATE With ∆t =12 zs = 1.2× 10−20 s,and using =6.582× 10−16 eV ⋅ s, we have 6.582×⋅ 10−16 eV s ∆≥E = =5.496× 104 eV ≈ 55 keV ∆×t 1.2 10−20 s ASSESS The uncertainty principle indicates that the minimum uncertainty of the cosmic-ray particle is about 55 keV for a time 12 zs.

55. INTERPRET We want to find the minimum uncertainty in the components of the velocity of an electron, given the size of the carbon nanotube. DEVELOP We consider the z (along the tube) and the radial (perpendicular to z) components of the velocity

separately. To find ∆vi, use the uncertainty principle, ∆∆xpii ≥ (Equation 34.15) with ∆pii=mv∆, ∆r = 1.2 nm and ∆z = 370 nm. EVALUATE (a) Along the tube, the above equation gives 1.055×⋅ 10−34 J s ∆pz () ∆vz =≥= =313 m/s mmz∆ ()()9.11×× 10−−31 kg 370 10 9 m (a) Along the radial direction, the uncertainty is 1.055×⋅ 10−34 J s ∆pr ()4 ∆vr =≥= =9.6× 10 m/s= 96 km/s mmr∆ ()()9.11×× 10−−31 kg 1.2 10 9 m

ASSESS The uncertainty in ∆vi is inversely proportional to ∆xi. In this case, since ∆rz<<∆, we have ∆vvrz>>∆.

56. INTERPRET We are to find the de Broglie wavelength of an electron that has been accelerated through a potential difference. DEVELOP We shall use Equation 34.14 λ = h/p for the de Broglie wavelength. Because the kinetic energy, K = eV = 4.5 keV is small compared to the electron’s rest energy mc2 = 511 keV, we can use the nonrelativistic expressions K = eV = p2/(2m), so the expression for the de Broglie wavelength becomes hh λ == . p 2meV EVALUATE For V = 4.5 kV, the de Broglie wavelength is 6.626×⋅ 10−34 J s λ ==18 pm 29.1110()××−−31 kg(1.610 19 C)4500V()

ASSESS The wavelength varies as 1/V . The greater the value of V, the smaller the wavelength.

57. INTERPRET An electron is confined in a channel with width ∆x = 6.6 nm. We are to find the minimum energy of the electron using the uncertainty principle. DEVELOP Using the same reasoning as given in Example 34.6, for an electron (0.511mc2 = MeV)confined to a channel of width ∆x =6.6 nm, the uncertainty principle requires that

2 p2 1 ⎛⎞ K =≥⎜⎟ 222mm⎝⎠∆ x EVALUATE From the above equation, we find the minimum kinetic energy to be

2 22 11c 1197.3⎛⎞ MeVfm ⎛⎞ ⎛⎞ ⎜⎟⋅ −−423 Kmin ==⎜⎟2 ⎜⎟ = =2.2× 10 eV= 3.5× 10 J 22mx∆∆ 2 mcx 2 20.511 MeV⎜⎟26.6106 fm ⎝⎠ ⎝⎠()⎝⎠()× ASSESS The minimum kinetic energy varies with 1/(∆x )2 .The value of ∆x here is 66 times that considered in Example 34.5, and therefore the minimum energy is 1/(66)2, or about 4000 times smaller than the 1 eV calculated there.

58. INTERPRET This problem involves the Bohr atom (i.e., a hydrogen atom). We are to find the highest possible energy for a photon emitted by such an atom. DEVELOP The energy of the photon emitted in a hydrogen atom transition between adjacent states

(1)nnn121→=− is

hc ⎡⎤−−22−−22 EhcRnnhcRnnn==H1() −−1211 1= H ( 1− )() 11 − λ ⎣⎦ (see Equations 34.6 and 34.9 and the discussion of the Bohr atom in the text).

EVALUATE (a) The maximum allowed energy occurs for n1 = 2, which gives 33 EhcRmax== H ()13.6 eV = 10.2 eV 44

(b) The upper (initial) energy level is n1 = 2 and the lower (final) energy level is n2 = 1. ASSESS This energy is in the ultraviolet portion of the electromagnetic spectrum.

