Radiation Protection and Safety Lecture Notes PDF

RADIATION PROTECTION AND SAFETY BY DR. ALI ALSHEHRI ASSOCIATE PROFESSOR DEPARTMENT OF MONDAY AND WEDNESDAY 16.30-18.00 Nuclear Medicine Radiation Oncology Medical Physics Radiation safety Radiology RADIOACTIVITY TRANSFORMATION MECHANISMS All radioactive transformations fall into one of the...

RADIATION PROTECTION AND SAFETY BY DR. ALI ALSHEHRI ASSOCIATE PROFESSOR DEPARTMENT OF MONDAY AND WEDNESDAY 16.30-18.00 Nuclear Medicine Radiation Oncology Medical Physics Radiation safety Radiology RADIOACTIVITY TRANSFORMATION MECHANISMS All radioactive transformations fall into one of the following categories: Radioactivity may be defined as ➢ Alpha emission spontaneous nuclear transformations in ➢Isobaric transitions (Given the atomic number of unstable atoms that result in the formation the parent nucleus is Z, that of the daughter nucleus is Z + 1, if a beta particle is emitted, or Z − of new elements. 1, if positron is emitted. The atomic mass number of the daughter is same as that of the parent.) ◦ Beta (negatron) emission ◦ Positron emission ◦ Orbital electron capture ➢Isomeric transitions (The atomic number and the atomic mass number of the daughter is same as that of the parent.) ◦ Gamma ray emission ◦ Internal conversion Alpha Emission An alpha particle is a highly energetic helium nucleus that is emitted from the nucleus of an unstable atom when the neutron-to-proton ratio is too low. It is a positively charged, massive particle consisting of an assembly of two protons and two neutrons Example )polonium) 210 8 4Po + 42He → 206 82Po Alph (𝜶) emission condition Generally, for alpha emission to occur, the following conservation equation must be satisfied: MP = Md + Mα + 2Me + Q where Mp, Md, Mα, and Me are respectively: the masses of the parent, the daughter, the emitted alpha particle, and the two orbital electrons that are lost during the transition to the lower atomic numbered daughter 𝑄: is the total energy release associated with the radioactive transformation For polonium (210 84Po) • 𝑀𝑝𝑜 = 210.04850 𝑎𝑚𝑢 • 𝑀𝑝𝑏 = 206.03883 𝑎𝑚𝑢 • 𝑀𝛼 = 4.00277 𝑎𝑚𝑢 • 𝑀𝑒 = 0.00055 𝑎𝑚𝑢 Q= MPo − MPb − Mα − 2Me Q= 210.04850 − 206.03883- 4.00277-2× 0.00055 = 0.0058 𝑎𝑚𝑢 =0.0058×931 =5.4MeV This Q value represents the total energy associated with the transformation of 210Po. The total released energy appears as kinetic energy and is divided between the alpha particle and the daughter, which recoils after the alpha particle is emitted. The exact energy division between the alpha and recoil nucleus depends on the mass of the daughter and may be calculated by applying the laws of conservation of energy and momentum. If M and m are the masses of the recoil nucleus and the 𝛼 particle respectively, and if V and v are their respective velocities, then 𝑄 = 12𝑀𝑉 2 + 12𝑚𝑣 2 According to the law of momentum conservation So, 𝑀𝑉 = 𝑚𝑣, 𝑚 𝑉= 𝑣 𝑀 1 𝑚 2 1 𝑄 = 2𝑀 𝑀𝑣 +2𝑚𝑣 2 let 𝐸 represents the kinetic energy of 𝛼 particle, 𝐸 = 12𝑚𝑣 2,so 𝑚 𝑄=𝐸 +1 𝑀 And , 𝐸= 𝑄 1+ 𝑚 𝑀 So, the kinetic energy of the 𝛼 particle emitted in the decay of 210 84Po is: 5.4 𝑀𝑒𝑉 𝐸= = 5.3 𝑀𝑒𝑉. 4 1+ 206 And the kinetic energy of the recoil nucleus is 0.1 MeV Isobaric Transitions • Beta emission A beta particle is a charged particle that is indistinguishable from an ordinary electron; it is ejected from the nucleus of an unstable radioactive atom whose neutron-to-proton ratio is too high. The particle has a single negative electric charge (−1.6 × 10−19 C) the beta particle is formed at the instant of emission by the transformation of a neutron into a proton and an electron according to the equation 1 0𝑛 + 11𝐻 + −10𝑒 For Beta (𝛽− ) emission to occur,, the exact nuclear mass of the parent must be greater The energy emitted by beta particle is : 𝑄 𝐸= 𝑚 1+ 𝑀 than the sum of the exact masses of the daughter nucleus plus the beta particle. 1.71 𝑀𝑒𝑉 𝐸= 1.000017 𝑀𝑃 = 𝑀𝑑 + 𝑀𝑒 + 𝑄 Example: Phosphorus decays 32 15𝑃 0 + 32 S + −1𝑒 + 1.71 MeV 15 The transformation energy is 1.71 MeV. 