MEDICINSK STRÅLNINGSFYSIK

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1 MEDICINSK STRÅLNINGSFYSIK TENTAMEN I MEDICINSK STRÅLNINGSFYSIK Kurs Joniserande strålnings växelverkan (7,5 hp) , Hjälpmedel: Physics handbook, Mathematical handbook, Tabellsammanställningar Alla ekvationer och approximationer skall motiveras. Ange svaren med rätt enheter. PROBLEMDEL Varje helt rätt löst problem ger 10 p 1. The function of an organ is investigated by injecting blood cells, marked with 203 Hg, that is taken up in the organ. For obtaining the best result it is important to know the position of the organ below the skin. This is obtained by measuring the ratio between the net count rates in the energy windows kev and kev with a NaI detector positioned 5.0 cm from the skin. The result was for an organ At what depth is the organ situated? The organ can be regarded as a point source. The NaI-detector as an efficiency of 0.90 in the energy window kev and 0.97 in the energy window kev. The photons pass through 1.0 cm of bone tissue beside the muscle tissue. The muscle tissue has the density 1.0 g/cm 3 and the bone tissue 1.8 g/cm 3. Assume that only primary photons are measured. Solution: Detector 5.0 cm Skin Bone tissue The count rate r at the detector is given by Source rr = GGGGGGGGee μμ ii xx ii G= geometry factor η= detector efficiency f= number of photons in the correct energy range per decay A= activity µ i = attenuation coefficient x i = attenuation distance

2 a) kev (279 kev) f=0.815 (from Table), µ tissue = cm -1, µ bone = = cm -1.(from Table), η=0.90 (given) b) kev (75 kev) f= (from Table), µ tissue = cm -1, µ bone = = cm -1.(from Table), η=0.97 (given) G and A are the same in both situations x bone =1,0 cm, x m is the muscle tissue thickness Attenuation in air can be neglected. Corresponds to less than 0.05 mm tissue rr 279 rr 75 = GG AA ee ee xx GG AA ee ee xx=10.3 Solving the equation gives x=5.51 cm The total thickness below skin includes also 1 cm of bone tissue. Answer: The organ is deposited at a distance of 6.5 cm. 2. A narrow parallel beam of photons with the energy 1.50 MeV impinges on a foil of Beryllium with the thickness 10 mg/cm 2. The beam area is 0.50 cm 2. A detector that only measures electrons is placed at a distance of 40 cm from the foil at an angle of 40 o to the impinging photon beam. The detector area is 1.25 cm 2. How many electrons hit the detector per second and what is the energy of the electrons? Self absorption in the foil may be neglected. The impinging photon beam has an energy fluence rate of 8.50 J s -1. Solution: 1.5 MeV photons Be-foil 40 o Electron detector If the electrons are emitted in angle Φ then the photons are emitted at an angle θ according to the relation cccccccc = 1 2 (1+αα) 2 tttttt 2 +1 α=hν/m o c 2 Φ=40 ο and α=1.5/0.511 MeV Data inserted in the equation gives θ=33.7 ο

3 This gives the photon energy from the compton equation hνν = hνν 1+hνν mm 0 cc 2 (1 cccccccc ) Data inserted gives hν = 0.67 MeV and the electron energy T= hν hν =0.83 MeV The number of electrons that are scattered in the angle 40 o is equal to the number of photons that are scattered at an angle of 33.7 o. Thus the number of electrons that hits the detector is given by the relation RR = ΦΦ dd ee σσ dddd NN ee SS ll 2 ΦΦ = pphoooooooo ffffffffffffff rrrrrrrr (ΦΦ = ΨΨ hνν ; = m -2 s -1 ) dd ee σσ dddd = KKKKKKKKKK NNNNNNhiiiiii cccccccccc ssssssssssssss ( cm 2 /(steradian electron)) N e =number of electrons in the foil (m Be ZN A /m a = /9.01= electrons) m Be =S fi d (beam area(0.5 cm 2 ) times foil thickness(10 mgcm 2 ) N A= Avogadros number ( ) Z= atomic number (4) M a =atomic mass (9.01) S=detector area (1.25 cm 2 ) l= distance foil-detector (0.40 m) Data inserted in the equation gives R=130 s -1. Answer: 130 electrons per second hit the detector.