59. INTERPRET This problem is about the wavelength and energy of the photon emitted when a Rydberg hydrogen atom undergoes a transition. DEVELOP The wavelength of the photon can be calculated using Equation 34.9: 111⎛⎞ =RH ⎜⎟22− λ ⎝⎠nn21 71− where RH =1.097× 10 m is the Rydberg constant. Once we know the wavelength, the energy of the photon may be found using E ==hf hc λ .

EVALUATE (a) The above equation gives

22 22 11nn12 ⎛⎞()()179 180 λ == =26.4 cm Rn22 n ⎜⎟1.097 10 71 m− 22 H1−× 2 ⎝⎠()()180− 179 (b) The energy of the emitted photon is hc 1.24×⋅ 10−6 eV m E == =4.70 μeV λ 0.264 m ASSESS The long wavelength corresponds to the radio region of the electromagnetic spectrum.

60. INTERPRET We are to find the maximum energy for a photon emitted by a transition in the Lyman series and in the Balmer series in a hydrogen atom. DEVELOP Apply Equation 13.12b, 13.6 eV E = 2 n2

where n2 is lowest quantum number for the given series. From Figure 34.11, we see that n2 = 1 for the Lynam

series and n2 = 2 for the Balmer series. To find the corresponding wavelength, use Equation 34.6 E = hf = hc/λ. EVALUATE (a) For the Lyman series, the highest energy photon is 13.6 eV E ==13.6 eV 12 The wavelength of this is hc 1240 eV⋅μm λ == =91.2 nm E 13.6 eV (b) For the Balmer series, the highest energy photon is 13.6 eV E ==3.40 eV 22 so the wavelength is hc 1240 eV⋅μm λ == =365 nm E 3.40 eV

ASSESS The Lyman series limit is in the ultraviolet. The Balmer series limit is at the very high-energy end of the visible spectrum.

61. INTERPRET The hydrogen atom undergoing a downward transition emits a photon. We are interested in the original state of the atom, given the energy of the photon and the quantum number of the final state of the atom. DEVELOP The wavelength of the photon can be calculated using Equation 34.9: 111⎛⎞ =RH ⎜⎟22− λ ⎝⎠nn21 71− where RH =1.097× 10 m is the Rydberg constant. Once we know the wavelength, the energy of the photon may be found using Equation 34.6: hc ⎛⎞11 Ehf== = hcRH ⎜⎟22 − λ ⎝⎠nn21

EVALUATE Solving for n1 gives

−1/2 nnEhcR⎡⎤−−−221/2[(225) (9.32 μeV 13.6 eV ] 229 12=⎣⎦−() H =−()=

Note that hcRH =13.6 eV is the ionization energy. ASSESS The wavelength of the emitted photon is 0.133 m, which falls into the radio wave spectrum.

62. INTERPRET This problem involves conservation of energy and the Bohr model of the atom. We can use these two concepts to find the energy of an electron ejected from the ground state of a hydrogen atom by a 48-eV photon.

DEVELOP Energy must be conserved, so the final kinetic energy is the original (ground state) energy plus the absorbed photon energy. EVALUATE The final kinetic energy is thus

KE=+=0 Eγ −13.6 eV+ 48 eV = 34 eV ASSESS This energy is much less than the electron’s rest mass, so relativistic effects do need to be considered.

63. INTERPRET This problem involves the Bohr model of the atom. We are to find the ionization energy of a hydrogen atom in its first excited state. DEVELOP Using Equation 34.12b, the energy of the first excited state (n = 2) is −13.6 eV E ==−3.40 eV 2 22 whereas an ionized atom (with zero electron kinetic energy) has energy zero.

EVALUATE Thus, we must supply an energy EI such that 0(3.40=+E − eV) I EI = 3.40 eV ASSESS The ionization energy here is only ¼ of the case where the atom is in the ground state. The higher the value of n, the smaller the ionization energy because the electron is less tightly bound to the nucleus.

