Phys 197 Homework Solution 38B Q12. In a photoelectric-effect experiment, which of the following will increase the maximum kinetic energy of the photoelectrons? (a) Use light of greater intensity; (b) use light of higher fre- quency; (c) use light of longer wavelength; (d) use a metal surface with a larger work function. In each case justify your answer. ————– (a) No. There are more photoelectrons, but the energy distribution, most specifically including the max energy, is unchanged. (b) Yes. Higher frequency means higher photon energy. Ephoton = hf. (c) No. Longer λ means lower frequency. (d) No. A larger work function means more photon energy goes into freeing the electron, so less is left over for its KE. Q15. Why must engineers and scientists shield against x-ray production in high-voltage equipment? ————– High voltages will lead to arcs which lead to bremsstrahlung x-rays. P16. (a) What is the minimum potential difference between the filament and the target of an x-ray tube if the tube is to produce x rays with a wavelength of 0.150 nm? (b) What is the shortest wavelength produced in an x-ray tube operated at 30.0 kV? ————– (a) The minimum voltage is such that an electron acquires the energy of a photon of the desired wavelength. Ephoton = hc/λ = (1240 eV nm)/(0.150 nm) = 8270 eV. The potential to give an electron· this energy is 8270 V. (b) This tube produces electrons of 30 keV, so the cutoff photon energy will be 30 keV. λ = (1240 eV nm)/(30 000 V) = 0.0413 nm. · P23. X rays with an initial wavelength of 0.900 10−10 m undergo Compton scattering. × For what scattering angle is the wavelength of the scattered x rays greater by 1.0% than that of the incident x rays? ————– Recognize a problem for the Compton formula. We will need: λ = 90.0 pm and λ′ = 90.9 pm. λ′ λ =(h/mc)(1 cos θ). With h/mc = 2.426 pm, (1 − cos θ) = (90.9 − 90.0)/2.426 = 0.3710. cos−θ = 0.6290 −θ = 51.0◦. ⇒ P25. An electron and a positron are moving toward each other and each has speed 0.500 c in the lab frame. (a) What is the kinetic energy of each particle? (b) The e+ and e− meet head-on and annihilate. What is the energy of each photon that is produced? (c) What is the wavelength of each photon? How does the wavelength compare to the photon wavelength when the initial kinetic energy of the and is negligibly small (see Example 38.6)? ————– (a) At 0.5 c, we will use relativistic. K = mc2(γ 1), with − γ = 1/p(1 v2/c2) = 1.1547 (keep extra sig dig). Since mc2 = 511− keV, K = (0.1547)(511 keV) = 79.05 keV. 2 2 Incidentally, KNewton = (1/2)m(c/2) = mc /8 = 63.9 keV. Off by 20%. (b) Since total energy is conserved, each photon gets the relativistic energy of one electron. 2 Ephoton = mc γ = (511 keV)(1.155) = 590.0 keV. (c) λ = 1240 eV nm/E = (1240 eV nm)/(590 000 eV) = 2.10 pm. · photon · In annihilation at rest, Ephoton = 511 keV, and the wavelength is proportionately longer. λann. at rest = (2.10 pm)(590/511) = 2.43 pm. P35. An incident x-ray photon of wavelength 0.0900 nm is scattered in the backward direction from a free electron that is initially at rest. (a) What is the magnitude of the momentum of the scattered photon? (b) What is the kinetic energy of the electron after the photon is scattered? ————– (a) Strategy: Calculate initial and final momenta of the photon, then use that to get pelectron. Use pphoton = h/λ to get momenta. λ′ = λ +(h/mc)(1 cos θ) = (90 pm)+(2.426 pm)(1 ( 1)) = 94.85 pm. − − − − −− p = h/λ = (6.626 10 34 Js)/(90 10 12 m) = +7.36 10 24 kg m/s. ′ −× × ′ × p =( )6.99 10 24 kg m/s. Since p is in the opposite direction of p, − × −24 ∆p = ( )14.35 10 kg m/s, which is equal to ( )∆pelectron. Since the electron started from rest,− × − p = 1.435 10−23 kg m/s. electron × (b) The energy lost by the photon appears as K of the electron. −24 8 −15 Ephoton = pc = (7.36 10 kg m/s)(3 10 m/s) = 2.21 10 J. ′ −×15 × × E photon = 2.10 10 J. No negative signs with energy! × − −16 ∆E = 1.1 10 16 J K = 1.1 10 J. Or 690 eV. photon × ⇒ electron × P39. Nuclear fusion reactions at the center of the sun produce gamma-ray photons with energies of about 1 MeV (106 eV). By contrast, what we see emanating from the sun’s surface are visible-light photons with wavelengths of about 500 nm. A simple model that explains this difference in wavelength is that a photon undergoes Compton scattering many times – in fact, about 1026 times, as suggested by models of the solar interior – as it travels from the center of the sun to its surface. (a) Estimate the increase in wavelength of a photon in an average Compton-scattering event. (b) Find the angle in degrees through which the photon is scattered in the scattering event described in part (a). (Hint: A useful approxi- mation is cos φ 1 φ2/2 which is valid for φ << 1. Note that φ is ≈ − in radians in this expression.) (c) It is estimated that a photon takes about 106 years to travel from the core to the surface of the sun. Find the average distance that light can travel within the interior of the sun without being scattered. (This distance is roughly equivalent to how far you could see if you were inside the sun and could survive the extreme temperatures there. As your answer shows, the interior of the sun is very opaque.) ————– (a) Calculate the change in wavelength and divide by 1026. 6 λ0 = 1240 eV nm/(10 eV) = 0.00124 nm. · −24 − ∆λ = (500 .00124) = 5 10 nm. Or5 10 33 m. − − × × (b) ∆λ = (1.2 10 12 m)(1 cos θ) × − − −⇒ − 1 cos θ = (5 10 33 m)/(1.2 10 12 m)=4 10 21. − × × × Making the suggested approximation, 1 cos θ = 1 (1 θ2/2) = θ2/2 − −− − − ⇒ −9 ◦ θ = √2 4 10 21 = 9 10 11 rad = 5 10 . · × × × (c) Calculate the total distance D the photon travels in 106 yr, then divide by number of collisions, given as 1026. D = (3 108 m/s)(3.15 107s/yr)(106 yr) = 9.5 1021 m. Divide by 1026 to get: d = 9.5 × 10−5 m = 10−4×m= 0.1 mm. × ×.