Transcript Nuclear Physics and Radioactivity

Nuclear Physics and
Radioactivity
Vocabulary
alpha particle - positively charged particle consisting of two protons and
two neutrons. (Helium nucleus)
atomic mass number (A) - the number of protons and neutrons in the
nucleus of an atom.
atomic mass unit - the unit of mass equal to 1/12 the mass of a carbon-12
nucleus; the atomic mass rounded to the nearest whole number is called the
mass number.
atomic number (Z) - the number of protons in the nucleus of an atom.
beta particle - high speed electron emitted from a radioactive element
when a neutron.
decays into a proton
binding energy – the energy required to completely separate the nucleus
into its individual protons and neutrons.
element - a substance made of only one kind of atom.
isotope - a form of an element which has a particular number of neutrons,
that is, has the same atomic number but a different mass number than the
other elements which occupy the same place on the periodic table.
Vocabulary
mass defect - the mass equivalent of the binding energy in the nucleus of
an atom by E = mc2
neutron - an electrically neutral subatomic particle found in the nucleus of
an atom
nuclear reaction - any process in the nucleus of an atom that causes the
number of protons and/or neutrons to change
nucleons - protons or a neutrons
strong nuclear force - the force that binds protons and neutrons together
in the nucleus of an atom
transmutation - the changing of one element into another by a loss of
gain of one or more protons
Equations and symbols
E  mc 2
1u  1.66x10 27 kg  931MeVc 2
where
ΔE = binding energy of the nucleus
Δm = mass defect of the nucleus
c = speed of light = 3 x 108 m/s
u = atomic mass unit
X = element symbol
A = atomic mass number (number of protons and
neutrons)
Z = atomic number (number of protons)
A
Z
X
Particle
Symbol
proton
1
1
neutron
electron
H
1
0
0
1
e
n
or e-
Relative mass
Charge
Location
1
+1
nucleus
1
0
nucleus
0
-1
electron
orbitals
around the
nucleus
Find the number of protons, electrons and
neutrons in a neutral atom of iron.
# protons + #neutrons
56
26
Fe
# protons
26 protons
30 neutrons
neutral → #protons = # electrons = 26
Before
M1
After
M2
M3
Eo <=
Mo >
=
Ef
Mf
Eo = Ef + (Mo-Mf)c2
Eo = Ef + (Δm)c2
MASS is transferred to ENERGY
14
7
N
n n n
n n n n p
p p p p p p
14
7
N
n n n
n n n n p
p p p p p p
Mass defect is responsible for the binding energy.
Ebinding = (Δm)c2
Calculate the binding energy of Nitrogen. The
atomic mass of Nitrogen is 14.003074 u.
Nucleon
Proton
Neutron
Mass (u)
1.00782
1.00866
14
7
N
7 protons
7 neutrons
Mass of individual protons and neutrons
7(1.00782u) + 7(1.00866u) = 14.11536u

Δm = 14.11536u – 14.003074u = .112286u
931MeVc 2
.112286u 
 105MeVc 2
1u
E  mc 2  105MeVc 2 c 2  105MeV
Transmutation of nitrogen into carbon
14
7
N n C  H
1
0
14
6
1
1
Nuclide/particle
14
N
neutron
Mass of products
14.003074u + 1.008664u = 15.011738u

14
C
proton

Mass of reactants
14.003241u + 1.007825u = 15.011066u

Δm = 15.011738u – 15.011066u = .000672u
931MeVc 2
.000672u 
 0.626MeVc 2
1u
E  mc 2  0.626MeVc 2 c 2  0.626MeV
Mass (u)
14.003074
1.008664
14.003241
1.007825
Nuclear fusion – The joining of two small nuclei to form
one large nucleus. The mass of the smaller nuclei is greater
than the mass of the large nucleus. High temperatures are
required for fusion.
Example: the sun
Nuclear fission – Splitting a large nucleus into two smaller
nuclei. The mass of the large nucleus is greater than the
two smaller nuclei.
Example: Nuclear power plant
In both cases the mass of the products is less
than the mass of the reactants, which results in a
mass defect. The mass is converted to energy
according to the equation E = (Δm)c2
Binding energy per nucleon vs. mass number
Fission
Iron (Fe)
staple
fusion
Binding energy
Per nucleon
(MeV)
Mass number
 