64. INTERPRET We are to find the final energy of electrons ejected from ground-state hydrogen atoms by ultraviolet radiation. We will apply conservation of energy and use the Bohr model of the atom. DEVELOP By conservation of energy, the final kinetic energy of the electron (actually, the ionized atom with the nucleus assumed to be at rest) is its initial ground state energy (−13.6 eV) plus the energy E = hf = hc/λ (see Equation 34.6) absorbed from the photon. EVALUATE Summing the two energies, we find the kinetic energy of the electron to be hc 1240 eV⋅ nm K =−13.6 eV+ =− 13.6 eV+ = 2.9 eV λ 75nm ASSESS The photon supplies a total energy of 13.6 eV + 2.9 eV = 15.5 eV.

65. INTERPRET He+ is a hydrogen-like atom with a nuclear charge +2e. We are to apply the Bohr model to this system to find the ground-state electron radius and the energy difference between the n =2 and n = 1 state. DEVELOP Modifying the treatment of the Bohr atom in the text (see derivation of Equations 34.11 and 34.12) for + 2 singly ionized helium (He ) by replacing the nuclear charge with 2e, one gets rkeEnn=−22() and E (2ke22 ) m 2 2 n 2 . Thus, n =− () 2 ke 2 ⎛⎞13.6 eV En =−4222=− ⎜⎟ 2na0 ⎝⎠ n and na2 r = 0 n 2 + EVALUATE (a) The radius of the ground state of He is a 0.0529 nm r ==0 =0.0265 nm 1 22 (b) The energy released in the transition from n = 2 to n = 1 is ⎛⎞1 ∆E =413.6() eV⎜⎟ 1−= 40.8 eV ⎝⎠4 Note that there is also a small change in m for helium, different from the correction in hydrogen, for the motion of the nucleus. ASSESS In general, replacing the nuclear charge with Ze gives results for any one-electron Bohr atom, where Z is the number of protons for the atom (i.e., the atomic number).

66. INTERPRET This problem involves the de Broglie wavelength. We are to find the kinetic energy needed for an electron to have a de Broglie wavelength the size of a hydrogen atom.

DEVELOP Since 0.1 nm=>>=λλC 2.43 pm, nonrelativistic expressions can be used to write the kinetic energy in terms of the de Broglie wavelength:

2 2 ph22⎛⎞1 h ()hc K ==⎜⎟ =222 = 2222mmmmc⎝⎠λλλ 2 ()1.24ke V⋅ nm ==2 150 eV 2511keV0.10nm()() EVALUATE Because the kinetic energy gained in an acceleration from rest equals the potential energy difference, K =eV ∆, and ∆V =150 V.

ASSESS Recall that λC = h/(mc) = 2.43 pm for an electron. The result of 150 V is an easily accessible potential difference.

67. INTERPRET The resolution of a microscope depends on the wavelength used. A smaller de Broglie wavelength will improve the resolution of an electron microscope. We are to find the minimum electron speed that will make its de Broglie wavelength less than 450 nm. DEVELOP The resolution of the electron microscope is better than the optical microscope with 450-nm light if the de Broglie wavelength λ of the electrons is less than 450 nm. Thus, h λ =<450 nm p Since p = mv (for nonrelativistic electrons), the above condition allows us to obtain the minimum electron speed. EVALUATE The above inequality gives h 6.626×⋅ 10−34 J s v >= =1.62 km/s m()450 nm ()9.11× 10−31 kg() 450 nm So the minimum speed is 1.62 km/s.

ASSESS Since 450 nm>>λC =hmc = 2.43 pm (Compton wavelength of the electron), our use of the nonrelativistic momentum was justified. The electron microscope can provide resolutions down to about 1 nm and magnifications of 106.

68. INTERPRET You want to see what energy of electrons is needed to resolve microtubules. You’ll need to consider the de Broglie wavelength of the particles. DEVELOP To resolve the microtubules, the electron microscope that you buy will need electrons with de Broglie wavelength less than or equal to the size of the microtubules (25 nm). This corresponds to a momentum of ph= /,λ and a kinetic energy of K = pm2 /2 . With these equations, you can calculate the minimum energy you need for an electron microscope. EVALUATE The minimum electron kinetic energy needed is

22 h2 ()hc ()1240 eV⋅ nm K == =2 =2.4 meV 22mmcλλ2222511()() keV25 nm where we have used the rest-energy of the electron, mc2 = 511 keV. The minimum kinetic energy is far below 40 keV, so you don’t need to buy the more expensive microscope, since the less expensive microscope will work. ASSESS By the above arguments, the 40 keV electron microscope should have resolution of around 6 pm, since that is the corresponding de Broglie wavelength of the electrons. However, there are complications involved in focusing a beam of electrons, so most electron microscopes have resolutions around a nanometer.

