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Inside the atom
1.
2.
3. • 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?
4.
5. • 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.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29. 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
30. Proton
- by Eugen Goldstein, 1886
-symbol “p or p +”
-relative electrical charge : + 1
-atomic mass unit = 1
-Location : inside the nucleus
31. Neutron
by - James Chadwick, 1932
-symbol “n or n ” 0
-relative electrical charge : 0
-atomic mass unit = 1
-Location : inside the nucleus
33. Proton is a color
combination of three
colored quarks. Quarks are
bound together by the
exchange of color gluons.
Emission or absorption of a
gluon causes the quarks to
make a transition from one
color to another.
34. There are six types of quarks
(up, down, charm, strange, top, and
bottom). The lightest quarks — called
up and down — are the most common.
38. TOTAL # OF PROTONS & NEUTRONS
TOTAL # OF PROTONS IN AN ATOM IN THE ATOM’S NUCLEUS
39. Atomic Number
-# of protons in the nucleus
-symbol Z, determines identity of
an element.
-equal to the # of protons, w/c is equal to
the # of electrons in an uncharged atom.
Z = number of p = number of e
+ -
40. Mass Number
- symbol A in elemental
notation, consists of the total
# of protons and neutrons in
the nucleus of the atom.
A = number of p + + number of n0
41. 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 47
ELEMENT NAME OF ELEMENT COMPLETE DESIGNATION OF ELEMENT
a
b
c
d
e
42. ELEMENT SYMBOL MASS NUMBER ATOMIC NUMBER NUMBER
NUMBER OF NUMBER OF OF
NEUTRONS PROTONS ELECTRONS
Sodium 11
15
35
Zn
Barium
How many protons, neutrons, and
electrons are present in
(a) 3 (b) 79 (C) 27
H Se Al
1 34 13
43.
44.
45. Isotopes
-Atoms of an element with the same
atomic # but different mass #
-different mass NEUTRONS
numbers but
identical atomic
numbers.
46. NUMBER OF NUMBER OF NUMBER OF
ISOTOPE
PROTONS ELECTRONS NEUTRONS
35
Cl 17 17
17
37
Cl 17 17
17
28
Si
14
29
Si
14
30
Si
14
47. ELEMENT SYMBOL MASS ATOMIC ISOTOPIC PERCENTAGE
NUMBER MASS (amu) MASS ABUNDANCE
HYDROGEN 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%
48. • 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
55. - 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).
57. • 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.
58. • 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.
59. • 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.
60. TYPE OF MASS ATOMIC
SYMBOL NUMBER NUMBER
CHARGE
RADIATION
ALPHA 4 2 2+
PARTICLE
BETA 0 0 1-
PARTICLE
GAMMA 0 0 0
RAY
PROTON 1 1 1+
NEUTRON 1 0 0
POSITRON 0 1 1+
61.
62. 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
63. 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 1m
INTENSITY OF RADIATION (1/2)2 = ¼ 1
64.
65. 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 the
sum of the atomic #s of the particles and the atoms on one side of
the equation are equal to their counterparts on the other side.
66. • 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.
67. 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)
68. 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)
69. • 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.
70. 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)
71. 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)
72. • 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.
73.
74. • 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
76. TYPES OF
ELEMENT RADIOISOTOPES HALF-LIFE
RADIATION
NATURALLY OCCURING
RADIOISOTOPES
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,γ
MEDICAL 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
77. 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.
78. • 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?
79. 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 μCi
NOTE: 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 left
in 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
80. • 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?
81. SOLUTION: (half-life of carbon-14 = 5730
1 half-life 5730 yrs.
2 half-lives X
1 half- life
= 11, 000 years
NOTE: We would estimate that the animal lived in
the tar pits about 11, 000 years ago, or about 9000
B.C.
82. • 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
83.
84. MEASUREMENT UNIT MEANING
ACTIVITY CURIE (Ci) 3.7 X 1010 disintegrations/s
ABSORBED DOSE Rad 10-5 J/g
BIOLOGICAL DAMAGE TO HUMANS Rem Rad X RBE
NOTE: RADIOISOTOPE ACTIVITY
The activity of sample is measured in terms of the number of
disintegrations or nuclear transformations produced by the sample
per second. The curie (Ci) is the unit used to express nuclear
disintegration. The curie was named for Marie Curie who discovered
radioactive elements radium and polonium together with her
husband Pierre curie.
