Radiation Biophysics Lecture 7 PDF

Radiation Biophysics For 3rd year Biophysics Lecture 7 By: Dr. Heba Mohamed Fahmy ,Biophysics Department ,Faculty of Science Cairo University Egypt  Interaction of radiation with matter :Beta particles :Attenuation of beta particles For th...

Radiation Biophysics For 3rd year Biophysics Lecture 7 By: Dr. Heba Mohamed Fahmy ,Biophysics Department ,Faculty of Science Cairo University Egypt  Interaction of radiation with matter :Beta particles :Attenuation of beta particles For the given setup, the count rate from a pure beta emitter will decrease rapidly with increase in absorber thickness and then slowly. A thickness will be reached that will stop all beta and the detector will be only counting the background. :The Range of beta particles It is the end point of the absorption curve where no further decrease in count Count rate rate is observed. It depends on the material of absorber and the energy of Absorber thickness beta particles. It is the maximum distance travelled by maximum energy beta particles. Range-energy relationship: The given figure shows that the required thickness of absorber for any given beta energy decreases as the density of absorber increases. :Areal density It is the number of absorber electrons in the path of beta particles (electrons/cm2). It is the factor that really affects the absorption of beta particles in an absorber (atomic number of absorber is of minor effect and is neglected in calculation of beta-particle shield). Density thickness of absorber, t (g/cm2) = d density, ρ (g/cm ) x linear thickness, tl (cm). 3 :Advantage of using density thickness It makes possible to specify absorbers.1.independent of absorber material Example: the density thickness of a sheet of Al, 1 cm thick is: T d = 2.7 g/cm3 x 1 cm = 2.7 g/cm2. For Plexiglas (1.18 g/cm3) to produce the same beta absorption properties of 1cm of Al, its thickness should be: Tl = td/ρ = (2.7 g/cm2) / (1.18 g/cm3) = 2.39 cm It allows the addition of thicknesses of different materials in a radiologically.2.meaningful way It is now possible to have a universal curve of beta ray range (in units of :density thickness) versus energy Example: What must be the thickness of a shield made of Plexiglas (ρ=1.18g/cm3) in order that no beta from Sr-90 source pass through? (Maximum Energy β = 0.54 MeV for Sr-90 and 2.27 for Y-90). Solution: One must stop the beta particles of maximum energy which is emitted from the daughter Yttrium-90. From the energy range relationship, the range of 2.27 MeV beta particles is 1.1g/cm2 therefore: Tl = td/ρ = (1.1 g/cm2) / (1.18 g/cm3) = 0.932 cm The range-energy relationship is used by health physicist as an aid to identify unknown beta-emitting containment. This is done by measuring the range of beta radiation, then finding the corresponding energy from the range-energy graph, then looking up in table of isotopes for the isotope that emits beta particle of that energy. Mechanism of energy loss a) Ionization and Excitation (inelastic collisions): It is due to the interaction between the electric field of beta particles and the orbital electrons of the medium.  The electron is held in the atom by electrical forces, and the energy is lost by beta particles in overcoming these forces. Since electrical forces cover long distances, the “collision” between beta particle and an electron occurs without the two particles coming into actual contact, as in the collision between like poles of two magnets. :The amount of energy lost by beta particles depends on.Its distance of approach to the electron -1.Its kinetic energy -2 :The kinetic energy of the ejected electron is given by Ek= Et- φ Et: the energy lost by beta particle during collision, φ: the ionization potential of.absorbing medium In many ionizing collisions, only one ion pair is produced, in other cases the ejected electron may have enough kinetic energy to make a small cluster of several collisions. Delta rays: are ejected electrons that have enough energy to travel a long distance and produce a trail of ionizations. Beta particles have the same mass as orbital electrons and are easily deflected during collisions.  The figure shows the path of beta particles through photographic emulsion. The ionizing events expose the film at the points of ionization, thus making them visible after development of the film. Electronic excitation: The average energy expended (used) in the production of an ion pair is found to be two to three times greater than the ionization potential. This is due to energy used in electronic excitation. Specific ionization: It is the number of ion pairs formed per unit distance travelled by beta particles. It represents the linear rate of energy loss, which is an important parameter in health physics instrumentation and the biological effects of radiation. Specific ionization is high for low energy beta- particles and decreases rapidly as beta energy increases until a broad minimum is reached at 1 MeV. b) Bremsstrahlung: They are x-rays that are emitted when high-speed beta particles suffer rapid deceleration. As beta particles pass close to a nucleus, the strong attractive coulomb forces causes beta particles to deviate sharply from its original path. For the purpose of estimating radiation hazard, the following approximate relation :may be used f 3.5 x 10  4 ZE The fraction of incident beta energy converted to x-ray photons Atomic number of absorber Maximum energy of.beta particles, MeV Since the likelihood of bremsstrahlung increases with atomic