Physics

Physics · optics

Problem

5.2. The radioactive source used by Cherenkov emits either α\alpha particles (i.e. helium nuclei) having a rest mass Mα=3.8GeV/c2M_{\alpha}=3.8 \mathrm{GeV} / c^{2} or β\beta particles (i.e. electrons) having a rest mass Me=0.51MeV/c2M_{\mathrm{e}}=0.51 \mathrm{MeV} / c^{2}. Calculate the numerical values of Tmin⁡T_{\min } for α\alpha particles and β\beta particles. Knowing that the kinetic energy of particles emitted by radioactive sources never exceeds a few MeV\mathrm{MeV}, find out which particles give rise to the radiation observed by Cherenkov. Given constants: CHERENKOV LIGHT AND RING IMAGING COUNTER Light propagates in vacuum with the speed cc. There is no particle which moves with a speed higher than cc. However, it is possible that in a transparent medium a particle moves with a speed vv higher than the speed of the light in the same medium cn\frac{c}{n}, where nn is the refraction index of the medium. Experiment (Cherenkov, 1934) and theory (Tamm and Frank, 1937) showed that a charged particle, moving with a speed vv in a transparent medium with refractive index nn such that v>Cnv>\frac{C}{n}, radiates light, called Cherenkov light, in directions forming with the trajectory an angle θ=arccos⁡1βn(1)\theta=\arccos \frac{1}{\beta n} \tag{1} where β=vc\beta=\frac{v}{c}. <image_1> 1. To establish this fact, consider a particle moving at constant velocity v>Cnv>\frac{C}{n} on a straight line. It passes A\mathrm{A} at time 0 and B\mathrm{B} at time t1t_{1}. As the problem is symmetric with respect to rotations around AB\mathrm{AB}, it is sufficient to consider light rays in a plane containing AB\mathrm{AB}. At any point C\mathrm{C} between A\mathrm{A} and B\mathrm{B}, the particle emits a spherical light wave, which propagates with velocity cn\frac{c}{n}. We define the wave front at a given time tt as the envelope of all these spheres at this time. Context question: 1.1. Determine the wave front at time t1t_{1} and draw its intersection with a plane containing the trajectory of the particle. Context answer: <image_2> Context question: 1.2. Express the angle φ\varphi between this intersection and the trajectory of the particle in terms of nn and β\beta. Context answer: φ=arcsin⁡1βn\varphi=\arcsin \frac{1}{\beta n} Context question: 2. Let us consider a beam of particles moving with velocity v>Cnv>\frac{C}{n}, such that the angle θ\theta is small, along a straight line IS. The beam crosses a concave spherical mirror of focal length ff and center C\mathrm{C}, at point S\mathrm{S}. SC makes with SI a small angle α\alpha (see the figure in the Answer Sheet). The particle beam creates a ring image in the focal plane of the mirror. Explain why with the help of a sketch illustrating this fact. Give the position of the center O\mathrm{O} and the radius rr of the ring image. This set up is used in ring imaging Cherenkov counters (RICH) and the medium which the particle traverses is called the radiator. Note: in all questions of the present problem, terms of second order and higher in α\alpha and θ\theta will be neglected. Context answer: <image_3> Extra Supplementary Reading Materials: 3. A beam of particles of known momentum p=10.0GeV/cp=10.0 \mathrm{GeV} / \mathrm{c} consists of three types of particles: protons, kaons and pions, with rest mass Mp=0.94GeV/c2M_{\mathrm{p}}=0.94 \mathrm{GeV} / c^{2}, Mκ=0.50GeV/c2M_{\kappa}=0.50 \mathrm{GeV} / c^{2} and Mπ=0.14GeV/c2M_{\pi}=0.14 \mathrm{GeV} / c^{2}, respectively. Remember that pcp c and Mc2M c^{2} have the dimension of an energy, and 1eV1 \mathrm{eV} is the energy acquired by an electron after being accelerated by a voltage 1 V1 \mathrm{~V}, and 1GeV=109eV,1MeV=106eV1 \mathrm{GeV}=10^{9} \mathrm{eV}, 1 \mathrm{MeV}=10^{6} \mathrm{eV}. The particle beam traverses an air medium (the radiator) under the pressure PP. The refraction index of air depends on the air pressure PP according to the relation n=1+aPn=1+a P where a=2.7×10−4 atm−1a=2.7 \times 10^{-4} \mathrm{~atm}^{-1} Context question: 3.1. Calculate for each of the three particle types the minimal value Pmin⁡P_{\min } of the air pressure such that they emit Cherenkov light. Context answer: Pmin⁡=16 atm for protons, Pmin⁡=4.6 atm for kaons, Pmin⁡=0.36 atm for pions. \begin{array}{ll} P_{\min }=16 \mathrm{~atm} & \text { for protons, } \\ P_{\min }=4.6 \mathrm{~atm} & \text { for kaons, } \\ P_{\min }=0.36 \mathrm{~atm} & \text { for pions. } \end{array} Context question: 3.2. Calculate the pressure P12P_{\frac{1}{2}} such that the ring image of kaons has a radius equal to one half of that corresponding to pions. Calculate the values of θκ\theta_{\kappa} and θπ\theta_{\pi} in this case. Is it possible to observe the ring image of protons under this pressure? Context answer: P12=6P_{\frac{1}{2}}=6 , θκ=1.6\theta_{\kappa}=1.6 , θπ=3.2\theta_{\pi}=3.2 Extra Supplementary Reading Materials: 4. Assume now that the beam is not perfectly monochromatic: the particles momenta are distributed over an interval centered at 10GeV/c10 \mathrm{GeV} / \mathrm{c} having a half width at half height Δp\Delta p. This makes the ring image broaden, correspondingly θ\theta distribution has a half width at half height Δθ\Delta \theta. The pressure of the radiator is P12P_{\frac{1}{2}} determined in 3.2. Context question: 4.1. Calculate ΔθκΔp\frac{\Delta \theta_{\kappa}}{\Delta p} and ΔθπΔp\frac{\Delta \theta_{\pi}}{\Delta p}, the values taken by ΔθΔp\frac{\Delta \theta}{\Delta p} in the pions and kaons cases. Context answer: ΔθκΔp=0.51\frac{\Delta \theta_{\kappa}}{\Delta p}=0.51 , ΔθπΔp=0.02\frac{\Delta \theta_{\pi}}{\Delta p}=0.02 Context question: 4.2. When the separation between the two ring images, θπ−θκ\theta_{\pi}-\theta_{\kappa}, is greater than 10 times the half-width sum Δθ=Δθκ+Δθπ\Delta \theta=\Delta \theta_{\kappa}+\Delta \theta_{\pi}, that is θπ−θκ>10Δθ\theta_{\pi}-\theta_{\kappa}>10 \Delta \theta, it is possible to distinguish well the two ring images. Calculate the maximal value of Δp\Delta p such that the two ring images can still be well distinguished. Context answer: 0.30.3 Extra Supplementary Reading Materials: 5. Cherenkov first discovered the effect bearing his name when he was observing a bottle of water located near a radioactive source. He saw that the water in the bottle emitted light. Context question: 5.1. Find out the minimal kinetic energy Tmin⁡T_{\min } of a particle with a rest mass MM moving in water, such that it emits Cherenkov light. The index of refraction of water is n=1.33n=1.33. Context answer: 0.517Mc20.517 M c^{2}

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