NUCLEAR CHEMISTRY

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NUCLEAR CHEMISTRY

• THE STUDY OF NUCLEAR REACTIONS

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When nuclei change spontaneously, by emitting
radiation, they are radioactive. USES OF
RADIOACTIVE ELEMENTS INCLUDE MEDICINE (FOR
DIAGNOSTICS AND CANCER TREATMENT), THE STUDY OF
CHEMICAL REACTION MECHANISMS, TO GENERATE POWER,
AND TO CREATE THE MOST DESTRUCTIVE MILITARY
WEAPONS.
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NUCLEAR ENERGY POSES THE PROBLEMS OF NUCLEAR
WASTE, OF WHICH THERE IS NO VIABLE WAY TO
DISPOSE, AND THE POSSIBILITY OF HUMAN ERROR OR
ACCIDENT THAT COULD RESULT IN CATASTROPHY.
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RADIOACTIVITY - DEFINITIONS NUCLEONS - TWO OF
THE PARTICLES IN THE NUCLEUS, THE PROTONS AND
NEUTRONS. ISOTOPES - ATOMS OF THE SAME ELEMENT
WITH DIFFERENT NUMBERS OF NEUTRONS. MASS NUMBER -
THE TOTAL NUMBER OF NUCLEONS IN THE
NUCLEUS. NUCLIDE - THE NUCLEUS OF A SPECIFIC
ISOTOPE OF AN ELEMENT. RADIONUCLIDES -
RADIOACTIVE NUCLIDES . RADIOISOTOPES - ATOMS THAT
CONTAIN RADIOACTIVE NUCLIDES.
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NUCLEAR STABILITY AND RADIATION EMISSION OF
RADIATION IS ONE WAY IN WHICH AN UNSTABLE
SUBSTANCE IS CHANGED INTO A LOWER ENERGY, MORE
STABLE SUBSTANCE.
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A GENERAL GUIDE TO NUCLEAR STABILITY THE ZONE
(OR BELT) OF STABILITY (DIAGRAM IN ZUMDAHL PG.
1023, BROWN LEMAY PG. 775. IT SHOULD BE
REALIZED THAT PROTONS PACKED TIGHTLY IN A SMALL
NUCLEUS NEED NEUTRONS PACKED AMONG THEM TO OFFER
SOME STABILITY AND BINDING FORCE TO THE NUCLEUS.
(NOTE DUE TO THE REPULSION BETWEEN
PROTONS). PLEASE NOTE THAT DUE TO THE SOFTWARE
LIMITATIONS, SUBSCRIPTS AND SUPERSCRIPTS DO NOT
APPEAR DIRECTLY ABOVE AND BELOW ONE ANOTHER IN
THIS PRESENTATION, BUT THEY SHOULD BE.
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FOR ELEMENTS UP TO ATOMIC NUMBER 20 THE RATIO OF
PROTONS TO NEUTRONS IS APPROXIMATELY 11. AS THE
NUMBER OF PROTONS INCREASE, THE NUMBER OF
NEUTRONS INCREASE EVEN MORE. IN THE BELT OF
STABILITY, THE RATIO IS APROXIMATELY 11, AND IT
ENDS AT BISMUTH WHICH HAS ATOMIC NUMBER 83. ALL
NUCLEI ABOVE ATOMIC NUMBER 83 ARE RADIOACTIVE.
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THE TYPE OF NUCLEAR DECAY CAN BE PREDICTED, TO AN
EXTENT, ACCORDING TO AN ELEMENTS NEUTRON TO
PROTON RATIO. NUCLEI WITH A HIGH NEUTRON TO
PROTON RATIO (ABOVE THE ZONE) - CAN EMIT A BETA
PARTICLE, (10n ? 0-1e 11p) WHICH IS A NEUTRON
-TO - PROTON CHANGE, SO THE RATIO LEVELS OUT
MORE. This increases the atomic number. NUCLEI
WITH A LOW NEUTRON TO PROTON RATIO (BELOW THE
ZONE) - CAN EMIT POSITRONS ( 11p ?
01e 10n )(EQUIVILENT TO PROTON - TO -NEUTRON
CHANGE). This decreases the atomic number OR
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ELECTRON CAPTURE (THIS IS MORE COMMON AS NUCLEAR
CHARGE INCREASES). (0-1e 11p ? 10n ), thereby
decreasing the atomic number. HEAVY NUCLEI
(ATOMIC NUMBER gt 84 CAN UNDERGO ALPHA EMISSION.