𝐸 =1.70997 MeV And the energy emitted by the recoil nucleus is 0.000029 MeV Positron Emission If the neutron-to-proton ratio is too low and alpha emission is not energetically possible, the nucleus may, under certain conditions, attain stability by emitting a positron, the positron is a positive beta particle 𝛽 + . Because of the fact that the nucleus loses a positive charge when a positron is emitted, the daughter product is one atomic number less than the parent. Example 22 11𝑁𝑎 → 22 10𝑁𝑒 + 01𝑒 + 𝜈 For positron emission, the following conservation equation must be satisfied: 𝑀𝑝 = 𝑀𝑝 + 𝑀+𝑒 + 𝑄. where Mp, Md, and M+e are the masses of the parent nucleus, daughter nucleus, and positron, respectively, and Q is the mass equivalent of the energy of the reaction. Since the daughter is one atomic number less than the parent, it must also lose an orbital electron, M−e, immediately after the nuclear transition. In terms of atomic masses, therefore, the conservation equation is: Mp = Md + M−e + M+e + Q. Orbital Electron Capture if a neutron-deficient atom is to attain stability by positron emission, it must exceed the weight of its daughter by at least two electron masses. If this requirement cannot be met, then the neutron deficiency is overcome by the process known as orbital electron capture or, alternatively, as electron capture or as K capture. In this radioactive transformation, one of the extra nuclear electrons is captured by the nucleus and unites with an intranuclear proton to form a neutron according to the equation 0 −1𝑒 + 11𝐻 → 10𝑛 + 𝜈 The energy conservation requirements for K capture are much less rigorous than for positron emission. It is merely required that the following conservation equation be satisfied: Mp + Me = Md + φ + Q, where Mp and Md are the atomic masses (not the atomic mass numbers) of the parent and daughter, Me is the mass of the captured electron, φ is the binding energy of the captured electron, and Q is the energy of the reaction Lecture3 Isobaric and Isomeric transitions Isobaric transitions (Given the atomic number of the parent nucleus is Z, - the daughter nucleus is Z + 1, if a beta particle is emitted, - or Z − 1, if a positron is emitted. Note : The atomic mass number of the daughter is same as that of the parent Isomeric transitions (The atomic number and the atomic mass number of the daughter is same as that of the parent. ◦ Gamma ray emission ◦Internal conversion Gamma ray - Gamma rays are monochromatic electromagnetic radiations - emitted from the nuclei of excited atoms following radioactive transformations. - The excited nuclei rides its excitation energy without affecting either the atomic number or the atomic mass number of the atom. - Internal conversion What is the difference in mechanism between the Gamma ray and internal conversion? Internal conversion may be thought of as an internal photoelectric effect, that is, an interaction in which the gamma ray collides with the tightly bound electron and transfers all of its energy to the electron. The energy of the gamma ray is divided into two parts: ✓The work done to overcome the binding energy of the electron. ✓The kinetic energy imparted to the electron. This may be expressed by the equation And can be expressed as following: 𝐸𝛾 = 𝐸𝑒 + 𝜙 𝐸𝛾 ∶is the energy of Gamma ray 𝐸𝑒 :is the kinetic energy of conversion electron 𝜙: is the binding energy of electron Example: 137Cs transformed by beta emission to 137mBa (metastable) It remains in the excited state for 3.7 minutes, then emits 0.661 Mev Gamma ray Half –life time It is The time required for any given radionuclide to decrease to one-half of its original This quantity is a measure of the speed with which it undergoes radioactive transformation From the definition of the half-life, it follows that the fraction of a radionuclide remaining after n halflives is given by the relationship 𝐴 1 = 𝑛 𝐴∘ 2 𝐴∘ :the origin quantity of activity A: the activity left after n half -lives example Cobalt-60, a gamma-emitting isotope of cobalt whose half-life is 5.3 years, is used as a radiation source for radiographing pipe welds. Because of the decrease in radioactivity with increasing time, the exposure time for a radiograph will be increased annually. Calculate the correction factor to be applied to the exposure time in order to account for the decrease in the strength of the source. Solution: 𝐴° = 2𝑛 𝐴 Take log for both sides, 𝐴° 𝑙𝑜𝑔 = 𝑛𝑙𝑜𝑔2 𝐴 where n, the number of 60Co halflives in 1 year, is 1/(5.3) = 0.189, 𝐴° 𝑙𝑜𝑔 = 0.189 × 𝐴 𝐴° So, =1.14 𝐴 0.301 = 0.0569 The ratio of the initial quantity of cobalt to the quantity remaining after 1 year is 1.14. The exposure time after 1 year, therefore, must be increased by 14%. It should be noted that this ratio is independent of the actual amount of activity at the beginning and end of the year. After the