1. • Three words have been combined to make this grid of letters.• How many times does each of these words appear?• Try to compare your performance while searching for just one word vs. two of them at the same time?• How many times is the word SUN shown? How many times is the word BUS shown? How many times is the word NONE shown?
2. • If you get all, It only means that you are a very focus person but if NOT…here are the reasons…• Dividing attention results in less attention power devoted to all the different tasks that you are trying to do at the same time.• The more tasks, the less attention can be devoted to each.• The result is more errors and waste of time. Although we all have the feeling that multi-tasking saves us time, it is often not the case.
3. Electron -by J. J. Thomson, 1897 -symbol “e or e -”-relative electrical charge : - 1-atomic mass unit = 5. 484 x 10 -4 -location : outside the nucleus
4. Proton - by Eugen Goldstein, 1886 -symbol “p or p +” -relative electrical charge : + 1 -atomic mass unit = 1-Location : inside the nucleus
5. Neutron by - James Chadwick, 1932 -symbol “n or n ” 0-relative electrical charge : 0-atomic mass unit = 1-Location : inside the nucleus
6. Quarks -fundamental particles of proton and neutron -found inside the proton and neutron-held together by “gluons”
7. Proton is a color combination of threecolored quarks. Quarks are bound together by the exchange of color gluons.Emission or absorption of agluon causes the quarks tomake a transition from one color to another.
8. There are six types of quarks (up, down, charm, strange, top, andbottom). The lightest quarks — calledup and down — are the most common.
9. Quarks -fundamental particles of proton and neutron -found inside the proton and neutron-held together by “gluons”
12. TOTAL # OF PROTONS & NEUTRONSTOTAL # OF PROTONS IN AN ATOM IN THE ATOM’S NUCLEUS
13. Atomic Number -# of protons in the nucleus-symbol Z, determines identity of an element.-equal to the # of protons, w/c is equal tothe # of electrons in an uncharged atom. Z = number of p = number of e + -
14. Mass Number - symbol A in elementalnotation, consists of the total# of protons and neutrons in the nucleus of the atom. A = number of p + + number of n0
15. ELEMENT ATOMIC MASS NUMBER OF NUMBER OF NUMBER OF NUMBER NUMBER PROTONS (p+) ELECTRONS (e-) NEUTRONS (n) a 20 40 b 84 48 c 82 125 d 52 76 e 108 47ELEMENT NAME OF ELEMENT COMPLETE DESIGNATION OF ELEMENT a b c d e
16. ELEMENT SYMBOL MASS NUMBER ATOMIC NUMBER NUMBER NUMBER OF NUMBER OF OF NEUTRONS PROTONS ELECTRONSSodium 11 15 35 ZnBarium How many protons, neutrons, and electrons are present in (a) 3 (b) 79 (C) 27 H Se Al 1 34 13
17. Isotopes-Atoms of an element with the same atomic # but different mass #-different mass NEUTRONS numbers butidentical atomic numbers.
18. NUMBER OF NUMBER OF NUMBER OFISOTOPE PROTONS ELECTRONS NEUTRONS 35 Cl 17 17 17 37 Cl 17 17 17 28 Si 14 29 Si 14 30 Si 14
19. ELEMENT SYMBOL MASS ATOMIC ISOTOPIC PERCENTAGE NUMBER MASS (amu) MASS ABUNDANCEHYDROGEN H 1 1.00794 1.007 8 99.985% D 2 2.014 1 0.015% T 3 3.016 1 0% BORON B 10 10.811 10.012 9 19.91% 11 11.009 8 80.09% OXYGEN O 16 15.9994 15.994 9 99.759% 17 16.999 3 0.037% 18 17. 999 2 0.204%NITROGEN N 14 14.00674 14.003 1 99.63% 15 15.000 1 0.37%MAGNESIUM Mg 24 24.305 23.985 0 78.99% 25 24.985 8 10.00% 26 25. 985 6 11.01% CHLORINE 35 35 35. 45 34.969 75.53% 37 37 36.966 24.47%
20. • SOLUTION: Step #1: Multiply the atomic mass of each isotope by its percentage abundance. Remember to convert the value to decimal equivalent. 34.969 x 0.7553 = 26.41 amu 36.996 x 0.2447 = 9.053 amu Step#2: Add the products obtained to get the relative atomic mass. 26.41 + 9.053 = 35.46 amu
22. - Radiation that carries more than 1216 kJ/mol of energy.• e.g. UVB rays (higher end of the UV spectrum), x- rays, gamma rays, cosmic rays. - Radiation that carries less than 1216 kJ/mol of energy.• e.g. radiowaves, microwaves, infrared, visible light, UVA rays (lower end of the UV spectrum).