 

Fusion Examples:
Nuclide
1
H
2
H
3
H
3
He
4
He
1
0
n
Mass (u)
1.007825
2.014101
3.016049
3.016029
4.002603
1.008664



2
1
3
1
4
2
1
0
H H He n
2
1
3
2
1
0
2
1
3
1
H H He n
2
1
2
1
2
1
1
1
H H H p
3
2
4
2
1
1
H He He p
2
1
3
1
4
2
1
0
H H He n
Nuclide
1
H
2
H
3
H
3
He
4
He
1
0
n
Mass (u)
Mass of reactants
2.014101u + 3.016049u = 5.03015u
1.007825
2.014101
3.016049
3.016029
4.002603
Mass of products
4.002603u + 1.008664u = 5.011267u
Δm = 5.03015u – 5.011267u = .018883u
1.008664
931MeVc 2
.018883u 
 17.5MeVc 2
1u
E  mc 2  17.5MeVc 2 c 2  17.5MeV

Fission Example:
236
92
and
U absorbs a neutron and splits into 100
42 Mo
126
, Write
the equation for the nuclear reaction and
50 Sn
calculate the energy released in this reaction.
Z
92
42
50
symbol
U
Mo
Sn
A
236
100
126
Mass (u)
236.045563
99.907476
125.907653
Fission Example:
236
92
and
U absorbs a neutron and splits into 100
42 Mo
126
, Write
the equation for the nuclear reaction and
50 Sn
calculate the energy released in this reaction.
U  n Mo 
236
92
1
0
100
42
126
50
Sn  neutrons
U  n Mo 
Sn  ? n 
U  n Mo 
Sn 11 n 
236
92
236
92
1
0
1
0
100
42
100
42
126
50
126
50
1
0
1
0
Z
92
42
50
symbol
U
Mo
Sn
A
236
100
126
U  n Mo 
236
92
1
0
100
42
126
50
Mass (u)
236.045563
99.907476
125.907653
Sn 11 n 
Mass of products
236.045563u + 1.008664u = 237.054227
Mass of reactants
125.907653u + 99.907476u + 11(1.008664u) = 236.910433u
Δm = 237.054227u – 236.910433u = .143794u
931MeVc 2
.143794u 
 134 MeVc 2
1u
E  mc 2  134 MeVc 2 c 2  134 MeV
1
0
Radioactivity – Particles are randomly
emitted from an unstable nucleus in order to
become more staple. These are the different
particles emitted.
Name
symbol
Alpha
α
Beta
β-
positron
β+
gamma

4
2
He
0
1
0
1
e
e
High Energy
Photons
Nuclear Equations:
U  Th  
238
92
Th Pa  ?
234
91
? Al  P  n
27
13
1
0
30
15
n p 
1
1
1
0
Nuclear Equations:
U Th  
238
92
238
92
234
4
2
U90Th He
Alpha decay
Beta decay
Nuclear Equations:
Th Pa -1?
234
90
90
234
234
91
234
91
0
0
1
Th Pa e

Th Pa    
90
234
234
91
 = anti - neutrino
Nuclear Equations:
4
2
4
2
? Al  P  n
27
13
27
13
30
15
30
15
1
0
1
0
He Al P n
  Al P n
27
13
30
15
Transmutation
1
0
Nuclear Equations:
1
0
1
0

1
0
n
1
1
0
p -1?
0
1
n p e
1
0

1
1
n p  
1
1

n p    
1
1

So a neutron is the combination of a proton and an electron