THE STUDY OF NUCLEAR REACTIONS NUCLEAR CHEMISTRY THE STUDY OF NUCLEAR REACTIONS

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.

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.

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.

NUCLEAR STABILITY AND RADIATION: EMISSION OF RADIATION IS ONE WAY IN WHICH AN UNSTABLE SUBSTANCE IS CHANGED INTO A LOWER ENERGY, MORE STABLE SUBSTANCE.

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.

FOR ELEMENTS UP TO ATOMIC NUMBER 20 THE RATIO OF PROTONS TO NEUTRONS IS APPROXIMATELY 1:1. AS THE NUMBER OF PROTONS INCREASE, THE NUMBER OF NEUTRONS INCREASE EVEN MORE. IN THE BELT OF STABILITY, THE RATIO IS APROXIMATELY 1:1, AND IT ENDS AT BISMUTH WHICH HAS ATOMIC NUMBER 83. ALL NUCLEI ABOVE ATOMIC NUMBER 83 ARE RADIOACTIVE.

THE TYPE OF NUCLEAR DECAY CAN BE PREDICTED, TO AN EXTENT, ACCORDING TO AN ELEMENT’S NEUTRON TO PROTON RATIO. NUCLEI WITH A HIGH NEUTRON TO PROTON RATIO (ABOVE THE ZONE) - CAN EMIT A BETA PARTICLE, (10n  0-1e + 1+1p) 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  0+1e + 10n )(EQUIVILENT TO PROTON - TO -NEUTRON CHANGE). This decreases the atomic number OR

ELECTRON CAPTURE (THIS IS MORE COMMON AS NUCLEAR CHARGE INCREASES) ELECTRON CAPTURE (THIS IS MORE COMMON AS NUCLEAR CHARGE INCREASES). (0-1e + 1+1p  10n ), thereby decreasing the atomic number. HEAVY NUCLEI (ATOMIC NUMBER > 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.

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).

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

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.

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.

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.

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

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) ________________________________________________

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

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

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.

PARTICLE ACCELERATORS (AKA 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.

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).

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.

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?

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

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 )

EX. IF WE START WITH A 1. 00 g SAMPLE OF 90Sr AND 0 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

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.

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.

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)

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.

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.

THE ENERGY / MASS CHANGES ASSOCIATED WITH NUCLEAR REACTIONS: USING EINSTEIN’S 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

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...

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

THE END