23. •ALPHA PARTICLES•BETA PARTICLES•GAMMA RAYS
24. • Contains two protons and two neutrons, which gives it a mass number of 4 and atomic number of 2.• Because of two protons, an alpha particle has a charge of 2+ that makes it identical to Helium nucleus.
25. • Is identical to an electron, has a charge of 1- and mass number of (0) zero.• Βeta particles are produced by unstable nuclei when neutrons are change into protons.
26. • GAMMA RAYS are high- energy radiation released as an unstable nucleus undergoes a rearrangement to give a more stable, lower-energy nucleus.• Since gamma rays are energy only, there is NO mass or charged associated with their symbols.
27. TYPE OF MASS ATOMIC SYMBOL NUMBER NUMBER CHARGERADIATION ALPHA 4 2 2+ PARTICLE BETA 0 0 1- PARTICLE GAMMA 0 0 0 RAY PROTON 1 1 1+NEUTRON 1 0 0POSITRON 0 1 1+
28. DISTANCE PARTICLE TRAVELS THROUGH INTO TYPE SYMBOL SHIELDING AIR TISSUE Paper,Alpha α 2 – 4 cm 0.05 mm clothing Heavy clothing,Beta β 200 –300 cm 4 – 5 mm lab coats, gloves Lead,Gamma γ 500 cm 50 mm concrete
29. Keep your distance! The greater the distance from the radioactive source, the lower the intensity of radiation received. If you double your distance from the radiation source, the intensity of radiation drops to (1/2)2 or one-fourth of its previous value.DISTANCE FROM THE SOURCE 2m 1mINTENSITY OF RADIATION (1/2)2 = ¼ 1
30. Process wherein the nucleus spontaneously breaks down by emitting radiation.Radioactive nucleus New nucleus + Radiation (α,β,γ)NOTE: N.E. is balanced when the sum of the mass #s and thesum of the atomic #s of the particles and the atoms on one side ofthe equation are equal to their counterparts on the other side.
31. • ALPHA emitters are radioisotopes that decay by emitting alpha particles. • EXAMPLE: - uranium-238 decays to thorium-234 by emitting alpha particles. +• NOTE: the ALPHA particle emitted contains 2 protons, which gives the new nucleus 2 fewer protons, or 90 protons. That means that the new nucleus has an atomic # of 90 and is therefore thorium (Th). Since the alpha particle has a mass # of 4, the mass # of the thorium isotope is 234, 4 less than of the original uranium nucleus.
32. EXAMPLE: COMPLETE THE NUCLEAR EQUATION - radium-226 emits alpha particles to form a new isotope. Determine the mass #, atomic # and the new isotope form. +• SOLUTION: the new isotope is RADON-222• 226 – 4 = 222 (mass number of the new isotope)• 88 – 2 = 86 (atomic number of the new isotope)
33. EXAMPLE: COMPLETE THE NUCLEAR EQUATION - radon-222 emits alpha particles to form a new isotope. Determine the mass #, atomic # and the new isotope form. +• SOLUTION: the new isotope is POLONIUM-218• 222 – 4 = 218 (mass number of the new isotope)• 86 – 2 = 84 (atomic number of the new isotope)
34. • BETA emitters is a radioisotope that decays by emitting beta particles.• EXAMPLE: - carbon-14 decays to nitrogen isotope by emitting beta particles. +• NOTE: the newly form protons adds to the number of protons already in the nucleus and increases the atomic number by 1. However, the mass number of the newly formed nucleus stays the same.
