Binding energy per nucleon, nuclear fusion and fission

1.

Today’s
Objectives
Binding energy per nucleon, Nuclear fusion and fission
(a) Sketch the variation of binding energy per nucleon with
nucleon number
(b) Explain the relevance of binding energy per nucleon to
nuclear fusion and fission
(c) Solve problems involving binding energy per nucleon and
other related questions.

2. Binding Energy Per Nucleon

The binding energy per nucleon of a nucleus is the binding energy divided by
the total number of nucleons in the nucleus
Binding energy per nucleon =
In the case of helium
total binding energy of nucleus
__________________________
number of nucleons in the nucleus
28.3
________
=
4
7.1 MeV
The great of nuclides have a value of around 8MeV per nucleon

3.

4.

The largest value occurs at around Fe-56.
This implies that Fe-56 is the most stable
nucleus.
Notice that uranium 235 a more massive
nucleus has a lower binding energy per
nucleon.
If you can “split” a nucleus of U-235 you will
release the excess binding energy, because
lower mass “fission products” with higher
binding energy per nucleon will be produced.
This releases a tiny amount of energy per
fission (less than 1 MeV) but this adds up to
18.7 million kWh per kg of U-235.)

5.

The key to this release is the
chain reaction produced by the
free neutrons realesed during
fission when the mass of U-235
exceeds a critical level (about
52kg- a sphere17cm in diameter

6.

Less massive nuclei
can release binding
energy if they are
fused together to
make larger nuclei.
Nuclear Fusion
powers the Sun
where hydrogen
nuclei are converted
into helium atoms in
a step by step
process known as
the proton-proton
chain

7.

Nuclear Binding Energy: per nucleon summary graph
Mass Number

8.

EXAMPLES
Units used for Nuclear Energy Calculations
electron volt - (ev)
The energy an electron acquires when it moves through
a potential difference of one volt:
1 ev = 1.602 x 10-19J
Binding energies are commonly expressed in units
of megaelectron volts (Mev)
1 Mev = 106 ev = 1.602 x 10 -13J
A particularly useful factor converts a given mass defect
in atomic mass units to its energy equivalent in electron
volts:
1 amu = 931.5 x 106 ev = 931.5 Mev

9. Binding Energy per Nucleon of Deuterium

Deuterium has a mass of 2.01410178 amu.
Hydrogen atom = 1 x 1.007825 amu = 1.007825 amu
Neutrons = 1 x 1.008665 amu = 1.008665 amu
2.016490 amu
Mass difference = Theoretical mass - actual mass
= 2.016490 amu - 2.01410178 amu = 0.002388 amu
Calculating the binding energy per nucleon:
Binding Energy
-0.002388 amu x 931.5 Mev / amu
=
Nucleon
2 nucleons
=

10.

Calculation of the Binding Energy per
Nucleon for Iron- 56
The mass of Iron -56 is 55.934939 amu, it contains 26 protons and
30 Neutrons
Theoretical Mass of Fe - 56 :
Hydrogen atom mass = 26 x 1.007825 amu = 26.203450 amu
Neutron mass = 30 x 1.008665 amu = 30.259950 amu
56.463400 amu
Mass defect =Actual mass - Theoretical mass:
55.934939 amu - 56.46340 amu = - 0.528461 amu
Calculating the binding energy per nucleon:
Binding Energy
- 0.528461 amu x 931.5 Mev / amu
=
nucleon
56 nucleons
=

11.

Calculation of the Binding Energy per
Nucleon for Uranium - 238
The actual mass of Uranium - 238 = 238.050785 amu, and it has
92 protons and 146 neutrons
Theoretical mass of Uranium 238:
Hydrogen atom mass = 92 x 1.007825 amu = 92.719900 amu
neutron mass = 146 x 1.008665 amu = 147.265090 amu
239.984990 amu
Mass defect = Actual mass - Theoretical mass:
238.050785 amu - 239.984990 amu = - 1.934205 amu
Calculating the Binding Energy per nucleon:
Binding Energy
-1.934205 amu x 931.5 Mev / amu
=
mucleon
238 nucleons
=

12.

Mass and Energy in Nuclear Decay - I
Consider the alpha decay of 212Po
212Po
211.988842 g/mol
208Pb
T1/2 = 0.3 s
+ + Energy
207.976627 g/mol + 4.00151 g/mol
Products = 207.976627 + 4.00151 = 211.97814 g/mol
Mass = Po - Pb + = 211.988842
211.97814
0.01070 g/mol
E = mC2 = (1.070 x 10-5 kg/mol)(3.00 x 108m/s)2
= 9.63 x 1011 J/mol
9.63 x 1011 J/mol
-12J/atom
=
1.60
x
10
6.022 x 1023 atoms/mol

13.

Mass and Energy in Nuclear Decay - II
The Energy for the Decay of 212Po is 1.60 x 10-12J/atom
1.60 x 10-12J/atom
1.602 x 10-19 J/ev
1.00 x 107 evx
atom
= 1.00 x 107 ev/atom
1.0 x 10-6 Mev = __________________ !!!!!
ev
The decay energy of the alpha particle from 212Po is = 8.8 Mev !!!!