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1- 25-10-1434 H 2-11-1434 H 3- 9-11-1434 H 4- 16-11-1434 H 5- 23-11-1434 H 6- 26-11-1434 H 7- 1—12--1434 H أجازة (8- 8-12-1434 - أجازة -9- 15-12-1434 10-

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Presentation on theme: "1- 25-10-1434 H 2-11-1434 H 3- 9-11-1434 H 4- 16-11-1434 H 5- 23-11-1434 H 6- 26-11-1434 H 7- 1—12--1434 H أجازة (8- 8-12-1434 - أجازة -9- 15-12-1434 10-"— Presentation transcript:

1 H H H H H H 7- 1— H أجازة ( أجازة H H H H ( (, 40% boys) ( Hfinal exam H عملى 20%

2 دورى1 15% دورى2 15»% بحث ومقال 10% حضور 5%؟ واجبات 5% عملى 20% امتحان نهائى 40%

3 Course Description, Ch2: Fundamental Scintifiuc. Chapter 3-1 –A Radiation Sources 1- Radioactivity 2- Transformation Mechanisms Ch 3-2 Transformation Kinetics.Chapter 3-3 –B Activity 1- Naturally Occuring Radiation Chapter 3-4 Serial Transformation Chapter 3-5, Machine Sources 1- Production of X Rays 2- Production of Accelerated Charged Particle 3- Production of Linear Accelerator Radiation Chapter 5 Interaction of Radiation with Matter, 1- Beta Particles 2- Alpha Particles 3- Gamma Rays 4- Neutrons

4 c Excess Chapter 6 Radiation Quantities and Units Chapter 7 Radiation Protection in Medicine

5 Radioactivity CH3, Cember, 2003 Charged particles:: alpha beta and positron emission, orbital electron capture Rays: gamma and x-rays

6 Transformation Mechanism Alpha emission: Z number of Daughter nuclides decrease by two and atomic mass number decrease by four compared of parent nuclides. Isobaric transition: Given the atomic number of parent nucleus is Z, that of the daughter nucleus is z+1, if a beta particle is emitted or Z-1, If a 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 Transition: The atomic number and the atomic mass number of the Daughter is same as that of the parent. Gamma ray emission Internal conversion \

7 Radioactive Decay Processes The result of an Alpha emission is a daughter with atomic number less by two and mass number less by four compared to parent; e.g.: Alpha (α) emission: Nature of Alpha (α) Particle: It is a highly energetic Helium Nucleus that is emitted from the nucleus of the radioactive isotope when the neutron to proton ratio is too low. It is positively charged, massive particle consisting of two protons and two neutrons. In this example the 210 Po has a neutron to proton ratio of 126/84 = 1.5; after decaying by α emission, a stable daughter nucleus 206 Pb is formed. The new ratio is now 124/82 = (i.e. increased).

8 With the exception of Sm (samarium), naturally occurring alpha emitters usually have atomic number greater than 82. Notice that, 210 Po is a pure alpha emitter, i.e. no other particles or gamma rays accompany the emission of alpha particle. Conditions for α-Particle emission: There are two main reasons for this fact: Electrostatic repulsive forces in heavy nuclei increase rapidly than the cohesive nuclear forces and its magnitude may exceed that of cohesive forces. The emitted particle must have enough energy to overcome the high potential barrier (about 25 MeV) at the surface of the nucleus resulting from the presence of positively charged nucleons. In addition to a minimum of 3.8 MeV of kinetic energy. 1 2

9 According to quantum mechanical theory, an alpha particle may escape from the potential well by tunnelling through the potential barrier. For alpha emission to occur, the following conservation equation must be Satisfied: The two orbital electrons are lost during transition to lower atomic number daughter, while Q, is the total energy release associated with the radioactive transformation.

