Definitions
Nucleons: Subatomic particles in the nucleus : protons and neutrons
Radionucleotides: Radioactive nuclei. Unstable nuclei that
spontaneously emit particles and electromagnetic radiation.
Radioisotopes: Atoms containing radioactive nuclei.
Types of Radioactive Decay
When unstable nuclei decay, the reactions they undergo generally
involve one or more of the following particles listed in the first
column. Some facts about these particles may be found in the next 5
columns. In the 7th column you will find an example of each type of
decay. Notice that for an alpha particle decay, the equation is
balanced with regard to atomic number (92, 90+2) and atomic
weight (238, 234+4). This type of balancing is true for all nuclear
reactions. In the last column are instructions for how to predict
when each type of emission or capture will occur. Fore example, if
the atomic number is greater than 83, alpha particle decay is most
likely.
particle
What is it?
symbol
charg
e
mass
relativ
e
penetr
ating
power
alpha
particles
helium nuclei
4
2He
or 2a4
+2
6.664 E-24 g
1
beta
particles
high speed
electrons
–1
9.11 E-28 g
100
o
-1e
or o
1B
Example
238
92U
131
53I
Applies to which particles
=> 90Th234 + 2He4
Atomic Numbers > 83; the 2 p+ 2n0
loss brings the atom diagonally back
to the belt of stability.
=> 54Xe131 + -1eo
Isotopes below the belt of stability
(high neutron : proton ratios). Causes
a loss of 1 neutron and a gain of 1
proton.
When a B-particle is emitted, the at. no. increases by 1. A neutron is converted into a p+
1
1
o
and e-:
on => 1p + -1e
gamma
Rays
high energy
photons
o
og
0
0
10000
Generally accompanies other radioactive radiation because it is the energy lost from other
nucleon changes. Gamma radiation is generally not shown in the nuclear equation.
positron
emission
positron
o
1e
+1
9.11 E-28 g
6C
11
=> 5B11 + 1eo
Causes the atomic number to decrease. It converts a proton to a neutron + positron
1
1
o
1p => on + 1e
electron
capture
inner shell
electron
o
-1e
–1
9.11 E-28 g
81
o
81
37Rb + -1e => 36Kr
The nucleus capture an inner shell electron; thereby converting a p+ to a no
1
o
1
1p + -1e => on
Isotopes above the belt of stability
(low neutron : proton ratios). Causes a
loss of 1 proton and a gain of 1
neutron.
Isotopes above the belt of stability
(low neutron : proton ratios). Causes a
loss of 1 proton and a gain of 1
neutron.
Stable n:p ratios:
Neutrons are needed to hold protons together in the nucleus by the strong force. At low atomic
numbers, the ratio of neutrons to protons in stable isotopes is generally very close to 1. As the
atomic number rises, so does the neutron/proton ratio, such that at atomic number 80, stable
isotopes have a ratio of about 1.5. This information is tabulated below.
atomic number
7
40
50
80
ratio n/p for stable isotopes
1
1.25
1.4
1.5
Problem
Solution
Predict the type of decay of the isotope 92U238
explanation
emission
z > 83
alpha
preliminary equation
238
92U
=> X + 2He4
balancing to solve for X
238
92U
=> 90Th234 + 2He4
Z > 83 so alpha particle emission
92U
238
=> 90Th234 + 2He4
unpredicted Beta particle emission
234
90Th
=>
234
91Pa
+ -1eo
234
92U
+ -1eo
unexpected Beta particle emission
234
91Pa
=>
Z > 83 so alpha particle emission
234
92U
=> 90Th230 + 2He4
90Th
230
=> 88Ra226 + 2He4
226
88Ra
=> 86Rn222 + 2He4
222
=> 84Po218 + 2He4
218
=> 82Pb214 + 2He4
Z > 83 so alpha particle emission
Z > 83 so alpha particle emission
Z > 83 so alpha particle emission
86Rn
Z > 83 so alpha particle emission
84Po
132/82 = 1.61 high so Beta particle emission
214
=> 83Bi214 + -1eo
214
=> 84Po214 + -1eo
214
=> 82Pb210 + 2He4
82Pb
131/83 = 1.59 high so Beta particle emission
83Bi
Z > 83 so alpha particle emission
84Po
128/82 = 1.57 high so Beta particle emission
210
=> 83Bi210 + -1eo
210
=> 84Po210 + -1eo
82Pb
127/83 = 1.53 high so Beta particle emission
83Po
Z > 83 so alpha particle emission
210
84Po
=> 82Pb206 + 2He4
Stable nuclear configurations:
Some configurations of protons and neutrons are particularly stable just
as some configurations of electrons (2, 10, 18, 36, 54, 86) are.
proton numbers: 2, 8, 20, 28, 50, 82
neutron number: 2, 8, 20, 28, 50, 82, 126
nuclei with even numbers of protons and neutrons are more stable than
those with odd numbers.