1 curie = 3.7 X 1010 disintegrations/s
85. MEASUREMENT UNIT MEANING
ACTIVITY CURIE (Ci) 3.7 X 1010 disintegrations/s
ABSORBED DOSE Rad 10-5 J/g
BIOLOGICAL DAMAGE TO HUMANS Rem Rad X RBE
NOTE: RADIATION ABSORBED DOSE
The rad (for radiation absorbed dose) is a unit that measures the
amount of radiation absorbed by a gram of material such as body
tissue. One rad is the absorption of 10-5 J of energy per gram of
tissue.
(1 cal = 4.18 J)
1rad = 10-5 J/g
86. MEASUREMENT UNIT MEANING
ACTIVITY CURIE (Ci) 3.7 X 1010 disintegrations/s
ABSORBED DOSE Rad 10-5 J/g
BIOLOGICAL DAMAGE TO HUMANS Rem Rad X RBE
NOTE: RADIATION EQUIVALENT IN HUMANS
The rem (for radiation equivalent in humans) is a unit that measures the
biological damage caused by the various kinds of radiation. The rem considers
the biological effects of alpha, beta and gamma radiation on tissue are not the
same. The alpha particles reach the tissues, they can cause more ionization and
therefore more damage than do beta particles and gamma rays. Radiation
biological effectiveness value of gamma = 1; beta = 10; alpha = 20
Rem = Rad X RBE
87. • 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?
88. 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 particles
NOTE:
We calculate the number of beta particles from a
radioisotope’s activity. Since 1 Ci is 3.7 X 1010
disintegrations/s, there must be 3.7 X 1010 beta particles
produced in a second.
89.
90. • 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.
91. SOURCE DOSE (mrem)
NATURAL
The ground 15
Air, water, food 30
Cosmic rays 40
Wood, concrete, brick 50
MEDICAL
Chest x-ray 50
Dental x-ray 20
Upper gastrointestinal tract x-ray 200
OTHER
Television 2
Air travel 1
Global fallout 2
Cigarette smoking 35
92. Lethal Doses of Radiation for Some Life-Forms
Life – Form LD50 (rem)
Insect 100, 000
Bacterium 50, 000
Rat 800
Human 500
Dog 300
NOTE:
Exposure to radiation of about 500 rem is expected to
cause death in 50% of the people receiving that dose. This
amount of radiation is called LETHAL DOSE for one-half the
population, or LD50. Radiation of about 600 rem would be
fatal to all humans within a few weeks.
93.
94. ELEMENT RADIOISOTOPE MEDICAL USE
CHROMIUM 51 Cr Spleen imaging, blood volume,
TECHNETIUM 99mTc Brain, Lung, Liver, Spleen, Bone and bone
marrow scans
GALLIUM 67Ga Treatment of lymphomas
PHOSPHORUS 32P Treatment of leukemia, polycythemia vera,
and lymphomas; detection of brain and
breast tumors
SODIUM 24Na Vascular disease, extra cellular and blood
volume
STRONTIUM 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
95. RADIATION DOSE USED FOR DIAGNOSTIC PROCEDURES
ORGAN DOSE (rem)
Liver 0.3
Thyroid 50.0
Lung 2.0
RADIATION DOSE USED FOR THERAPEUTIC PROCEDURES
CONDITION DOSE (rem)
Lymphoma 4500
Skin Cancer 5000 – 6000
Lung Cancer 6000
Brain Tumor 6000 – 7000
96.
97. • 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.
98. • When a nonradioactive isotope such as boron-10
is bombarded by an alpha particle, it is converted
to nitrogen-13 a radioactive isotope.
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• TRANSMUTATION – The process of changing
one element into another resulting to the
formation of a radioactive isotope by means of
nuclear bombardment.
99. All of the known elements that have atomic numbers
greater than 92 have been produced by bombardment
and none of these elements occurs naturally. Most
have been produced in only small amounts and exist
for such a short time that it is difficult to study their
properties.
• An example is element 105, unnilpentium, which is produced
when californium-249 is bombarded with nitrogen-15.
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100. • Gallium-67 is used in the treatment of lymphomas. It
is produced by the bombardment of Zinc-66 by a
proton.
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• Write the equation of the bombardment of
Aluminum-27 by an alpha particle to produce the
radioactive isotope Phosphorus-30 and one neutron.
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101. + +
• 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).
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• APPLICATION IN NUCLEAR MEDICINE
Technetium-99 is a radioisotope used in nuclear
medicine for several diagnostic procedures,
including the detection of brain tumors and
examination of liver spleen. How to produce Tc-99?
103. • The source of technetium-99 is molybdenum-99, which is produced
in nuclear reactor by neutron bombardment of molybdenum-98.
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• Molybdenum-99 decays to give Technetium-99m
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• Technetium-99m has a half-life of 6 hours and decays by emitting gamma
rays
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