Examination Radiation Physics - 8N120 5 November 2014, 13:30-16:30 Four general remarks: This exam consists of 8 assignments on a total of 3 pages. There is a table on page 4 listing the maximum number of points that can be achieved in each assignment. You are not allowed to use books, notes, lecture notes, graphical calculators, notebooks, telephones, tablets etc during the exam. It is allowed to use a ordinary calculator, not a graphical calculator. You can use the appended list of constants and formulas during the exam. All answers should be formulated and motivated clearly. The answers may be given in english or in dutch. 1. Consider the thermal radiation of a human body with a temperature of 37 C and a surface area of 2 m 2. You may assume that the human body is a blackbody. (a) Compute the wavelength λ max at which the radiancy of the human body is maximal. (b) Compute the total power radiated by the human body. (c) Compute the total power radiated by the human body by waves with wavelengths λ > 10λ max. Hint: For these long wavelengths the radiancy is very well described by the Rayleigh Jeans formula. 2. Consider the Compton effect where an incoming photon with an energy of 50.00 kev interacts with a free electron at rest. The trajectory of the scattered photon has an angle θ = 45 with respect to the trajectory of the incoming photon. (a) Compute the wavelength and energy of the scattered photon. (b) Compute the angle φ between the trajectory of the scattered electron and incoming photon. 3. The energy of a hydrogen atom in state n is given by E n = A n 2, where A = 13.6 ev. (a) Give the energy and wavelength of a photon that is emitted when a hydrogen atom goes from state n to state m. The Paschen series describes the wavelength λ n of photons that are emitted when a hydrogen atom goes from state n to state 3. This wavelength is given by λ n = λ limit n 2 n 2 9, where λ limit is called the series limit of the Paschen series. (b) Express λ limit in A and other physical constants. (c) Compute from part (b) the numerical value in nm of λ limit. 1
4. In this exercise we consider a particle with mass m in an infinitely deep, three-dimensional potential well ( particle in box ) with dimensions L L L. The well has coordinates 0 < x < L, 0 < y < L and 0 < z < L. Inside the well the potential is 0, outside. (a) Formulate the Schrödinger equation for the wave function Ψ(x, y, z) for this case. (b) Which boundary conditions on Ψ(x, y, z) must hold in this case? We now try to find solutions of this Schrödinger equations using the separation of variables method. Therefore suppose Ψ(x, y, z) = f(x)g(y)h(z). (c) Which equations for f(x), g(y) and h(z) result from this approach? (d) Give the general solutions of the Schrödinger equation for the considered case and the corresponding values of the energy E. This solution does not have to be normalized. (e) What is the lowest possible value of the energy E in this case? 5. Consider the following part of the energy level diagram of hydrogen. Figure 1: Part of the energy level diagram of hydrogen. (a) Indicate with arrows which transitions are allowed by the selection rule when the hydrogen atom goes from one state to a lower state. 6. Consider the generation of X-rays with a silver ( 47 Ag) target. The measured X-ray energies in silver are for K α : 21.990 kev and for K β : 25.145 kev. The binding energy of the K shell electron (n = 1 electron) is 25.514 kev. (a) Compute the energy of the L α X-ray. (b) What is the binding energy of an L shell electron (n = 2 electron)? 2