Nuclear Transmutations: occur when nuclei are struck by neutrons or
other nuclei. These reactions are useful in creating new radioisotopes.
Notation: The reaction in which N-14 is bombarded with an alpha
particle: 7N14 + He => O17 + 1H1 is written as
14 (a, p) O17 in the following order: target nucleus, (bombarding
N
7
8
particle, ejected particle), product nucleus.
2
4
8
Nuclear Transmutation Notation Worksheet
Part 1: Write balanced equations for the following transmutations: in other words, from the
information in column "Problem" you should be able to write the answers in column
"Solution."
Problem
7N
13
(a, p) 8O16
9
(a,n) 6C12
4Be
27
13Al
8O
16
(n,a) 11Na24
(p,a) 7N13
23
26
11Na (a,p) 12Mg
14
17
7N (a, p) 8O
10
5B
(a,n) 7N13
Solution
7N
13
9
4Be
+ 2He4 => 8O16 + 1H1
+ 2He4 => 6C12 + on1
27
13Al
8O
16
23
11Na
7N
+ 0n1 => 11Na24 + 2He4
+ 1H1 => 7N13 + 2He4
+ 2He4 =>
14
10
5B
12Mg
26
+ 1H1
+ 2He4 => 8O17 + 1H1
+ 2He4 => 7N13 + on1
Part 2: Complete the following nuclear equations: Note that the equation must
balance on each side with respect to atomic number and atomic mass.
Problem
7N
13
+ 2He4 => 8O16 + ?
1H
9
4Be
3
30
=>
213
83Bi
13
14S
30
+?
12Mg
26
9
4Be
15P
=> 2He4 + ?
+ 2He4 =>
+ 2He4 => 8O16 + 1H1
1H
30
3
+?
+ 2He4 => 7N13 + ?
23
11Na
=>
14S
30
+ 1e0 (positron)
=> 2He4 + 81Tl209
+ 2He4 =>
47Ag
10
5B
=> 2He3 + -1e0
+ 2He4 => 6C12 + on1
213
83Bi
Ag106 => Cd106 + ?
10
5B
7N
=> 2He3 + ?
+ 2He4 => 6C12 + ?
15P
23
11Na
Solution
106
=>
12Mg
48Cd
106
26
+ 1H1
+ -1e0
+ 2He4 => 7N13 + on1
Part 3: Write the following reactions using transmutation notation. Note that this
is the reverse of Part 1.
Problem
7N
4
13
Be9
13
+ 2He4 => 8O16 + 1H1
+2
Al27
8
O16
23
11Na
7
10
5B
He4
+0
n1
+1
=>
=> 11
H1
=> 7
+ 2He4 =>
N14
Solution
+2
He4
6
C12
Na24
N13
+o
+2
+2
=> 8
He4
He4
26
+ 1H1
+1
H1
12Mg
O17
n1
+ 2He4 => 7N13 + on1
13
(a, p) 8O16
9
4Be
(a,n) 6C12
7N
27
13Al
8O
16
23
11Na
7N
(n,a) 11Na24
(a,p) 12Mg26
14
10
5B
(p,a) 7N13
(a, p) 8O17
(a,n) 7N13
Radioactive Emission Worksheet
For the following particle, what type of emission is expected, why, and what is the equation
describing that emission?
238
131
11
81
226,
212, Tl209, Ba140
92U , 53I , 6C , 37Rb , 88Ra
84Po
81
56
answers immediately below
isotope
explanation
emission
238
92U
z > 83
alpha
ratio = 78/53=1.47 high
beta
11
6C
ratio = 5/6 = 0.83 low
positron emission or
electron capture
81
37Rb
ratio = 44/37 = 1.19 low
positron emission or
electron capture
226
z > 83
alpha
226
88Ra
=> 86Rn222 + 2He4
212
z > 83
alpha
84Po
212
=> 82Pb208 + 2He4
209
ratio = 128/1.58 high
beta
81Tl
209
=> 82Pb209 + -1eo
140
ratio = 84/56 = 1.5 high
beta
140
56Ba
=> 57La140 + -1eo
53I
131
88Ra
84Po
81Tl
56Ba
equation
238
92U
=> 90Th234 + 2He4
131
53I
=> 54Xe131 + -1eo
11
11
o
6C => 5B + 1e
11
o
11
6C + -1e => 5B
37Rb
81 =>
81+
37Rb
81
o
36Kr + 1e
o
81
-1e => 36Kr
Nuclear Transmutation Notation Worksheet
Part 1: Write balanced equations for the following
transmutations: in other words, from the information in column
"Problem" you should be able to write the answers in column
"Solution."