4 TEORIDEL 1. Why are there for Tungsten two values for the mass attenuation coefficient at the energy MeV? (1 p) Answer: At the K-edge there are two values of the attenuation coefficient, one just below the K-edge and one just above, where there is a possibility to emit an electron from the K-shell through photoelectric effect. 2. If the stopping power for a 2.5 MeV proton in water is MeVcm 2 /g, what is approximately the stopping power of a 10 Mev He ++ -ion? (1 p) Answer: A 10 Mev He ++ -ion has about the same velocity as a 2.5 MeV proton. Then the stopping powers are related by the Z 2. Thus the stopping power for the He ++ -ion is 4*134.3=537 MeVcm 2 /g 3. Which the maximal energy a 20 MeV 4 He ++ -ion can transfer to a free electron in a collision? (2 p) Answer. The maximum energy transfer is given by the relation T=4Tm e /M, where T is the energy of the incoming particle, m e =the mass of an electron and M is mass of the incoming particle. This can be obtained using momentum and energy equations. The mass of a helium ion is around 4*1830. m e. Thus T=204/(4*1830)=0.011 MeV 4. The collision mass stopping power increases with energy for energies over around 1 MeV. Why? The collision mass stopping powers for air and graphite are nearly the same for energies between 0.5 and 1 MeV, but then is the increase faster for air than for graphite. Why? (2 p) Answer: The mass collision stopping power increases with energy due to relativistic effects. The Lorentz contraction makes all distances smaller and the coulomb force larger thus increasing the stopping power.. The density effect increases with increasing energy, as it is then possible to interact with atoms at larger distances and there may be more atoms between the incoming particle the interacting atom. The effect is more important for graphite, with its larger density, than for air and thus the increase in stopping power with energy is larger for air. 5. To calculate the transport of particles through matter the Monte Carlo method is often used. When simulating photon transport it is possible to follow each independent interaction of the photons. This is often not realistic when simulating transport of electrons. Why? How is the problem solved when simulating electron transport. (2 p) Answer. Electrons interact very often and to stop an electron often several thousand interactions are needed in a short distance. Also some of the cross sections for single interactions are uncertain. This means that to calculate all interactions will take a long computation time and may be uncertain. To perform Monte Carlo simulations for electrons one then let the electrons perform short steps including many interactions. After each step the energy loss and the change in direction is obtained through a Monte Carlo simulation. Then a new step with a new energy and a new direction is performed from the new position. This continues until the electron energy reaches a specified low energy value, which is then forced to be absorbed on the spot. 6. In radiation physics the photons are often regarded as particles with certain energy. As they are also waves it is possible to describe the photon by its wavelength. Show that for Compton scatter the change in wave length for a certain scattering angle is independent of the photon energy. (2 p)

5 Answer: The relation between the scattered and the primary photon energy is given by hνν = This equation can be rewritten as hνν 1+hνν mm 0 cc 2 (1 cccccccc ) 1 hνν = 1 + hνν mm 0cc 2 (1 cccccccc) hνν 1 hνν = 1 (1 cccccccc) + hνν mm 0 cc 2 The relation between wave length and frequency gives λ=c/ν Thus and λλ hcc = λλ (1 cccccccc) + hcc mm 0 cc 2 λλ h(1 cccccccc) λλ = mm 0 cc The difference in wave length is thus independent on the photon energy. 7. According the text book (Podgorsak) the mass scattering power may be defined as TT = Θ2 ρρ ρρρρ This means that the mean square scattering angle Θ 2 increases linearly with the absorber thickness t and it is possible to obtain very large values of Θ 2. This is however correct. Why? What happens with Θ 2 for large values of the absorber thickness tt Answer: When the particle is transported through the material the energy is decreased and thus the mass scattering power can t be constant. Besides, when the particles are getting more and more out scattered there is a chance that a particle that is at a large scattering angle can be scattered into a smaller angle. Thus after a certain depth there is diffusion so the out scattered particles are compensated by in scattering and there will be a constant value of ΘΘ What is moderation of a neutron beam? Which interaction process is the most important for moderation? Explain why material that contains hydrogen is a better moderator than a material that only contains atoms with a Z as e g Pb. Give example of a construction of a neutron radiation protection barrier that can completely stop neutrons with a mean energy of 1 MeV. (2)