(AXZ ? 42He A-4X-2Y) THIS LOWERS BOTH
ATOMIC NUMBER AND MASS NUMBER (PROTONS AND
NEUTRONS). THESE TYPES OF EMISSIONS WILL BE
ADDRESSED IN DETAIL NOW, FOR BETTER UNDERSTANDING.
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NUCLEAR EQUATIONS ALPHA PARTICLES (?), ARE 4He
PARTICLES EMITTED IN SOME NUCLEAR DECAY
REACTIONS. FOR EXAMPLE, WHEN 238 U EMITS AN ? -
PARTICLE IT LOSES A 42He WHICH, WHEN
SUBTRACTING THE 4 FROM THE MASS OF 238 U AND
THE ATOMIC NUMBER OF He (2) FROM THE ATOMIC
NUMBER OF 238 U (92) , LEAVES THE ISOTOPE WITH
MASS 234 AND ATOMIC 90, WHICH IS THORIUM
-234. 23892 U ? 23490 Th 42He NOTICE THE
SUM OF THE MASS ON THE RIGHT SUM OF MASS S
ON THE LEFT (ALSO NOTE SAME WITH ATOMIC S).
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BETA DECAY ( ? - DECAY ) - IS WHERE AN ELECTRON
IS LOST. 13153I ? 13154Xe 0-1e THESE ? -
PARTICLES, EVEN THOUGH THEY ARE ELECTRONS, DO NOT
ORIGINATE FROM ORBITALS. THEY ARE EMITTED FROM
THE NUCLEUS. THEY ARE NOT NORMALLY RESIDING IN
THE NUCLEUS AND COME INTO BEING ONLY WHEN THE
NUCLEUS UNDERGOES A NUCLEAR REACTION. THIS IS
EQUIVILENT TO A NEUTRON - TO - PROTON
CHANGE 10n ? 11p 0-1e
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GAMMA RADIATION (? ) - IS ANOTHER COMMON TYPE OF
NUCLEAR DECAY. ? - RADIATION CONSISTS OF VERY
SHORT ? s, THUS HIGH ENERGY, ELECTROMAGNETIC
RADIATION. (i.e., PHOTONS). THEY ARE OF HIGHER
ENERGY THAN X-RAYS. THE MASS AND ATOMIC NUMBER
DO NOT CHANGE. ? - RADIATION ACCOMPANIES OTHER
RADIOACTIVE EMISSION AND IS GENERALLY NOT SHOWN
IN EQUATIONS. IT REPRESENTS THE ENERGY LOST
DURING THE STABLILIZING REORGANIZATION OF THE
NUCLEUS.
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POSITRON EMISSION - WHEN A PARTICLE THE SAME SIZE
OF AN ELECTRON BUT OPPOSITE IN CHARGE IS EMITTED.
( 01e ). THEY ARE DESTROYED QUICKLY UPON
COLLISION WITH ELECTRONS TO MAKE ? -
RAYS. 01e 0-1e ? 2 00? EX. 2211Na ?
01e 2210Ne EX. 116 C ? 01e 115B THIS
IS THE EQUIVILENT OF CHANGING THE PROTON TO A
NEUTRON, THUS THE DECREASE IN ATOMIC . THE MASS
STAYS THE SAME.
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ELECTRON CAPTURE IS WHEN AN INNER - ORBITAL
ELECTRON IS CAPTURED BY THE NUCLEUS. IT
SIMILARLY HAS THE EFFECT OF CONVERTING A PROTON
TO A NEUTRON. EX. 20180Hg 0 -1e ?
20179Au 00? EX. 8137Rb 0 -1e ?
8136Kr 00? NOTICE THAT THE ELECTRON IS
CAPTURED BY THE NUCLEUS AND IS THEREFORE WRITTEN
ON THE REACTANT SIDE OF THE EQUATION.
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SAMPLE PROBLEMS WRITE EQUATIONS FOR THE
FOLLOWING A. CARBON - 11 PRODUCES A POSITRON B.
BISMUTH - 214 PRODUCES A ? - PARTICLE C.
NEPTUNIUM - 237 PRODUCES AN ? - PARTICLE D.
MERCURY - 201 UNDERGOES e- CAPTURE. E. THORIUM -
231 DECAYS TO PROTACTINIUM - 231 F. SUPPLY THE
CORRECT PARTICLE 19579Au _________ ?