10 Example: Calculate the total energy release for the given alpha decay, Knowing that M Po = , M Pb = , M α = and M e = Solution: Q = M Po - M Pb – M α – 2 x = amu In energy units, Q = x 931 (MeV/amu) = 5.4 MeV Notice that since, there is no gamma rays emitted in the last reaction, the released energy appears as kinetic energy for alpha particle. Notice that the value Q = 5.4 MeV is greater than 3.8 MeV, the condition for α- emission. Example: Calculate the total energy release for the given alpha decay, Knowing that M Po = , M Pb = , M α = and M e = Solution: Q = M Po - M Pb – M α – 2 x = amu In energy units, Q = x 931 (MeV/amu) = 5.4 MeV

11 The dead outer layer of skin is sufficiently thick to absorb all alpha radiations. Thus alpha radiations from outside the body are not considered a radiation hazard. Alpha-emitting isotopes are extremely toxic when irradiating from inside the body, where, the energy of alpha particle is dissipated directly in living tissues. Hazard of alpha radiation:

12 Division of energy between alpha-particle and recoil nucleus: The exact energy division between alpha particle and recoil nucleus depends on the ratio of their masses (m/M) and can be calculated from laws of conservation of energy and momentum. The total energy release associated with the radioactive transformation Q is given by: According to conservation of momentum: Substitute for V in the equation for total energy release : Let E represent the K.E. of Alpha-particle you will get: Q = ½ mv2(m/M +1) Q= E (m/M +1)

13 Example: Calculate the kinetic energy of the alpha particle and recoil nucleus ( 206 Pb) emitted in the decay of 210 Po. Solution: The kinetic energy of the emitted alpha particle is given by: Substitute for the values of Q, m and M: The kinetic energy of recoil nucleus is therefore:

14 Beta (β) emission: Nature of beta (β) Particle: It is an ordinary electron ejected from the nucleus of beta-unstable radioactive atom. It has a single negative electrical charge (1.6 x C) and a very small mass amu. How is (β) Particle formed? It is formed at the instant of emission, where a neutron is transformed to a proton and electron: The transformation shows that beta decay takes place among isotopes having surplus of neutrons. Conditions for β-Particle emission: The exact nuclear mass of the parent must be greater than the mass of daughter nucleus and beta particle. Thus:

15 Example: The daughter nucleus is one atomic number higher than parent. The transformation energy (in this case 1.71 MeV) is equal to the difference in mass between 32 P and the sum of masses of 32 S and β-particle. And appears as kinetic energy of beta particle Example: Calculate the total energy release for the previous beta decay, Knowing that M P = , M S = and M e = Solution: We have, When neutral atomic masses are used, the mass of electron is not considered since it is included in the extra-nuclear electronic structure of 32 S, thus: Q = M P - M d = – = amu In energy units, Q = x 931 (MeV/amu) = 1.71 MeV

16 Energy spectrum of emitted beta-particles: As an example, the given figure shows the beta ray energy spectrum of 32 P decay. The energy spectrum of beta particles is characterized by: Most of beta particles have energies less than the maximum K.E. The average energy for beta- radioactive isotopes is about 30-40% of maximum energy. For 32 P, it is 41% of maximum energy = 0.7 MeV. When we mention the energy of beta particles, we mean the maximum energy. The reason that beta particles are not monoenergetic is that their emission is accompanied by a neutrino which energy is equal to the difference between the maximum energy and the energy of certain beta particle

17 Types of beta emitters Pure beta emitter beta-gamma emitter Emits no gamma rays, e.g. 32 P, 3 H, 14 C, and 90 Sr Emits beta particles, followed instantaneously by gamma rays, e.g. 203 Hg, 42 K, and 131 I. The daughter nucleus is left in an excited state and gets rid of excess energy by emission of gamma rays. Types of beta decay emits only one group of beta rays, e.g. 32 P & 203 Hg Simple emits more than one group of beta rays, e.g. 42 K & 131 I Complex


19 Hazard of beta radiation It is able of penetrating tissues depending on its energy, thus it may be an external hazard The exact degree of hazard depends on the beta-emitting isotope activity & beta energy spectrum Beta rays with energies less than 200 keV have very limited penetrability and are not considered an external hazard, e.g. Tritium 3 H, 35 S & 14 C Any beta particle deposited internally is a hazard unless in amounts thought to be safe Beta rays can produce highly penetrating x-rays (bremsstrahlung) when stopped by shielding. A proper shield must attenuate emitted x-rays as well

20 Nature of positrons It is a beta particle whose charge is positive. ejected from the nucleus of beta- unstable radioactive atom. It has a single positive electrical charge (+1.6 x C) and a very small mass amu. It is ejected from the nucleus which neutron to proton ratio is too low and alpha emission is not energetically possible. After positron emission the daughter nucleus is one atomic number less than the parent. The mass number remains unchanged It results from a transformation within the nucleus of a proton into neutron Example

21 Positron’s life time: Unlike electrons, positrons do not occur freely in nature, they have only a transitory existence. They appear in nature only as a result of the interaction of cosmic rays and the atmosphere, and disappear in the matter of microseconds after formation. Annihilation: The positron combines with an electron and the two particles are annihilated, giving rise to two gamma-ray photons whose energies are equivalent to the mass of positron and electron. Conditions for positron emission For positron emission to take place the following conservation equation must be satisfied: Since the daughter is one atomic number less than the parent, it must also lose an orbital electron immediately after the nuclear transformation.