35. EXAMPLE: COMPLETE THE NUCLEAR EQUATION - cobalt-60, a radioisotope used in the treatment of cancer decays by emitting a beta particle. Write the nuclear equation for its decay. +• SOLUTION: the new isotope is NICKEL• 27 + 1 = 28 (atomic number of the new isotope)• 60 (mass number of the new isotope)
36. EXAMPLE: COMPLETE THE NUCLEAR EQUATION - iodine-131, a beta emitter, is used to check thyroid function and to treat hyperthyroidism. Write its nuclear equation. +• SOLUTION: the new isotope is XENON• 53 + 1 = 54 (atomic number of the new isotope)• 131 (mass number of the new isotope)
37. • There are very few pure GAMMA emitters, although gamma radiation accompanies most alpha and beta radiation.• EXAMPLE:- unstable form of technetium-99 most commonly used gamma emitter by emitting gamma rays the unstable nucleus becomes stable. Nuclear equation for Tc-99m. +• NOTE: (m) state or metastable means - a high-energy excited stage by emitting energy in the from of gamma rays, the nucleus becomes stable.
38. • The time it takes for one-half of a radioactive sample to decay.• EXAMPLE:- iodine-131, a radioactive isotope of iodine used in diagnosis and treatment of thyroid disorders, has a half-life of 8 days. If we began with sample containing 1000 atoms of iodine-131, there would be 500 atoms remaining after 8 days and so on… TIME ELAPSED 0 8 DAYS 16 DAYS 24 DAYS # of half-lives 0 1 2 3 elapsed Quantity of (I-131) 1000 atoms 500 atoms 250 atoms 125 atoms remaining
40. TYPES OF ELEMENT RADIOISOTOPES HALF-LIFE RADIATIONNATURALLY OCCURINGRADIOISOTOPES CARBON 14 C 5730 yrs. β POTASSIUM 40K 1.3 X 109 yrs. β,γ RADIUM 226Ra 1600 yrs. α,γ URANIUM 238U 4.5 X 109 yrs. α,γMEDICAL RADIOISOTOPES CARBON 11 C 20 min β+ CHROMIUM 51Cr 28 days γ IODINE 131I 8 days β,γ IODINE 125I 60 days γ IRON 59Fe 46 days β,γ
41. TYPES OF ELEMENT RADIOISOTOPES HALF-LIFE RADIATION MEDICAL RADIOISOTOPES PHOSPOROUS 32P 14 days β OXYGEN 15O 2 min β+ POTASSIUM 42K 12 hours β,γ SODIUM 24Na 15 hours β,γ STRONTIUM 25Sr 64 days γ TECHNETIUM 99mTc 6.0 hours γ• NOTE: technetium-99m emits half-life of its radiation in its 6 hr. This means that a small amount of the radioisotopes given to patient is essentially gone within 2 days. The decay products of technetium-99m are totally eliminated by the body.
42. • Nitrogen-13, which has a half-life of 10 min. is used to manage organs in the body. For diagnostic procedure the patient receives an injection of a compound containing radioisotopes. Originally, the nitrogen-13 has an activity of 40 microcuries (μCi). If the procedure requires 30 min, what is the remaining activity of the radioisotopes?