69. INTERPRET This problem involves the uncertainty principle. We want to find the minimum velocity of an electron based on its uncertainty in position. DEVELOP Using the uncertainty principle given in Equation 34.15 with ∆x = 23 nm (the width of the well), we have 197.3 eV⋅ nm/c ∆≥p = =8.58 eV/c ∆x 23 nm

This is small compared to mc= 511 keV/ c , so nonrelativistic formulas are sufficient (see Example 34.6). Therefore, ∆vpmv=∆=2. EVALUATE The above conditions lead to

∆pc8.58 eV/ 8 vc=≥=8.39× 10−−66= 9.65× 10 3.00 × 10 m/s= 2.5× 10 3 m/s= 2.5 km/s 2 ()() 2m 2511() keV/c

Thus, the minimum speed is vmin = 2.9 km/s. ASSESS Quantum wells have important applications in the field of semiconductor fabrication. Note that the result is given to two significant digits, as warranted by the data.

70. INTERPRET This problem involves the uncertainty principle applied to the variable energy and time. We shall use this to find the uncertainty in energy in a typical atomic transition given the lifetime of the excited atomic state. DEVELOP The uncertainty principle (Equation 34.16) for energy and time is ∆∆tE ≥ (note that the units of are time-energy). Apply this to find the minimum energy given the uncertainty in time of ∆t = 10−8 s. EVALUATE The uncertainty in energy is 6.582×⋅ 10−16 eV s ∆≥E = =710eV× −8 ∆t 10−8 s ASSESS This energy uncertainty is called the natural linewidth.

71. INTERPRET This problem involves energy-time uncertainty. We are interested in the minimum measurement time needed to measure the energy with the desired precision. DEVELOP The electron is nonrelativistic (13001),vc=<< so we can use the nonrelativistic expression K = mv2/2 for kinetic energy. The desired uncertainty in the kinetic energy is

22 ⎛⎞11242−− ⎛⎞v 4 ⎛⎞ − 4 ∆Emvmc=20.01%()×⎜⎟== 10 ⎜⎟ 10() 511 keV ⎜⎟ = 5.6810× eV ⎝⎠2300 ⎝⎠c ⎝⎠ The minimum time can then be calculated using Equation 34.16, ∆∆≥Et . EVALUATE An energy measurement of this precision requires a time 6.582×⋅ 10−16 eV s ∆≥t = =1 ps ∆×E 5.68 10−4 eV to a single significant figure. ASSESS The energy-time uncertainty principle implies that the minimum measurement time must necessarily go up in order to achieve a greater accuracy in energy measurement.

72. INTERPRET We are to derive the classical Rayleigh–Jeans law (Equation 34.5) for blackbody radiation from Planck’s law (Equation 34.3). DEVELOP For λ >> hc kT , the exponent in Planck’s law (Equation 34.3) is much, much less than unity, so we can express the exponential function as ⎛⎞hc hc exp⎜⎟≈− 1 ⎝⎠λλkT kT Insert this result into Planck’s law. EVALUATE With the above approximation, Planck’s law takes the form 222πππhc22 hc ckT RT(,) λ ≈5 ==54 λλ⎣⎦⎡⎤11−−hc() kT λλhc() kT λ which is the Rayleigh–Jeans law. ASSESS Thus, the Rayleigh–Jeans law is an approximation of the more accurate Planck law.

73. INTERPRET In this problem, we want to show that if a photon’s wavelength is equal to a particle’s Compton wavelength, then the photon’s energy is equal to the particle’s rest energy.

DEVELOP From Equation 34.8, we see that the Compton wavelength of a particle is λC = hmc. The rest energy of a particle is E = mc2 (see discussion preceding Equation 33.9). From Equation 34.6, we see that the energy of a photon is E = hf = hc/λ.