7. (a) Compute for a 135 56 Ba (Barium 135) nucleus the nuclear binding energy and the nuclear binding energy per nucleon. (b) Do you expect that the binding energy per nucleon of 135 57 La (Lanthanum 135) is more 135 or less than that of 56 Ba? Explain your answer. Hint: Look at Figure 2. Figure 2: Part of the chart of nuclides 8. Consider the part of the chart of nuclides that describes the decay of Figure 3. 213 At 85 (Astatine 213) in Figure 3: Part of the chart of nuclides (a) What type of radioctive decay is possible for equation. 213 At 85? Give the corresponding reaction (b) The chart of nuclides indicates a reaction product with an energy of 9.08 MeV. Compute the theoretical energy value of this reaction product. 3
Points: (total 100) Question 1a: 4 points Question 2a: 5 points Question 3a: 5 points Question 1b: 6 points Question 2b: 8 points Question 3b: 5 points Question 1c: 6 points Question 3c: 3 points Question 4a: 4 points Question 5a: 4 points Question 6a: 6 points Question 4b: 4 points Question 6b: 6 points Question 4c: 5 points Question 4d: 5 points Question 4e: 5 points Question 7a: 5 points Question 8a: 3 points Question 7b: 4 points Question 8b: 7 points 4
Radiation Physics - 8N120 Formulas and constants - 2014/2015 physical constants: h = 6.626 10 34 Js = 4.136 10 15 ev s (Planck s constant) k B = 1.381 10 23 J/K (Boltzmann s constant) c = 2.998 10 8 m/s (speed of light) e = 1.602 10 19 C (elementary charge) σ = 5.670399 10 8 W/(m 2 K 4 ) (Stefan-Boltzmann constant) N a = 6.022 10 23 mole 1 (Avogadro s constant) V m = 22.4 liter (at 1 atm, 0 C) (Molair volume R = 1.097 10 7 m 1 (Rydberg constant) µ B = 9.274 10 24 J/T (Bohr magneton) ɛ 0 = 8.854 10 12 F/m (vacuum permittivity) a 0 = 0.0529 10 9 m = 0.0529 nm (Bohr radius) hydrogen ionization energy= 13.6 ev unit conversions: 1 ev = 1.602 10 19 J (electron-volt to Joule) 1 J = 6.242 10 18 ev (Joule to electron-volt) 1 u= 1.66056 10 27 kg = 931.5 MeV/c 2 (atomair mass unit to kg to energy) hc = 1240 ev nm (hc in electronvolt nanometer) masses: m e = 9.109 10 31 kg = 0.000549 u =511 kev/c 2 m p = 1.672 10 27 kg = 1.0072766 u m n = 1.675 10 27 kg = 1.0086654 u (mass of electron) (mass of proton) (mass of neutron) M( 1 1 H) = 1 1M = 1.007825 u (mass of hydrogen atom) M( 4 2 He) = 4 2M = 4.00260325 u (mass of helium atom) M( 135 56 M( 213 85 M( 211 84 M( 209 83 Ba) = 135 56M = 134.905689 u (mass of barium isotope) At) = 213 85M = 212.992937 u (mass of astatine isotope) Po) = 211 84 Bi) = 209 83 M = 210.986653 u (mass of polonium isotope) M = 208.980399 u (mass of bismuth isotope) Photons and radiation laws: λ ν = c E = hν λ max T = 2.898 10 3 mk R(λ) = 8π λ 4 kt c 4 R(λ) = c 8π hc 1 4 λ 4 λ e hc/(λk BT ) 1 I = σt 4 (relation wave length frequency) (energy of photon) (Wien s displacement law) (radiation law of Rayleigh Jeans) (Planck s radiation law) (law of Stefan Boltzmann) 1
Fotoelectric effect and Compton effect: hf = ev s + φ λ λ = h (1 cos(θ)) m e c (foto electric equation) (Compton scattering) Relativistic energy and impuls (E is the total energy, including the rest energy): E 2 = (mc 2 ) 2 + (p c) 2 E = p = mc 2 1 v 2 /c 2 mv 1 v 2 /c 2 Atom model of Bohr: f = cr ( 1 n 2 1 m 2 ) (Rydberg formula) r n = a 0 n 2 = 4πɛ 0 h 2 m e e 2 n2 (radius nth orbit) E n = 1 (4πɛ 0 ) 2 m e e 4 2 h 2 1 n 2 = 13.6 n 2 ev (energy of nth orbit of hydrogen E n = 13.6Z2 n 2 ev (idem, for other atoms (ions) with one electron) Wave mechanics: λ = h p h2 d 2 Ψ + UΨ = EΨ 2m dx2 x p x h E t h (De Broglie wave length) (1-dim. Schrödinger equation) (uncertainty relationship) (uncertainty relationship) Nuclear physics: R = r 0 A 1/3 B tot (A, Z) = (ZM H + (A Z)M n A ZM)c 2 Q = (m initial m final ) c 2 T α = A 4 A (radius of nucleus) (binding energy) (Q-value) Q (kinetic energy of alpha particle) Some codings on the Chart of Nuclides: β + : beta plus decay, ɛ : electron capture, β : beta minus decay, α : alpha decay γ : gamma decay, e : internal conversion 2