second year, the ratio of the cobalt at the beginning of the second year to that at the end will be 1.14. The same correction factor, 1.14, therefore, is applied every year to the exposure time for the previous year. The illustrative example given above could have been solved graphically with the aid of the curve in the above Figure. The ordinate at which the time in units of half-life, 0.189, intersects the curve shows that 87.7% of the original activity is left. The correction factor, therefore, is the reciprocal of 0.877: Correction factor 1 = 0.877 = 1.14 he fact that the graph of activity versus time, when drawn on semilog paper, is a straight line tells us that the quantity of activity left after any time interval is given by the following equation: 𝐴 = 𝐴° 𝑒 −𝜆𝑡 , where 𝐴° is the initial quantity of activity, A is the amount left after time t, λ is the transformation rate constant (also called the decay rate constant, or simply the decay constant), and e is the base of the system of natural logarithms. The transformation rate constant is the fractional decrease in activity per unit time and is defined as: Δ𝑁 Τ𝑁 lim ∆𝑡→0 ∆𝑡 = −𝜆, where N is a number of radioactive atoms and Δ𝑁 is the number of these atoms that are transformed during a time interval Δ𝑡. The fraction Δ𝑁Τ𝑁is the fractional decrease in the number of radioactive atoms during the time interval Δ𝑡. A negative sign is given to λ to indicate that the quantity N is decreasing with the time In this case: ΔN is 3.7 × 104, Δt is 1 s, and N, the number of radium atoms per microgram, may be calculated as follows: Example: A mass of 1-μg radium is found to emit 3.7 × 104 alpha particles per second. If each of these alphas represents a radioactive transformation of radium, what is the transformation rate constant for radium? 6.02 × 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙 𝑁= × 𝑊𝑔 𝐴 𝑔/𝑚𝑜𝑙 where A is the atomic weight and W is the weight of the radium sample 6.02 × 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙 −6 𝑁= × 10 𝑔 2.26 × 102 𝑔/𝑚𝑜𝑙 = 2.66 × 1015 𝑎𝑡𝑜𝑚𝑠 The transformation rate constant, therefore, is Δ𝑁Τ𝑁 3.4 × 104 𝑎𝑡𝑜𝑚𝑠/2.66 × 1015 𝑎𝑡𝑜𝑚𝑠 𝜆= = Δ𝑡 1 = 1.27 × 10−11 1𝑠 The average life time is defined simply as sum of the lifetimes of the individual atoms divided by the total number of atoms originally present. The instantaneous transformation rate of a quantity of radioisotope containing N atoms is λN. During the time interval between t and t + dt, the total number of transformations is λN dt. Each of the atoms that decayed during this interval, however, had existed for a total lifetime t since the beginning of observation on them. The sum of the lifetimes, therefore, of all the atoms that were transformed during the time interval between t and t + dt, after having survived since time t = 0, is tλN dt. The average life, τ , of the radioactive species is 1 ∞ 𝜏 = න 𝑡𝜆𝑁𝑑𝑡 𝑁° 0 where 𝑁° is the number of radioactive atoms in existence at time t = 0 𝑁 = 𝑁° 𝑒 −𝜆𝑡 , we have, ∞ 1 𝜏 = න 𝑡𝜆𝑁⋄ 𝑒 −𝜆𝑡 𝑑𝑡 𝑁⋄ 0 Integrate by parts, so we have 1 𝜏= 𝜆 If the expression for the transformation constant in terms of the half-life, 𝑇1 of the radioisotope , we have 2 0.693 𝜆= 𝑇1 2 the relationship between the half-life (𝑇1 ) and the mean life (𝜏) is found to be: 2 𝑇1 2 𝜏= = 1.44𝑇1 0.693 2 Activity Uranium-238 and its daughter 234Th each contain about the same number of atoms per gram—approximately 2.5 × 1021. Their half-lives, however, are greatly different; 238U has a half-life of 4.5 × 109 years while 234Th has a halflife of 24.1 days (or 6.63 × 10−2 year). Thorium-234, therefore, is transforming 6.8 × 1010 times faster than 238U. Also, Another example of greatly different rates of transformation that may be cited is 35S and 32P. These two radionuclides, Both have about the same number of atoms/g, but have half-lives of 87 and 14.3 days, respectively. The 32P, therefore, is decaying about 6 times faster than the 35S. In this context, therefore, 1/6 g of 32P is about equivalent to 1 g of 35S in radioactivity, while 15 mg of 234Th is equivalent in activity to 1 g of 238U. So, if the interest is radioactivity, the gram unit is not useful. To be meaningful, the unit for quantity of radioactivity must be based on the number of radioactive decays occurring within a certain time in the radioactive material. Activity: It is the number of decays within a given time. There are Two units are used to measure the