Problem
Solution
13
16
13
16
1
7N (a, p) 8O
7N + He => O + 1H
9
12
4Be (a,n) 6C
9 + He => C12 + n1
Be
4
o
4
2
4
2
27 (n,a) Na24
Al
13
11
8
11
O16 (p,a) 7N13
Na23 (a,p) 12Mg26
14 (a, p) O17
N
7
8
10 (a,n) N13
B
5
7
27 +
Al
13
16
8O
0n
6
1
=> 11
Na24 +
+ 1H1 => N13 +
7
23 +
Na
11
14
7N
8
4
2He
+
10 +
B
5
4
2He
4
2He
=>
12
4
2He
4
2He
Mg26 + 1H1
=> 8
O17 + 1H1
=>
N13 + on1
7
Part 2: Complete the following nuclear equations: Note that the
equation must balance on each side with respect to atomic
number and atomic mass.
Problem
Solution
13 + He => O16 + ?
13 + He => O16 + H1
N
N
7
7
1
3 => He3 + ?
3 => He3 + e0
H
H
1
1
-1
9 + He => C12 + ?
9 + He => C12 + n1
Be
Be
4
4
o
30 => S30 + ?
30 => S30 + e0 (positron)
P
P
15
15
1
213 => He + ?
213 => He + Tl
Bi
Bi
83
23 + He => Mg26 + H1
Na23 + He => Mg26 + ?
Na
11
1
Ag106 => Cd + ?
Ag106 => Cd + e0
10 + He => N13 + ?
10 + He => N13 + n1
B
B
5
5
o
4
2
4
2
8
2
2
4
2
4
4
4
7
2
83
2
12
106
2
6
14
2
2
4
2
6
14
11
8
4
2
4
81
12
106
48
47
4
7
-1
209
Part 3: Write the following reactions using transmutation
notation. Note that this is the reverse of Part 1.
Problem
Solution
13
16
1
13
16
7N + He => O + 1H
7N (a, p) 8O
Be9 (a,n) 6C12
4
9
12
1
+ on
4Be + He => C
4
2
4
2
13
Al27
16
8O
+
6
1
0n
=> 11
Na24
+
+ 1H1 => N13 +
7
23 +
Na
11
14
7N
8
4
2He
+
10 +
B
5
4
2He
4
2He
=>
12
13
4
2He
8
4
2He
Mg26 + 1H1
=> 8
O17 + 1H1
=>
N13 + on1
7
Al27 (n,a) 11Na24
O16 (p,a) 7N13
23 (a,p) Mg26
Na
11
12
14 (a, p) O17
N
7
8
10 (a,n) N13
B
5
7
Problems calculating mass defect
What is the mass defect for 3Li6, the lithium isotope with 3 neutrons and
an atomic weight of 6.01513?
2. Calculate the mass defect for 17Cl35 whose mass is 34.9689 AMU.
3. The actual mass of 29Cu63 is 62.9298 AMU. What is its mass defect?
4. The actual mass of 35Br81 is 80.9163 AMU. What is its mass defect?
5. The actual mass of 52Te120 is 119.9045 AMU. What is its mass defect?
1:
answers with solutions
isoto
mass of protons
pe
mass of
neutrons
mass of
electrons
act mass of
atom
mass
defect
6.0497
6.01513
.0346
3 * 1.0073
= 3.0219
3 * 1.0087=
3.0261
17 * 1.0073
= 17.124
18 *
17 * .00055
1.0087=18.157 = .0093
35.290
34.9689
.321
29Cu
63
29 * 1.0073
= 29.212
34 * 1.0087=
34.296
29 * .00055
= .016
63.523
62.9298
.594
35Br
81
35 * 1.0073
= 35.256
46 * 1.0087=
46.400
35 * .00055
= .019
81.675
80.9163
.759
52Te
120
52 * 1.0073
= 52.380
68 *
52 * .00055
1.0087=68.592 = .029
121.001
119.9045
1.096
6
3Li
17Cl
5
3
3 * .00055
= .0017
calc mass of
atom
Problems calculating energies associated with mass defects
2.
Calculate the energies associated with the mass defects for one mole of the atoms from the five problems above using E = mc
answers with solutions
1
.0345 AMU* 6.02E23 atoms * [1g/6.023E23AMU] [3.00E8m/s]2*[1kg/1000g] = 3.11E12 kg
m2/s2 = 3.11E12 J
2
.321 AMU* 6.02E23 atoms * [1g/6.023E23AMU] [3.00E8m/s]2*[1kg/1000g] = 2.89E13 kg
m2/s2 = 2.89E13 J
3
.594AMU* 6.02E23 atoms * [1g/6.023E23AMU] [3.00E8m/s]2*[1kg/1000g] = 5.35E13 kg
m2/s2 = 5.35E13 J
4
.759AMU* 6.02E23 atoms * [1g/6.023E23AMU] [3.00E8m/s]2*[1kg/1000g] = 6.84E13 kg
m2/s2 = 6.84E13 J
5
1.095AMU* 6.02E23 atoms * [1g/6.023E23AMU] [3.00E8m/s]2*[1kg/1000g] = 9.87E13 kg
m2/s2 = 9.87E13 J