22 The decay scheme of 22 Na: 22 Na is a useful isotope for biomedical research. It decays to 22 Ne by two competing mechanisms: β + emission 89.8% K (electron) capture 10.2% Notice that both modes of decay result in 22 Ne in an excited state of energy MeV which instantly appears as γ-rays. Calculation of the exact atomic mass of Neon:

23 Since positrons are electrons they have the same hazard as electrons. All positron emitters are considered external radiation hazard as they all emit gamma rays. Hazard of positrons: Orbital electron capture “orbital electron capture” or “k capture”. One of the extra-nuclear electrons is captured by the nucleus, units with an intra-nuclear proton and forms a neutron: The probability that the captured electron be from the k shell is much greater than any other shell, that is why the process is usually called k capture.

24 In k capture, as in positron emission, the atomic number of the daughter is one less than parent, while the atomic mass remains unchanged. Energy conservation requirements for electron capture: For electron capture to happen the following conservation equation must be satisfied: Example: Calculate the total energy of the decay of 22 Na by electron capture into 22 Ne, Knowing that M Na = , M Ne = , and M e = knowing that the binding energy of Sodium K electron is 1.08 keV. Solution:

25 In the last example, if we subtract the energy of the emitted gamma rays (1.277 MeV) from the total energy Q (3.352 MeV), the difference is MeV. Only very small part of this energy is taken by the recoil nucleus, the rest of it is emitted in the form of neutrino. Rule: In all types of radioactive decay involving either the capture or emission of an electron, a neutrino must be emitted in order to conserve energy. In contrast to positron and negatron (ordinary beta) decay, in which the neutrino carries off the difference in actual K.E. of the particle and the maximum observed kinetic energy (i.e. continuous energy distribution ), the neutrino in orbital electron capture is monoenergetic. *Simply because there is no ejected particle other than neutrino.

26 Auger Electrons: They are monoenergetic electrons which are sometimes observed when the x-ray yield (fluorescent radiation) accompanying electron capture is absent or reduced. While escaping from the atom, x-rays may interact with one of the L or M electrons and eject it from the atom. The energy of Auger electron is equal to the difference in energy between the x-ray energy and L shell energy (E k -E L ). X-ray emission in case of electron capture: The vacancy in the k-shell of the daughter nucleus after electron capture is filled by an orbital electron from a higher energy level. This results in the emission of x-rays characteristic of the daughter nucleus. In the medical field, if isotopes which decay by electron capture are used internally, the dose due to such x-rays should be accounted for.

27 Gamma Rays: They are monoenergetic electromagnetic radiation that are emitted from the nuclei of excited atoms following radioactive transformation. They provide a mechanism for ridding excited nuclei from their excitation energy. Internal conversion: It is a mechanism by which an excited nucleus of gamma-emitting radioisotope may rid itself of the excitation energy. A tightly bound electron interacts with the nucleus, takes its excitation energy (which is essentially emitted in the form of gamma rays) and is ejected from the atom. Internally converted electrons appear in monoenergetic groups. The energy of internally converted electron is equal to the difference in energy between the gamma ray photon emitted by the radioisotope and the binding energy of the electron. E e = E γ - φ. ( Explain the phenomena of breast Imaging, Mamogram using Emission of gamma rays). One can say that the gamma photon emitted from the nucleus collides with the tightly bound electron and transfers all of its energy to the electron. Emission of gamma rays:

28 Example: Cesium Cs is transformed by beta emission to an excited state of Barium 137m 56 Ba. Barium then emits MeV photon which undergoes internal conversion in 11% of the transitions. The conversion electrons are found superimposed on the beta spectrum of Cs-137. After internal conversion, characteristic x-rays are emitted as outer orbital electrons fill the vacancies left in the deeper energy levels by the conversion of electrons. These x-rays may themselves interact with orbital electrons causing the ejection of Auger electrons.

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