NUCLEAR CHEMISTRY: INTRO
1. Kinetic Stability : probability that an unstable nucleus will decompose
into more stable species through radioactive decay.
2. All nuclides with 84 or more protons are unstable and will decay.
• Light nuclides where Z = A-Z (neutron/proton ratio is 1).
• Nuclides with even numbers of neutrons and protons are unusually
stable.
• Especially stable nuclides exhibit magic numbers,
2,8,20,28,50,82,126 of neutrons or protons.
• A nuclide is a unique atom of the type.
A
X
•
Z
Alpha decay, emits a
4
2
He
NUCLEAR CHEMISTRY: STABILITY GRAPH
#n
120
110
100
90
80
70
60
50
40
30
20
10
20 p
40 p
60 p
80 p
100 p
NUCLEAR CHEMISTRY: BETA DECAY
14
6
1.
2.
3.
4.
5.
C

14
N
7
0
+
e
BETA PARTICLE
-1
BETA DECAY
THE ATOMIC NUMBER OF THE PRODUCT INCREASES.
NUCLIDES ABOVE THE PENNINSULA (ZONE) OF STABILITY DECAY
WITH BETA DECAY(SEE GRAPH ON OTHER SLIDE).
PENETRATING RADIATION.
THE BETA PARTICLE CONES FROM THE DECOMPOSITION OF A
NEUTRON TO A PROTON AND BETA PARTICLE. THE BETA PARTICLE
IS AN ELECTRON “BORN” IN THE NUCLEUS.
BETA DECAY IS SPONTANEOUS, NOTICE ONLY ONE REACTANT.
1
n
0
1

p
1
0
+
-1
e
NUCLEAR CHEMISTRY: ALPHA DECAY
238
U
92
4

He
2
234
+
Th
90
ALPHA PARTICLE
1.
2.
3.
4.
5.
6.
7.
ALPHA DECAY
THE ATOMIC NUMBER OF THE PRODUCT DECREASES BY 2.
THE MASS NUMBER OF THE PRODUCT DECREASES BY 4.
ALPHA RADIATION IS NON PENETRATING TO HUMAN SKIN,
HOWEVER IT CAN BE INGESTED.
COMMON MODE OF DECAY FOR HEAVY RADIOACTIVE NUCLIDES.
NEUTRON/PROTON RATIO INCREASES.
THE ALPHA PARTICLES ARE POSITIVE, THESE HIGH ENERGY He
ATOMS HAVE LOST THE ELECTRONS, ARE REPELLED BY POSITIVE
ELECTRODES AND ARE AFFECTED BY MAGNETIC FIELDS.
ALPHA DECAY IS SPONTANEOUS.
NUCLEAR CHEMISTRY: POSITRON EMISSION
22
11
Na
0

e
+1
22
+
Ne
10
POSITRON
1.
2.
3.
4.
POSITRON EMISSION
DECAY MODE FOR NUCLIDES BELOW ZONE OF STABILITY.
CHANGES A PROTON TO A NEUTRON.
PRODUCT HAS A HIGHER NEUTRON TO PROTON RATIO.
THE POSITRON IS THE ANTIPARTICLE TO AN ELECTRON THE
REACTION OF A POSITRON WITH A BETA PARTICLE PRODUCES
GAMMA RADIATION .
0
e
+1
0
+
e
-1

0
GAMMA
0
NUCLEAR CHEMISTRY: ELECTRON CAPTURE
201
80
Hg
0
+
e
-1
201
Au
79

+
0 GAMMA
0
INNER ORBITAL SHELL
ELECTRON
ELECTRON CAPTURE
1. AN INNER SHELL ELECTRON IS CAPTURED BY THE NUCLEUS.
0
e
+1
0
+
e
-1

0
GAMMA
0
NUCLEAR CHEMISTRY: ELECTRON CAPTURE
ln (N0/N) = kt
Log
N0
N
( )
t1/2 =0.693/k
E = c2
m
= kt
2.303
Example problem, binding energy
OBJECTIVE calculate the
binding energy per nucleon of
E = c2 m
Mass defect equation
N –14, nuclear mass is
13.999234
mass of proton
mass of neutron
mass nucleus
m = (7(1.0072765) + 7(1.0086649)) – 13.999234 = 0.112536 amu
E = 0.112536 amu
Avagodro’s #
1g
6.0221 * 1023 amu
x
1kg
1000g
x
8.987551*1016 m2
s2
= 1.67682 * 10-11 kg m2/s2 = 1.67682 * 10-11 J
For 14N, A=14
Binding energy/nucleon = 1.67682 * 10-11 J/ 14 = 1.19773 * 10-12 J/nucleon
Example problems:Half Life
Calculate the mass of
Co-60 that remains from a
0.0100 g sample after 1.00
year has elapsed.
k= 0.693 = 0.693 = 0.132y
t1/2
5.27/y
1-FIRST FIND k FROM
HALF LIFE.
Co-60 Half life,from tables
2-FIND N0/N RATIO
Log
N0
N
( )
= kt
2.30
Log
N0
N
= (0.132/y)(1.00y) = 0.0570, ANTILOG OF 0.0570 IS 1.14
2.303
( )
N0 = 1.14 = 0.0100g
N
N
; N = 0.00877g OF
60
C
27
AFTER 1 YEAR