19578Pt
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SAMPLE PROBLEMS WHAT TYPE OF DECAY? SHOW
THE REACTION. A. CARBON - 14 (HAS A HIGH n0 TO
p RATIO) ________________________________________
_______ B. Xenon - 118 (A HEAVY
PARTICLE) ________________________________________
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• ANSWERS
• 116C ? 01e 115B
• 21483Bi ? 0-1e 21484Po
• 23793Np ? 42He 23391Pa
• 20180Hg 0-1e ? 20179Au
• 23190Th ? 0-1e 23191Pa
• 19579Au 0-1e ? 19578Pt
• 146C ? 147N 0-1e
• 11854Xe ? 11452Te 42He

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NUCLEAR TRANFORMATIONS THE CHANGE OF ONE ELEMENT
INTO ANOTHER CAN BE CAUSED BY BOMBARDING ITS
NUCLEUS WITH ANOTHER NUCLEUS OR WITH A
NEUTRON. THESE ARE INDUCED REACTIONS WRITTEN AS
FOLLOWS TARGET NUCLEI BOMBARDING PARTICLE
? EJECTED PARTICLE PRODUCT NUCLEI
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THESE EQUATIONS HAVE BOTH A SHORT AND LONG HAND
REPRESENTATION. SHORT HAND 2713Al (n , ? )
2411Na LONG HAND 2713Al 10 n ? 42He
2411Na WRITE THE SHORT HAND FOR 168O 11
H ? 42He 137N ANSWER 168O (p, ?) 137N THERE
ARE ALSO OTHER TYPES OF NUCLEAR TRANSMUTATIONS.
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PARTICLE ACCELERATORS (AKA. ATOM SMASHERS) ARE
USED TO MOVE PARTICLES FAST ENOUGH TO OVERCOME
THE REPULSION FORCES OF THE TARGET NUCLEUS. THE
PARTICLES ARE PROJECTED ON A SPIRAL PATHWAY.
LINEAR ACCELERATORS USE CHANGING ELECTRIC FIELDS
TO ACHIEVE THE HIGH VELOCITIES NEEDED AND PROJECT
THE PARTICLES ON A LINEAR PATHWAY.
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RADIOACTIVE DATING (Using 14C to determine the
age of substances like rocks, fossils, bones,
etc.) RADIOACTIVE DECAY IS A FIRST ORDER PROCESS
IN WHICH THE HALF LIFE IS CALCULATED AS
FOLLOWS t1/2 0.693 / K THIS REFERS TO THE
TIME IT TAKES FOR HALF OF A GIVEN QUANTITY TO
REACT (OR IN THIS CASE, DECAY).
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EX. THE HALF LIFE OF 60Co IS 5.3 YEARS. HOW MUCH
OF A 1.00 g SAMPLE WILL BE LEFT AFTER 15.9
YEARS? SOLN. 15.9 3 x 5.3 (THREE HALF
LIVES). IN ONE HALF LIFE, 0.500 g ARE LEFT IN TWO
HALF LIVES, 0.250 g ARE LEFT IN THREE HALF LIVES,
0.125 g ARE LEFT.
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EX. A ROCK CONTAINS 0.257 mg OF 206Pb FOR
EVERY 1mg OF 238U. THE HALF LIFE OF 238U
? 206Pb IS 4.05 X 109 YEARS. HOW OLD IS THE
ROCK?
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Soln. 1. Find the original quantity of 238U
because it is what we know the half life for. 2.
