Transcript NUCLEAR CHEMISTRY

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). (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.
238 U
92

234
90 Th +
4 He
2
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.
131 Xe +
I

53
54
131
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
1 n  1 p
0 e
CHANGE:
+
0
1
-1
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.
0
0 e  20 
e
+
1
-1
0
EX.
22 Na
11
 01e + 2210Ne
EX.
11
 01e + 115B
6C
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.
201 Hg
80
EX.
81 Rb
37
+
+
0
0
-1e
-1e

201

81 Kr
36
79Au +
+
0
0
0 
0
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:
195 Au
79
+ _________ 
195 Pt
78
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:

1.
11 C
6
2.
214
3.
237
4.
201
5.
231 Th
90
6.
195
7.
14
8.
118
0
1e
83Bi 
0 e
-1
93Np 
4
79Au +
201
79Au
0 e
-1
+
231 Pa
91
-1e

195 Pt
78
0
14 N
7
54Xe 
214 Po
84
233 Pa
He
+
2
91
-1e 

5B
+
0
80Hg +
6C 
11
+
114
+
52Te +
0
-1e
4
2He
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 :
27 Al
13
LONG HAND:
27 Al
13
(n ,  ) 2411Na
+ 10 n 
4
24 Na
He
+
2
11
WRITE THE SHORT HAND FOR:
16
1 H  4 He + 13 N
O
+
8
1
2
7
ANSWER: 168O (p, ) 137N
THERE ARE ALSO OTHER TYPES OF NUCLEAR
TRANSMUTATIONS.
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
238U
EVERY 1mg OF 238U. THE HALF LIFE OF
 206Pb IS 4.05 X 109 YEARS. HOW OLD IS THE
ROCK?
Soln. 1. Find the original quantity of
what we know the half life for.
238U
because it is
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
at present. So, to find the original quantity,
0.257 mg 206Pb x 238 mg 238U
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
AND 0.953 g REMAINS AFTER 2.00 yrs.
90Sr
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  – 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
1.00g x 1 mole
235 g
1.00 g OF 235U
x 6.02 x 1023 atoms x 3.20 x 10-11 J =
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:
2
1H
+ 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  DECAY OF 23892U ?
238
234 Th + 4 He
U

92
90
2
WE MUST FIRST CALCULATE THE NET CHANGE IN
MASS. THE ATOMIC MASS OF EACH PARTICLE IS:
238
92U
= 238.0508 u,
234
90Th
= 234.0437 u,
4
2He
= 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
1u
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
=
1.602 x 10-13 J
4.2 MeV
THE END