activity: Becquerel(Bq), and is defined as follows The Becquerel is that quantity (SI) of radioactive material in which one atom is transformed per second 1Bq= 1tps= 1dps 1kilobecquerel (kBq) = 103 Bq 1megabecquerel (MBq) = 106 Bq 1gigabecquerel (GBq) = 109 Bq 1terabecquerel (TBq) = 1012 Bq. The Curie Specific Activity: The curie (Ci), is the unit for quantity of It is defined as The concentration radioactivity that was used before the adoption of radioactivity, or the of the SI units and the becquerel. The curie, relationship between the mass of which originally was defined as the activity of 1 g radioactive material and the of 226Ra, is now more explicitly defined as: activity. curie is the activity of that quantity of Specific activity (SA)is the radioactive material in which 3.7 × 1010 atoms are transformed in one second. number of becquerels (or curies) per unit mass or volume. The curie is related to the Becquerel by 1 Ci = 3.7 × 1010 Bq. 1 millicurie (mCi) = 10−3 Ci 1 microcurie (μCi) = 10−6 Ci 1 nanocurie (nCi) = 10−9 Ci 1 picocurie (pCi) = 10−12 Ci 1 femtocurie (fCi) = 10−15 Ci. 𝑆𝐴 = 𝜆𝑁 = 𝜆×60.2×1023 𝐵𝑞 𝐴 𝑔 A is the atomic weight of the nuclide With the half-life time, we have 0.693 6.02 × 1023 𝑆𝐴 = × 𝑇1 𝐴 2 𝑆𝐴 = 4.18×1023 𝐵𝑞 𝐴×𝑇1 𝑔 2 X-ray production • X-rays were discovered by Roentgen in 1895 while studying cathode rays in a gas discharge tube. • He observed that another type of radiation was produced that could be detected outside the tube. • This radiation could penetrate opaque substances, produce fluorescence, blacken a photographic plate, and ionize a gas. He named the new radiation x-rays. • Later, this classified as one form of electromagnetic radiation The above figure is a schematic representation of a conventional x-ray tube. The tube consists of a glass envelope which has been evacuated to high vacuum. At one end is a cathode (negative electrode) and at the other an anode (positive electrode), both are sealed in the tube. The cathode is a tungsten filament which when heated emits electrons, a phenomenon known as thermionic emission. The anode consists of a thick copper rod, at the end of which is placed a small piece of tungsten target. When a high voltage is applied between the anode and the cathode, the electrons emitted from the filament are accelerated toward the anode and achieve high velocities before striking the target. The x-rays are produced by the sudden deflection or acceleration of the electron caused by the attractive force of the tungsten nucleus. The x-ray beam emerges through a thin glass window in the tube envelope. PHYSICS OF X-RAY PRODUCTION There are two different mechanisms by which x-rays are produced. One gives rise to bremsstrahlung x-rays and the other characteristic x-rays: the process of bremsstrahlung (braking radiation) is the result of radiative “collision” between a high-speed electron and a nucleus. The electron while passing near a nucleus may be deflected from its path by the action of Coulomb forces of attraction and lose energy as bremsstrahlung, a phenomenon predicted by Maxwell’s general theory of electromagnetic radiation. According to this theory, energy is propagated through space by electromagnetic fields. As the electron, with its associated electromagnetic field, passes in the vicinity of a nucleus, it suffers a sudden deflection and acceleration. As a result, a part or all of its energy is dissociated from it and propagates in space as electromagnetic radiation. The mechanism of bremsstrahlung production is illustrated in Characteristic x-rays Electrons incident on the target produce characteristic x-rays. The mechanism of their production is illustrated in the figure. An electron, with kinetic energy E0 , may interact with the atoms of the target by ejecting an orbital electron, such as a K, L, or M electron, leaving the atom ionized. The original electron will recede from the collision with energy E0 – ΔE, where ΔE is the energy given to the orbital electron. A part of ΔE is spent in overcoming the binding energy of the electron and the rest is carried by the ejected electron. When a vacancy is created in an orbit, an outer orbital electron will fall down to fill that vacancy. In so doing, the energy is radiated in the form of electromagnetic radiation. This is called characteristic radiation, i.e., characteristic of the atoms in the target and of the shells between which the transitions took