43. SOLUTION: 1 half-life Number of half-lives = 30 min X 10 min =3 The activity of the radioisotopes in 3 half-lives is: 40 μCi 10 min 20 μCi 10 min 10 μCi 10 min 5 μCiNOTE: Another way to calculate the activity of radioactive nitrogen-13 left in sample is to construct a chart to show the number of half-lives, elapsed time, and the amount of radioactive isotope that is leftin the sample. Time elapsed 0 10 min 20 min 30 min Number of half-lives elapsed 0 1 2 3 Activity of N-13 remaining 40 μCi 20 μCi 10 μCi 2μCi
44. • In Los Angeles, the remains of ancient animals have been unearthed at the La Brea tar pit. Suppose a bone sample from the tar pits is subjected to the carbon-14 dating method. If the sample shows about two half-lives have passed, about when did the animal live in the tar pits?
45. SOLUTION: (half-life of carbon-14 = 5730 1 half-life 5730 yrs. 2 half-lives X 1 half- life = 11, 000 yearsNOTE: We would estimate that the animal lived inthe tar pits about 11, 000 years ago, or about 9000B.C.
46. • Iron-59, used in the determination of bone marrow function, has a half-life of 46 days. If the laboratory receives a sample of 8.0 g of iron- 59, how many grams are still active after 184 days? ANSWER : 0.50 g
47. MEASUREMENT UNIT MEANINGACTIVITY CURIE (Ci) 3.7 X 1010 disintegrations/sABSORBED DOSE Rad 10-5 J/gBIOLOGICAL DAMAGE TO HUMANS Rem Rad X RBENOTE: RADIOISOTOPE ACTIVITYThe activity of sample is measured in terms of the number ofdisintegrations or nuclear transformations produced by the sampleper second. The curie (Ci) is the unit used to express nucleardisintegration. The curie was named for Marie Curie who discoveredradioactive elements radium and polonium together with herhusband Pierre curie. 1 curie = 3.7 X 1010 disintegrations/s
48. MEASUREMENT UNIT MEANINGACTIVITY CURIE (Ci) 3.7 X 1010 disintegrations/sABSORBED DOSE Rad 10-5 J/gBIOLOGICAL DAMAGE TO HUMANS Rem Rad X RBENOTE: RADIATION ABSORBED DOSEThe rad (for radiation absorbed dose) is a unit that measures theamount of radiation absorbed by a gram of material such as bodytissue. One rad is the absorption of 10-5 J of energy per gram oftissue. (1 cal = 4.18 J) 1rad = 10-5 J/g
49. MEASUREMENT UNIT MEANINGACTIVITY CURIE (Ci) 3.7 X 1010 disintegrations/sABSORBED DOSE Rad 10-5 J/gBIOLOGICAL DAMAGE TO HUMANS Rem Rad X RBENOTE: RADIATION EQUIVALENT IN HUMANSThe rem (for radiation equivalent in humans) is a unit that measures thebiological damage caused by the various kinds of radiation. The rem considersthe biological effects of alpha, beta and gamma radiation on tissue are not thesame. The alpha particles reach the tissues, they can cause more ionization andtherefore more damage than do beta particles and gamma rays. Radiationbiological effectiveness value of gamma = 1; beta = 10; alpha = 20 Rem = Rad X RBE
50. • In the treatment for leukemia, phosphorus-32, which has an activity of 2 millicuries (mCi), is used. If phosphorus-32 is a beta emmiter, how many beta particles are emitted in 1s?
51. SOLUTION: 1 Ci = 3.7 X1010 disintegrations/s 1 Ci 3.7 X1010 β particles 2 mCi X X1s 1000 mCi s Ci = 7.4 X107 beta particlesNOTE:We calculate the number of beta particles from aradioisotope’s activity. Since 1 Ci is 3.7 X 1010disintegrations/s, there must be 3.7 X 1010 beta particlesproduced in a second.
52. • The larger the dose of radiation received at one time, the greater the effect on the body. Exposure to radiation under 25 rem usually cannot be detected. Whole body exposure of 100 rem produces a temporary decrease in the number of white blood cells. If the exposure to radiation is 100 rem higher, the person suffers the symptoms of radiation sickness: nausea, vommiting, fatigue, and a reduction in white blood cells count. A whole- body dosage greater than 300 rem can lower the whote blood cell count to zero. The patient suffers diarrhea, hair loss and infection.