number, the minimum possible atomic number materials are used for beta-ray shielding. Atomic numbers more than 13 (Aluminium) are seldom if ever used. Alpha Rays Range-energy relationship: Extrapolated range: Is obtained by extrapolating the absorption curve to zero alpha particles transmitted Mean range: Is the range most accurately determined and corresponds to the range of the average alpha particle Alpha particle absorption curve is flat because alpha radiation is essentially monoenergetic. Increasing the thickness of alpha absorber serves only to reduce the energy of Alphas that pass through it, the number of alphas is not reduced until the approximate range is reached. At this point there is a sharp decrease in the number of alphas that pass through the absorber :The approximate range in Air for Alpha particles For E  4 MeV : R(cm) 0.56E ( MeV ) For 4 MeV  E  8 MeV : R(cm) 1.24 E ( MeV )  2.62 :The approximate range for Alpha particles in any other medium 2 1 A: Atomic wt. of medium, R: range of Alpha Rm (mg / cm ) 0.56 A R (cm)3 particle in Air, cm Example: What is the thickness of Aluminium foil, density 2.7 g/cm 3, required to stop alpha particles from 210Po (Ealpha = 5.3 MeV, atomic wt. = 27)? Solution: First, calculate the range of 210 Po alpha particle in Air: R = (1.24 x 5.3) - 2.62 = 3.95 cm, Second, calculate the range of alpha in Al: R = 0.56 x (27)1/3 x 3.95 = 6.64 mg/cm2 The thickness of Al in cm is calculated from: Tl = td/ρ = (6.64 mg/cm2) / (2.7 g/cm3) = 0.00246 cm  :A relation to calculate the range of alpha particle in tissue Ra: Range in air (cm), Rt: Range in Tissue (cm), ρa: density Ra  a Rt  t in air (g/cm3), ρt: density in tissue (g/cm3). Example: In the previous example, calculate the range of alpha particle in tissue (ρt=1g/cm3), where ρa=1.293 x 10-3 g/cm3. Solution: 3.95cm x 1.293 x 10  3 g/cm 3 3 Rt  3 5.1 x 10 cm 1g/cm :Energy Transfer The only significant energy loss for alpha particle is electronic excitation and ionization. In moving through air or soft tissue, an alpha particle loses about 35 eV per ion pair it produces. The specific ionization of an alpha particle is very high because, (a) Its high electric charge and (b) relatively slow velocity due its great mass. Gamma Rays Gamma rays are different from beta or alpha particles in that they cannot be completely absorbed, they can only be reduced in intensity by increasing the thickness of absorber  If gamma ray attenuation measurement is done under conditions of good geometry and the results are plotted we will get the following curves for normal and semi-log papers respectively. The equation of a straight line in the semi-log graph is: ln I  t  ln I 0 thus ln I  t , then I e  t I0 I0 I0 = gamma ray intensity at zero.absorber thickness.t = absorber thickness I = gamma ray intensity transmitted through absorber thickness, t. e = base of natural logarithm.system μ = slope of the absorption curve. = the attenuation coefficient  The total attenuation coefficient: it is the fraction of gamma ray beam attenuated per unit thickness of the absorber. Since the exponent in an exponential must be dimensionless, μ and t must be in reciprocal dimensions. Linear Attenuation Coefficient: If the absorber thickness is measured in cm, then the attenuation coefficient is called linear attenuation coefficient μl in dimension of cm-1. Mass Attenuation Coefficient: If the absorber thickness is measured in g/cm2, then the attenuation coefficient is called mass attenuation coefficient μm in dimension of cm2/g. The relation between linear and mass attenuation coefficients: 1 2 3 l (cm )  m (cm / g ). ( g / cm ) The atomic attenuation coefficient, μa : it is the fraction of gamma ray beam 2 l (cm  1 ) attenuated by a single atom. It is the  a (cm / atom)  N (atoms / cm3 ) probability that an absorber atom will interact with one of the photons inNthe is the number of absorber atoms per cm3 Microscopic Cross section, σ (cm2/atom): Since the μa is in cm2, (units of area), it is always referred to as “cross section” or “microscopic cross section “of absorber. Macroscopic Cross section Σ (cm-1): the linear attenuation coefficient is called the macroscopic cross-section 1 2 3 (cm )  (cm / atom) x N (atoms / cm ) With the aid of atomic cross sections, one can compute the attenuation coefficient of an alloy or biological sample. Example: Aluminium bronze, an alloy of 90% Cu (atomic wt. =63.57) and 10% Al (atomic wt. = 27) by weight, has a density of 7.6 g/cm 3. What are the linear and mass attenuation coefficients for 0.4 MeV gamma rays, if the cross-sections for Cu and Al for this energy are 9.91 and 4.45 barns? Solution: The linear attenuation coefficient of Al bronze is: μl = (μa)Cu x NCu + (μa)Al x NAl The number of Cu atoms in the alloy: NCu = [(6.03 x 1023 ) atoms/mole x (7.6 x 0.9) g/cm3 ] / 63.57g/mole = 6.49 x 1022 atoms/cm3. Similarly, NAl = 1.7 x 1022 atoms/cm3 μl = 9.91 x 10-24 x 6.49 x 1022 + 4.45 x 10-24 x 1.7 x 1022 = 0.705 cm-1 μm = μl / ρ = 0.705 / 7.6 = 0.0927 cm2/g  The given graph shows that:. The attenuating properties of matter vary systematically with atomic number of the absorber and with energy of incident gamma radiation For energies between 0.7 and 5 MeV, almost all materials have, on a mass basis (μm) the same gamma ray attenuating properties. For this range of energies, shielding properties are approximately proportional to the density of shielding material. For lower and higher gamma ray energies, absorbers of high atomic number are more effective than those of low atomic number.