place. With higher atomic number targets and the transitions involving inner shells such as K and L, the characteristic radiations emitted are of energies high enough to be considered in the x-ray part of the electromagnetic spectrum. the major characteristic radiation energies produced in a tungsten target is given in the table. It should be noted that, unlike bremsstrahlung, characteristic x-rays are emitted at discrete energies. If the transition involved an electron descending from the L shell to the K shell, then the photon emitted will have energy hv = EK – EL, where EK and EL are the electron-binding energies of the K shell and the L shell, respectively. The threshold energy that an incident electron must possess in order to first strip an electron from the atom is called critical absorption energy. These energies for some elements are given in Table 3.2. X-RAY ENERGY SPECTRA X-ray photons produced by an x-ray machine are heterogeneous in energy. The energy spectrum shows a continuous distribution of energies for the bremsstrahlung photons superimposed by characteristic radiation of discrete energies. A typical spectral distribution is shown in the figure Kramer’s equation : IE = KZ (Em – E) Kramer’s equation: IE = KZ (Em – E) where IE is the intensity of photons with energy E, Z is the atomic number of the target, Em is the maximum photon energy, and K is a constant KEY POINTS The x-ray tube: ▪ X-ray tube is highly evacuated to prevent electron interactions with air ▪ Choice of tungsten for filament (cathode) and target (anode) is based on its having a high melting point (3,370°C) and a high atomic number (Z = 74), which is needed to boost the efficiency of x-ray production. ▪ Heat generated in the target must be removed to prevent target damage, e.g., using a copper anode to conduct heat away, a rotating anode, fans, and an oil bath around the tube. The function of the oil bath is to provide electrical insulation as well as heat absorption. ▪ The function of a hooded anode (tungsten + copper shield around target) is to prevent stray electrons from striking the nontarget components of the tube and absorbing bremsstrahlung as a result of their interactions. ➢ Apparent focal spot size a is given as follows: a = A sin , where A is the side of actual focal spot presented at an angle with respect to the perpendicular to the direction of the electron beam (Fig. 3.2). The apparent focal spot size ranges from 0.1 × 0.1 to 2 × 2 mm2 for imaging, and 5 × 5 to 7 × 7 mm2 for orthovoltage therapy tubes. ➢ Peak voltage on an x-ray tube = √2 · line voltage · transformer turn ratio. ➢ Rectifiers conduct electrons in one direction only and can withstand reverse voltage up to a certain magnitude. Full-wave rectification increases effective tube current> ➢ X-ray output per mAs can be substantially increased by applying three-phase power to the x-ray tube. A three-phase, six-pulse generator delivers high-voltage pulses with a voltage ripple of 13% to 25%. ➢ A three-phase, 12-pulse generator is capable of providing high-voltage pulses to the x-ray tube with much less ripple (3% to 10%). ➢ • A high-frequency generator provides nearly constant high-voltage potential (with a ripple of less than 2%). Consequently, it generates higher x-ray output per mAs and shorter exposure times. X-ray production: ❖ X-rays are produced by two different mechanisms: bremsstrahlung and characteristic x-ray emission. ❖ Bremsstrahlung x-rays have a spectrum of energies. The maximum energy is numerically equal to the peak voltage. ❖ Average energy is about one-third of the maximum energy. ❖ Characteristic x-rays have discrete energies, corresponding to the energy level difference between shells involved in three electrons transition. ❖ The higher the energy of electrons bombarding the target, the more forward the direction of x-ray emission. ❖ The efficiency of x-ray production is proportional to the atomic number Z of the target and the voltage applied to the tube. The efficiency is less than 1% for x-ray tubes operating at 100 kVp (99% of input energy is converted into heat). The efficiency improves considerably for high-energy accelerator beams (30% to 95%, depending upon energy). Operating characteristics: ❖ Output (exposure rate) of an x-ray machine is very sensitive to the filament current. The output increases proportionally with tube current and approximately with the square of the voltage.