solutions 03
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solutions 03

Course Number: PHY 251, Spring 2001

College/University: SUNY Stony Brook

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phy251_hw03 PHY251 Homework Set 3 Reading: Chapter 4 Homework: Chapter 4, Questions 3,5,7 Problems 2,8,13,16,32 Hints and Solutions Note: SHOW ALL WORK! Question IV.3 How would you estimate the temperature of a red-hot piece of steel, given that the wavelength of red light is in the neighborhood of 700 nm? What assumption would you have to make for your estimate? Hints: How do you know that the red wavelength is...

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Homework phy251_hw03 PHY251 Set 3 Reading: Chapter 4 Homework: Chapter 4, Questions 3,5,7 Problems 2,8,13,16,32 Hints and Solutions Note: SHOW ALL WORK! Question IV.3 How would you estimate the temperature of a red-hot piece of steel, given that the wavelength of red light is in the neighborhood of 700 nm? What assumption would you have to make for your estimate? Hints: How do you know that the red wavelength is a maximum, and not a tail of the distribution in the visible part of the spectrum? Solution: If it were the tail of a shorter wavelength distribution, the color would be more yellow-blue; it could, however, be the high-end tail of the distribution, with a maximum at the near infrared; the radiation would be weaker ("dull-red") and the temperature would be somewhat less. Assuming we indeed see the maximum at = 700 nm, the temperature is simply derived from: maxT = 2.910-3 mK; T = 2.910-3 mK / 700 nm = 4.1103 K. Question IV.5 Suppose you have a source that emits a beam of light at some frequency f that impinges on a metal plate. What happens to the energy of the photoelectrons that are emitted when you are moving the source closer to the plate? Hints: What happens to the frequency when the source approaches the observer (electron)? Solution: The Doppler effect will make the photon's frequency to the electrons a bit higher, and therefore more energetic. The photoelectrons are thus slightly more energetic on average. Question IV.7 Without working out the algebra in detail, why is the process + e e forbidden by the energy-momentum conservation laws, while the process e + p + H is allowed? Here, p stands for the proton and H for hydrogen. The allowed process is known as the radiative capture of electrons. Hints: Consider the energy and momentum before, and after the interaction. Solution: file:///C|/DOCUME~1/PHY_CO~1/PHY251/PHY251_hw03_solutions.html (1 of 3) [2/26/2001 11:39:09 AM] phy251_hw03 Energy and momentum cannot both be conserved in the first process: in it the final electron has an energy and momentum related by its mass, and thus only three degrees of freedom exist in the final state. In the initial state, I can vary both energy and momenta freely (four degrees of freedom..), i.e. a mismatch exists. In more detail, choosing the center-of-mass system (in which the final electron is at rest): pi,e = hf, Ef = me = Ei = hf + (pi,e2 + me2) = hf + (pi,e2 + me2) > me, i.e. a contradiction! Problem IV.2 Chemical processes typically involve energies on the order of 1 eV. What, then, is the typical wavelength of electromagnetic radiation emitted in the course of chemical reactions? Nuclear processes involve energies on the order of 1 MeV. Where in the spectrum of electromagnetic radiation are the photons that may be emitted in a nuclear reaction? No Hints: hints Solution: Chemistry: = c/f = hc/hf = 1240 eVnm / 1 eV = 1240 nm = 1.24 m; i.e. in the infrared (heat) Nuclear: = c/f = hc/hf = 1240 MeVfm / 1 MeV = 1240 fm = 1.2410-3 nm; i.e. gamma rays Problem IV.8 Light of frequency 8.51015 Hz falls on a metal surface. If the (maximum) energy of the resulting photoelectrons is 1.7 eV, what is the workfunction of the metal? Hints: No hints Solution: Conservation of energy: hf = Kmax + W; W = hcf/c - Kmax = 1240 eVnm8.51015/3108 m - 1.7 eV = 35.1 - 1.7 eV = 33.4 eV (= 5.310-18 J). Problem IV.13 In a Compton scattering experiment, the wavelength of the incident X-rays is 7.07810-2 nm, while the wavelength of the outgoing X-rays is 7.31410-2 nm. At what angle was the scattered radiation measured? Hints: No hints Solution: = hc/mc(1 - cos) = (7.314-7.078)10-2 nm = 0.23610-2 nm; hc/ = 1240 eVnm / 511103 eV = 2.4310-3 nm. mc Then: (1 - cos) = 2.3610-3 nm / 2.4310-3 nm = 0.9725; = 88.4 Problem IV.16 The function u(f,T) is the distribution of blackbody radiation in terms of frequency; u(f,T)df is the file:///C|/DOCUME~1/PHY_CO~1/PHY251/PHY251_hw03_solutions.html (2 of 3) [2/26/2001 11:39:09 AM] phy251_hw03 energy contained in the frequency interval fron f to f+df. Use the relationship between frequency and wavelength to find the function Y(,T) that describes the distribution in wavelength; Y(,T)d is the energy contained in a wavelength interval from to +d. Hints: Use Planck's distribution from the book or the notes. Then: f = c/; and thus df = -(c/)d. Of course, you get the best insight if you plot and study both functions. Solution: u(f=c/,T)df = (8hf/c)[exp(hf/kT) - 1]-1(c/)d = (8h/)[exp(hc/kT) - 1]-1(c/)d = (8hc/5)[exp(hc/kT) - 1]-1d = Y(,T)d Problem IV.32 Neutrons (m = 1.6710-27 kg) pass through a crystal and exhibit an interference pattern. If the neutrons have a kinetic energy of 1.7 eV, and the separation between successive maxima in the interference pattern is 6.410-2 rad, what is the separation of the crystal planes that produce the interference pattern? Hints: Make a sketch. Note that the process is identical to the interference phenomenon with light, when you use the wavelength of 1.7 eV neutrons as = h/p, with p the momentum of 1.7 eV (kinetic) energy neutrons. Note, that the simplest units to use are "nuclear" units: eV for energy, eV/c for momentum, and eV/c for mass (e.g. the neutron's mass is 939 MeV/c). In these units the only quantity you need to memorize is hc = 1240 eVnm. Solution: The momentum of K = p2/(2mn) = 1.7 eV neutrons is: p = (2mnK) = (2 939106 1.7 eV2/c2) = 56.5103 eV/c. Then, use Eq. (4-27): = 2a sin; a = /(2sin) = hc/(2pcsin) = 1240 eVnm / (2 56.5103 eV 6.410-2) = 0.171 nm file:///C|/DOCUME~1/PHY_CO~1/PHY251/PHY251_hw03_solutions.html (3 of 3) [2/26/2001 11:39:09 AM]

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