EVALUATE When the wavelength of the photon is λ = λC, its energy is hc hc E == =mc2 λC hmc() which is the same as the particle’s rest energy. ASSESS This result is not surprising because photons have zero rest mass, so their energy is completely kinetic. Have you ever seen a photon that is not moving at the speed of light?

74. INTERPRET This problem involves the frequencies (i.e., energies) of the electronic transitions in hydrogen. We are to show that transitions to a level n from all higher levels cover the given frequency range. DEVELOP From Equation 34.9, we see that the frequency of a transition in which the electron is initially in state

n1 and ends up in state n2 = n is c ⎛⎞⎛⎞11 11 fcRcR==HH⎜⎟⎜⎟22 −= 22− λ ⎝⎠⎝⎠nn21 nn 1

The possible values for n1 are n + 1, n + 2, …, so the lowest possible frequency occurs for n1 = n + 1 and the

highest possible frequency occurs for n1 =∞. EVALUATE The frequency range is

1111⎛⎞ ⎛⎞⎜⎟ ∆ffn=()(11 =∞− fnn =+=10 ) cR H22 −− cR H −22 = cR H ⎜⎟nn⎜⎟ ⎝⎠⎝⎠()nn++11() ASSESS This agrees with the formula given in the problem statement.

75. INTERPRET This problem involves Compton scattering of a photon off an electron that is initially at rest (zero kinetic energy). We are to find an expression for the initial photon energy given the final total energy (kinetic energy plus rest energy) of the electron.

DEVELOP For Compton scattering at 90°, Equation 34.8 reduces to λλ=+0C λ. In terms of the photon energy (Equation 34.6) E = hf = hc/ and the electron’s Compton wavelength [Equation 34.8, hc m c2 ], this can λ λC = ()e be written as 11 1 =+ 2 E Emc0 e or 2 E0mce E = 2 E0 + mce The recoil electron’s kinetic energy is

2 2 2 Emc00e E KmcEEEee=()γ −1 =00−=−22= E00++mcee E mc This is a quadratic equation in E , namely EmcEmc2 10.22 The positive solution corresponds to the 0 00−−()γ () +e = initial photon energy that we seek.

EVALUATE The positive solution for E0 is

11⎡⎤2242422 ⎡ ⎤ Emcmcmcmc0 =()γγ−1141113ee+ ()−+ () γ−= e()()() γγγ−+−+ 22⎣⎦⎢⎥⎣ ⎦ ASSESS With some algebra, the kinetic energy of the recoiled electron can be written as

2 2 ⎡()()()γγγ−113+−+⎤ 22E0 1 ⎣ ⎦ Kmcmcee=()γ −1 =2 = e Emc 2 0 + e ()()()γγγ++113 −+

1 22 1 2 In the nonrelativistic limit where γ ≈1+2 vc, the above expression reduces to the expected result Kee≈ 2 mv.

76. INTERPRET We are to derive Wien’s law (see Section 34.2 on blackbody radiation) from Planck’s law. DEVELOP If we introduce the dimensionless variable x = hc/(λkT) into Planck’s law, R(λ, T) is proportional to xe5 ()x −1 . This can be a maximum when its derivative with respect to x is zero, which leads to the equation exx =55()−. EVALUATE For a maximum, this condition is satisfied by a value of x nearly equal to 5 (since x = 0 corresponds

to a minimum radiance). The value xmax can be found numerically to be about 4.965, so hc λ T ==2.898 mm ⋅ K max 4.965k which is Equation 34.2. ASSESS We have shown the desired relationship. Notice that Wien’s law is not a classical approximation, as is the case for the Rayleigh–Jeans law.