6 9. The figure below shows the fraction of the photon energy that on average is transferred to electrons at interaction. f τ, f σ and f κ represent photoelectric effect, compton scattering and pair production respectively. Explain the variation with atomic number and the photon energy. Comment in particular that it is only for photoelectric effect there is a dependence of the atomic number (3 p). Answer: The energy transferred to the electrons are given according to the relations. A) Photoelectric effect: T hν Β ( Β is the electron binding energy) In this equation the energy lost to Auger electrons is neglected. Β is dependent on the atomic number. At the indicated dashed lines are the binding energies of the K-electrons. It seems that the figure neglects the energies below the K- edge. For graphite, C, B is close to zero and nearly all energy is transferred to the electron and and f is close to 1.0. For e.g. Pb just above the K-edge hν is close to B and f is small. With increasing photon energy transferred to the electron increases and thus f increases as well to a value close to 1.0. B) Compton scattering Compton scattering is a process with a free electron and independent of the atomic number. Analyzing the Compton equation one finds that at high photon energies a large fraction of the photon energy is transferred to the electrons. At low photon energies most energy goes to the scattered photon. Thus f is increasing with increasing energy. C) Pair production The energy transferred to the electron is given by T=hν-2m o c 2 Thus with increasing photon energy the relative energy transferred to the electron increases from 0 at 1 MeV to 1 at very high energies. 10. The photon spectrum produced in a linear accelerator has often a somewhat higher mean energy centrally as compared to the beam edges. Besides, the fluence rate is higher at the centre. To obtain uniform beams a beam flattening filter, that is thicker at the centre, is positioned in the beam. How will this filter affect the spectral distribution and the mean energy of the photons? Discuss the differences in the spectral distributions centrally for an Al-filter and a Pb-filter respectively, using the knowledge on which interaction processes that are important at different energies and atomic numbers. (3 p) The figure below shows the spectrum before the filter.

7 10,00 Energispektrum för 21 MV röntgenstrålning 1,00 Utjämningsfilter Al eller Pb Tjocklek centralt 25 gcm -2 0,10 0, Fotonenergi/MeV Answer: It is mainly the transmitted photons that are of interest. The secondary photons may contribute somewhat. This is mainly when there is a Compton interaction. At low energies where photoelectric effect could be probable, the probability that the K-x-rays will be absorbed in the filter is high. At very high energies where there is a chance of pair production, the annihilation photons are emitted isotropical and the probability that they will reach the patient is low. When Compton scattering dominates there are some Compton scattered photons. As we are interested in photons in the forward direction, the scattering angles are small and the scattered photons will have energies only slightly less than the primary ones. Thus the discussion can concentrate on the transmitted photons. Al-filter: µ/ρ decreases continuously with increasing energy in the energy range up to 20 MeV. This is due to the reason that the Compton process dominates up to high energies and he probability for Compton scattering decreases with energy. Thus there is a larger attenuation for the lower energies. This means that there will be an increase in the mean energy. Pb-filter: µ/ρ decreases first but has a minimum at around 4 MeV. Then µ/ρ increases as the pair production will become dominant and the probability for pair production increases with energy. This means that both high and low photon energies will decrease more than middle energies. This may imply that the mean energy will not increase but it may even decrease with a filter made of lead.