53. SOURCE DOSE (mrem)NATURAL The ground 15 Air, water, food 30 Cosmic rays 40 Wood, concrete, brick 50MEDICAL Chest x-ray 50 Dental x-ray 20 Upper gastrointestinal tract x-ray 200OTHER Television 2 Air travel 1 Global fallout 2 Cigarette smoking 35
54. Lethal Doses of Radiation for Some Life-Forms Life – Form LD50 (rem) Insect 100, 000 Bacterium 50, 000 Rat 800 Human 500 Dog 300NOTE:Exposure to radiation of about 500 rem is expected tocause death in 50% of the people receiving that dose. Thisamount of radiation is called LETHAL DOSE for one-half thepopulation, or LD50. Radiation of about 600 rem would befatal to all humans within a few weeks.
55. ELEMENT RADIOISOTOPE MEDICAL USECHROMIUM 51 Cr Spleen imaging, blood volume,TECHNETIUM 99mTc Brain, Lung, Liver, Spleen, Bone and bone marrow scans GALLIUM 67Ga Treatment of lymphomasPHOSPHORUS 32P Treatment of leukemia, polycythemia vera, and lymphomas; detection of brain and breast tumors SODIUM 24Na Vascular disease, extra cellular and blood volumeSTRONTIUM 85Sr Bone imaging for diagnosis of bone damage and disease IODINE 125I Thyroid imaging; plasma volume, fat absorbtion IODINE 131I Study of thyroid; treatment of thyroid conditions such as hyperthyrodism
56. RADIATION DOSE USED FOR DIAGNOSTIC PROCEDURES ORGAN DOSE (rem) Liver 0.3 Thyroid 50.0 Lung 2.0RADIATION DOSE USED FOR THERAPEUTIC PROCEDURES CONDITION DOSE (rem) Lymphoma 4500 Skin Cancer 5000 – 6000 Lung Cancer 6000 Brain Tumor 6000 – 7000
57. • Today, more than 1500 radioisotopes are produced by converting stable, nonradioactive isotopes into radioactive ones.• To do this, a stable atom is bombarded by fast-moving alpha particles, protons, or neutrons. When one of these particles is absorbed by the stable nucleus, the nucleus becomes unstable and the atom is now a radioactive isotopes.
58. • When a nonradioactive isotope such as boron-10 is bombarded by an alpha particle, it is converted to nitrogen-13 a radioactive isotope. + +• TRANSMUTATION – The process of changing one element into another resulting to the formation of a radioactive isotope by means of nuclear bombardment.
59. All of the known elements that have atomic numbersgreater than 92 have been produced by bombardmentand none of these elements occurs naturally. Mosthave been produced in only small amounts and existfor such a short time that it is difficult to study theirproperties.• An example is element 105, unnilpentium, which is produced when californium-249 is bombarded with nitrogen-15. + +
60. • Gallium-67 is used in the treatment of lymphomas. It is produced by the bombardment of Zinc-66 by a proton. +• Write the equation of the bombardment of Aluminum-27 by an alpha particle to produce the radioactive isotope Phosphorus-30 and one neutron. + +
61. + +• SOLUTION: The sum for the mass #s for nickel and hydrogen is 59. Therefore, the mass # of the new isotope must be 59 minus 4, or 55. The sum of the atomic #s is 29. the atomic # of the new isotope is 29 minus 2, or 27. The element that has an atomic number of 27 is cobalt (Co).
62. + +• APPLICATION IN NUCLEAR MEDICINETechnetium-99 is a radioisotope used in nuclearmedicine for several diagnostic procedures,including the detection of brain tumors andexamination of liver spleen. How to produce Tc-99?
63. • The source of technetium-99 is molybdenum-99, which is produced in nuclear reactor by neutron bombardment of molybdenum-98. + • Molybdenum-99 decays to give Technetium-99m + • Technetium-99m has a half-life of 6 hours and decays by emitting gamma rays +