Find K, and from that, find t from ln (x / xo )
- k t Assume we take a piece of rock that has
1.00 mg of 238U at present. So, to find the
original quantity, 0.257 mg 206Pb x 238 mg
238U 0.295 mg 238U 207 mg
206Pb Therefore 1.00 mg 0.295 mg 1.295
mg original 238U k 0.693 / 4.5 x 10 9 yrs
1.6 x 10 -10 yrs-1 t ( - 1 / k) ln (x / xo
) (- 1 / 1.6 x 10 -10 ) ln ( 1 / 1.295) t
1.6 x 10 9 yrs
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IN SUMMARY 1. ASSUME A CURRENT SAMPLE ( X ) OF
1.00 mg. 2. CALCULATE THE ORIGINAL AMOUNT OF
THE SUBSTANCE ( XO ) BY USE OF DIMENSIONAL
ANALYSIS. 3. CALCULATE K 4. FIND t FROM
t (-1 / K) ln ( X / XO )
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EX. IF WE START WITH A 1.00 g SAMPLE OF 90Sr
AND 0.953 g REMAINS AFTER 2.00 yrs. A. WHAT IS
t1/2 ? B. HOW MUCH WILL REMAIN AFTER 5
YEARS? SOLN ln X / XO - Kt ln 0.953 / 1.00g
- K (2.00yr) , THEREFORE K 0.0241 yr
-1 t1/2 0.693 / 0.0241 yr-1 28.8 yrs B. ln X
/ XO - K t , ln X - ln 1.00g - (0.0241yr-1)
(5) ln X - 0.121 , X 0.886 g
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FISSION AND FUSION
IN A FISSION PROCESS, WHEN A NUCLEUS IS BOMBARDED
WITH A NEUTRON A b PARTICLE IS EMITTED, BUT THE
REMAINING PARTICLE BREAKS UP INTO FRAGMENTS. FOR
235U THE PARTICLE FIRST CHANGES TO THE UNSTABLE
236U. THIS BREAKS INTO A HEAVY AND A LIGHT
FRAGMENT AND AN AVERAGE OF 2.5 NEUTRONS ARE
RELEASED.
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THESE RELEASED NEUTRONS CAN PRODUCE AN AVERAGE OF
TWO MORE FISSION PROCESSES, WHICH CAN PRODUCE
FOUR OR FIVE MORE AND THUS A CHAIN REACTION CAN
OCCUR. IF NOT CONTROLLED, THIS CAN LEAD TO AN
EXPLOSION.
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THE AVERAGE ENERGY INVOLVED FOR THE FISSION
PROCESS OF 235U 235U n ? 236U ?
FRAGMENTS NEUTRONS 3.20 x 10 11 J /
atom THEREFORE FOR 1.00 g OF 235U 1.00g x 1
mole x 6.02 x 1023 atoms x 3.20 x 10-11 J
235 g 1 mole 1 atom 8.20 x 1010 J
8.20 x 107 kJ THIS IS AN ENORMOUS AMOUNT OF
ENERGY (EQUIVALENT TO THE COMBUSTION OF
APPROXIMATELY 3 TONS OF COAL)
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NUCLEAR FUSION THIS IS THE PROCESS IN WHICH
ENERGY IS PRODUCED ON THE SUN. THE HYDROGEN BOMB
IS A FUSION REACTION WHICH IS STARTED BY THE
EXPLOSION OF AN ATOMIC BOMB. A GENERAL FUSION
REACTION IS ONE WITH DEUTERIUM AND TRITIUM 21H
31H ? 42He 10n IF THIS PROCESS COULD BE
DEVELOPED IT WOULD SUPPLY AN UNLIMITED SOURCE OF
ENERGY.
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UNFORTUNATELY, THE TEMPERATURES REQUIRED FOR SUCH
A REACTION ARE OVER 40,000,000 K, IN ORDER TO
ALLOW FOR THESE NUCLEI THAT REPELL EACH OTHER TO
BE IN THE VERY CLOSE PROXIMITY NECESSARY FOR THE
REACTION TO OCCUR.
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THE ENERGY / MASS CHANGES ASSOCIATED WITH NUCLEAR
REACTIONS USING EINSTEINS EQUATION, E
mc2 THE ENERGY OF A NUCLER REACTION CAN BE
CALCULATED. HERE, m net change in mass (kg), c
velocity of light (meters / second), E energy
(joules or MeV which are mega electron volts) 1
MeV 1.602 x 10-13 J
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EX. WHAT IS THE ENERGY ASSSCOCIATED WITH THE a
DECAY OF 23892U ? 23892U ? 23490Th
42He WE MUST FIRST CALCULATE THE NET CHANGE IN
MASS. THE ATOMIC MASS OF EACH PARTICLE
IS 23892U 238.0508 u, 23490Th 234.0437 u,
42He 4.0026 u PRODUCTS REACTANTS
234.0437 4.0026 238.0508 -0.0045
u INDICATING A NET LOSS IN MASS. 1 u 1.66 X
10-24g (1/12 THE MASS OF 12C ) continued...
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Changing the mass to grams 0.0045 u x 1.66 x
10-24g 7.47 x 10-27 g 1 u E
7.47 x 10-30 Kg (3.00 x 108m/s)2 6.7 x 10-13
J 6.7 x 10-13 J x 1 MeV 4.2
MeV 1.602 x 10-13 J
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THE END