77. INTERPRET We shall use conservation of energy and conservation of momentum (with relativistic expressions for both energy and momentum) to derive equations related to Compton’s equation, and then derive the equation for the Compton shift. DEVELOP The energy of a photon is E ==hf hc λ (Equation 34.6) and for a particle it is E = γ mc2 (Equation ˆ ˆ 33.9). The relativistic momentum is p = γ mu for the electron and p = ih λ0 for the photon, where i is the direction of the photon’s motion. We will use conservation of energy to obtain one of the desired equations, and conservation of momentum in two dimensions to obtain the other two equations. EVALUATE The initial energy is the energy of the photon plus the rest energy of the electron:

2 Ei0=+hcλ mc . The final energy is the energy of the new photon plus the relativistic energy of the moving electron:

2 Ef =+hcλγ mc Equating these two energies (by conservation of energy) gives us the first of the three desired equations: hc hc +=+mc22λ mc . λλ0 ˆ The next two equations come from the initial momentum, pi0= ih λ and the components of final momentum h h pmufx =+λ cosθ γφ cos and pmufy ==0sinsinγφ −λ θ . By conservation of momentum, we can equate the initial and final momentum in each direction, which leads to hh pp= ⇒ =+cosθ γφ mu cos ifxx λλ 0 h pp==00sinsin⇒ =θ −γφ mu ifyy λ

These are the second two of the desired relationships we were to derive. Solving these three equations for λλ0 − directly is a lengthy algebraic process. An easier approach is to start with the momentum in vector form and use the law of cosines:

222 pppγγ=+′′ee→ ppp ′=+ γγ ′−2cos pp γγ ′θ 2 2 2 ⎛⎞hhhh⎛⎞ pe′ =+⎜⎟⎜⎟−2cosθ ⎝⎠λλλλ00⎝⎠ We now use conservation of energy in the form hc+=+mc22422 hc m c + p c , and solve for p2 to obtain λλ0 e′ e′ 2 hc− hc +mc224− m c 2 ()λλ0 p = e′ c2

2 We equate the two equations for pe′ to obtain 2 hc− hc +mc224− m c 2 2 ()λλ0 ⎛⎞hhhh⎛⎞ 2 =+⎜⎟⎜⎟−2cosθ c ⎝⎠λλλλ00⎝⎠

2 2 2 2 ⎛⎞hc⎛⎞ hc 222chm3322 chm ch⎛⎞ hc⎛⎞ hc hc hc ⎜⎟+ ⎜⎟−+−=+⎜⎟⎜⎟− 2cosθ ⎝⎠λλ0 ⎝⎠ λλλλλλ000⎝⎠⎝⎠ λλ0 mc mc h −=()1cos−θ λλλλ00 h λλ−0 =()1cos− θ mc ASSESS We have derived the equation for the Compton shift, using conservation of energy and momentum.

78. INTERPRET We are to find the peak of blackbody radiation (i.e., Wien’s law, Equation 3.42a), looking at frequency intervals rather than wavelength intervals. DEVELOP We start with the radiance curve (Equation 34.3, Planck’s law) 2π hc2 RT(,)λ = λ 5 ()ehc/()λ kT −1 use λ =c/f and differentiate with respect to f to find the maximum. We’ll have to solve dR df = 0 numerically. EVALUATE 22ππhc25 hf RfT(,)==5 3/hf kT c (1)ehf/ kT ce(1)− ()f − 425 dR10ππ hf 2 h f hf/ kT ==0 3/hf kT − 3 hf / kT e df c(1)(1) e−− c kT e hf 5 = ehf/ kT kT ⎛⎞5kT hf ln⎜⎟= ⎝⎠hf kT We solve this numerically for T/f to obtain T/f = 3.617 × 1011 K·s. 14 ASSESS If we use this formula to find the peak frequency for the Sun (T = 5800 K), we obtain f = 1.6 × 10 Hz. This corresponds to a wavelength of λ = 1.87 mm. Compare this with the wavelength peak found in Problem 34.19—it’s quite different!

79. INTERPRET We are to integrate the radiance equation (Equation 34.3, Planck’s law) over all wavelengths and show that the resulting total power radiated per unit area is equivalent to the Stefan-Boltzmann law (Equation 34.1). DEVELOP The Stefan-Boltzmann law gives the power per area as PA= σ T4, and the radiance equation is 2π hc2 RT(,)λ = λ 5 ()ehc/()λ kT −1 Substituting hc/(λkT) by the integration variable x gives 2π kT55⎛⎞ x 5 RxT(, )= 43 ⎜⎟x hc⎝⎠ e −1 The differentials dx and dλ are related by hc ddxλ =− xkT2 We will integrate R(x, T)dλ over all λ.

