Transcript Balancing Nuclear Equations

Natural Radioactivity –
Unstable Nuclei Emit Radiation



Spontaneous nuclear change to attain
good n/p ratio.
Form a new kind of atom.
Each isotope or nuclide decays in a
certain manner to get a better n/p
ratio. The decay mode is named for
the particle emitted. See Table N.
Balancing Nuclear
Equations
Nuclear Equations



Describe the decay process.
reactant or starting side (left) 
product or ending side (right).
 separates two sides
Nuclear Equations - tasks


Have to identify type (out of 4)
Have to balance to find 1 unknown
term.
Natural Transmutation –
I.D.



1 term on the reactant side –
starting isotope.
2 terms on the product side – ending
isotope and emitted particle.
Type of particle emitted is
characteristic of the isotope – look
up particle in Table N.
Nuclear Equations

Use conservation of atomic number
& conservation of mass number to
balance them.

Mass number = left superscript.

Atomic Number = left subscript.
Writing Equations



Write the equation for the decay of
Thorium-232.
Use Table N to find the decay mode: 
Write the initial equation:
232Th
90
 42He + X
 Have to figure
out what element
it turned into.
Alpha decay, Th-232
232Th
90

4He
2
+
YX
Z
Alpha decay, Th-232
232 = 4 + Y so Y = 228
232Th
90

4He
2
+
Y
Z
X
Conservation of Mass Number:
The sum of the mass numbers on the
left side must equal the sum of the
mass numbers on the right side.
Alpha decay, Th-232
232Th
90
 4He + 228X
2
Z
90 = 2 + Z
so Z = 88
Conservation of Atomic Number:
The sum of the atomic numbers on the
left side must equal the sum of the
atomic numbers on the right side.
Alpha decay, Th-232
232Th
90
 4He +
2
228X
88
Use the P.T. to find X:
232Th
90
 4He +
2
228Ra
88
X = Ra
Nuclear Equations



If there is only 1 unknown term you
can figure out what it is.
Doesn’t matter which one isn’t
known.
Don’t forget – you can look up the
decay mode in Table N. Decay mode
means what particle is emitted.
Write an equation for the 
decay of Am-241
241 = 4
241
95
+ Y so Y = 237
Am  4He +
95
2
= 2
What’s X?
YX
Z
+ Z so Z = 93
X = Np
Write equations for α decay
222Ra
88

218Rn
+ 4He
208Po
84

204Pb
82
+ 4He
256Lr

252Md
+ 4He
103
86
101
2
2
2
Writing equations for α decay
231Pa

91
225Ac

89
211Fr
87
185Au
79


227Ac
+ 4He
221Fr
87
+ 4He
89
2
2
207At
85
+ 4He
181Ir
+ 4He
77
2
2
α decay
229Th + 4He
233U 
2
90
92
149Gd  145Sm + 24He
62
64
232Th  228Ra + 4He
2
88
90
175Pt  171Os + 4He
76
2
78
237Np  233Pa + 4He
91
2
93
α decay
234Th  4He + 230Ra
90
88
2
144Nd  4
140Ce
He
+
60
2
58
146Sm  4
142Nd
He
+
62
2
60
151Ho  4
147Tb
He
+
67
2
65
192Pt  4
188Os
He
+
78
2
76
Radioactive Decay Series


Sometimes 1 transmutation isn’t
enough to achieve stability.
Some radioisotopes go through
several changes before they achieve
stability (and are no longer
radioactive).
Decay series for U-238.
Decay series for Thorium-232
Decay series for U-235
Alpha Decay
238
238 == 44 ++ 234
234
92 = 2 + 90
Positron Emission
1p
1
 1n + +10e
0
Beta minus emission
1n
0
 11p + 0e
-1
14C
6
18F
9
0e
 14
N
+
7
-1
 188 O + +10e
How does the mass number or atomic
number change in  (or  or ) decay?


Just go to Table N, find an isotope that
decays by alpha decay, write the equation,
and see how the mass number (or atomic
number) changes.
226Ra  4 + X so X has to be 222X
88

2
86
X is Ra-222. The mass number decreases by
4 and the atomic number decreases by 2.