EVALUATE Performing the integration gives PkTxhckTx∞∞22ππ55⎛⎞ 5⎛⎞ 44 ∞ ⎛⎞ 3 ==R()λλ,Td43⎜⎟xx⎜⎟ − 2 dx=− 32 ⎜⎟ dx AhcexkThce∫∫−−11 ∫ 00⎝⎠⎝⎠ 0 ⎝⎠ 2k 45π = T 4 15ch23 ASSESS This is equivalent to the Stefan-Boltzmann law, with σπ= 215.khc45() 32 80. INTERPRET We are to numerically verify the median wavelength as given by Equation 34.2b by numerically integrating the radiance equation from zero to the value given by 34.2b—the result should be 1/2. DEVELOP We shall numerically integrate Equation 34.3 2π hc2 RT(,)λ = λ 5 ()ehc/()λ kT −1

from λ = 0 to λmedian = 0.00411/T. If the value we get is half the value obtained when numerically integrating to

“infinity,” then we will have verified that λmedian is indeed the median wavelength. EVALUATE With most numeric software packages or simple integration routines, it is easiest to substitute some

value of T. For example, for T = 1000 K, the integral to λmedian is 28,379 and the integral to some approximately

“infinite” number is 56,704. Other values of T give different values, but in each case the integral to λmedian is approximately half the integral over all values of λ. ASSESS The median wavelength is given to only three significant figures. The numeric value of the integral taken

to λmedian is half the numeric value of the entire integral to more than three significant figures, so we have verified that the median wavelength given is correct.

81. INTERPRET We use conservation of momentum to find the recoil angle of the electron in Compton scattering. DEVELOP The momentum conservation equations are hh=+cosθ γφmu cos λλ 0 0sinsin=h θ −γφmu λ h The Compton scattering equation is λλ−0 =mc (1− cos θ ). We will solve the momentum equations for sinφ and cosφ , and then take the ratio to see if we can get the desired equation. EVALUATE First solving for sinφ and cosφ , we get 1 sinφθ= h sin γ mu ( λ )

1 ⎛⎞hh cosφ =⎜⎟− cosθ γ mu ⎝⎠λλ0

Taking the ratio and multiplying the numerator and denominator by λλ0 , we arrive at the desired result: λθsin tanφ = 0 λλ− 0 cos θ ASSESS To remove the dependence on the final wavelength, λ, we use the Compton shift equation: λθsin sin θ tanφ ==0 λλ−00+ λ()1cos− θ()() 1+ λ C0 / λ 1cos− θ

where λC = hmc/ is the Compton wavelength. This shows that as you decrease λ0 , i.e., as you go toward higher energy photons, the recoil angle, φ, of the electron decreases (for a given recoil angle, θ, for the photon).

82. INTERPRET We are to show that the correspondence principle (see Section 34.7) holds for the Bohr model in that the frequency of a photon emitted in a nn+1→ transition for large n equals the orbital frequency of the electron. We shall do this by taking the limit of both the transition energy and the orbital frequency as n →∞.

DEVELOP The energy of the photon emitted in a transition from n to n is Ekea222⎡⎤21 n 1 n , where 1 2 ∆=⎣⎦02()− 1 n2 = n and n1 = n + 1. The frequency of the photon is ∆E/h. The orbital frequency can be calculated from the orbital radius rkeEkemv2222 and the orbital velocity vn mr using f vr2 . =−()= () = () O = ()π

We will use the binomial expansion to approximate fγ and compare the result with the orbital frequency fO. EVALUATE From vn= () mr and rkemv= 22() we obtain the orbital radius n22 r = emk2 The orbital frequency is vemkek42 2 2 fO ==33 = 3, where a0 = 2 22ππrnhan 0 emk The frequency of the photon emitted is

ke22⎛⎞⎛⎞11 ke 1 1 f =⎜⎟⎜⎟−=1− γ 22ha⎜⎟⎜⎟ n2222 ha n 00⎝⎠⎝⎠()nn++111() so for 11n = ke222⎡⎤⎛⎞22 ke ⎛⎞ e k fγ ≈−−223⎢⎥11⎜⎟= ⎜⎟ = 22ha000 n⎣⎦⎝⎠ n ha n ⎝⎠ n ha n

ASSESS For large values of n, the optical frequency fγ is the same as the orbital frequency fO.

83. INTERPRET This problem involves a photoelectric effect experiment. Given the data of the stopping potential as a function of wavelength, we are to determine the Planck’s constant, work function, and the material. DEVELOP The electron’s maximum kinetic energy is expended in crossing the stopping potential, so

eVsmax= K Using Equation 34.7, we obtain hc hc φ eV= hf−φφ=− ⇒ V =− ssλλee

Thus, plotting Vs vs. 1/λ will give a straight line with slope equal to hc/, e which allows us to experimentally determine h. Similarly, the y-intercept corresponds to −φ /,e from which we deduce the work function φ. EVALUATE (a) The plot is shown below.

The slope of the best-fit line is 1.245×⋅ 10−6 V m which allows us to deduce the value of h: e(1.245×⋅ 10−−−6196 V m) (1.6 × 10 C)(1.245 ×⋅ 10 V m) h == =6.65×⋅ 10−34 J s c 3.0× 108 m/s (b) The y-intercept is −φ /2.314V,e =− which implies a work function of φ ≈ 2.31eV. (c) From Table 34.1, we determine the material to be potassium. ASSESS Work function φ is the minimum energy required to eject an electron, and the value is typically a few eV for most metals.

84. INTERPRET We investigate the relation between particle lifetimes and uncertainty in their rest energy measurements. DEVELOP From Equation 34.16, the uncertainties in the energy and time are constrained by ∆∆≥Et . Therefore, the lifetime of a given particle will be inversely proportional to the uncertainty in its rest energy: τ ∝∆1/E . EVALUATE The shortest lifetime will correspond to the curve with the largest uncertainty in its rest energy. In the graph, this is particle C. The answer is (c). ASSESS The distribution width shown in the graph is called the natural line width and is denoted by Γ. It is called “natural” to signify that this uncertainty is inherent to the particle and does not, like other uncertainties, come simply from the imperfect instruments used to collect the data.

85. INTERPRET We investigate the relation between particle lifetimes and uncertainty in their rest energy measurements. DEVELOP As argued above, the lifetime is related to the uncertainty in its rest energy by τ ≈∆ /.E EVALUATE For an uncertainty of 1 MeV, the lifetime must be roughly h 6.63×⋅ 10−34 J s⎡⎤ 1 eV 710s−22 τ ≈= =⎢⎥−19 =× ∆⋅∆EE221ππ() MeV1.610J⎣⎦ × The answer is (b). ASSESS Although this seems like an incredibly short amount of time, there are many particles that have lifetimes in this range.

86. INTERPRET We investigate the relation between particle lifetimes and uncertainty in their rest energy measurements. DEVELOP The inverse relation between energy uncertainty and lifetime is ∆≈E /.τ EVALUATE A longer lifetime leads to a narrower range in the energy measurement. By Einstein’s mass-energy equivalence, this corresponds to a narrower range in the mass, as well. The answer is (d). ASSESS Some particles, like the proton and the electron, appear to have infinite lifetimes, so we’d expect the uncertainty in their mass to be near to zero.

87. INTERPRET We investigate the relation between particle lifetimes and uncertainty in their rest energy measurements. DEVELOP Using Einstein’s mass-energy equation ()E = mc2 , we can write the time-energy uncertainty inequality as ∆≈mc /.2τ EVALUATE Plugging in the lifetime, the mass range is 6.63×⋅ 10−34 J s⎡⎤ 1 u m 710u−18 ∆≈227 =2 ⎢⎥− =× c τ 2310m/s10sπ ()()× 87− ⎣⎦1.661× 10 kg The answer is (c). ASSESS In particle physics, 10−7 s is a relatively long lifetime. A particle with roughly this long of a lifetime is the charged pion with τ =2.6× 10−8 s and a mass of 0.15 u.