nuclear binding
energy
emre özyetiş 10-U
The binding
energy is the
energy required
to decompose
the nucleus
into protons
and neutrons.
In chemical reactions,
the nucleus remains the
same; no new atoms are
formed. The number and
kinds of atoms are
conserved. There is no
extra energy given off
from nucleus besides the
energy given off or taken
in by the breakage or
forming of bonds.
However, in
nuclear
reactions,
divisions and
combinations of
nuclei occur.
New atoms are
formed; rays and
particles with
energy are
emitted.
When new atoms are formed, binding energy is
given off to keep the neutrons and protons
together and to become stable.
This energy can be calculated by Einstein’s
famous formula.
E=mc2
energy=mass×(speed of light) 2
According to The
Relativity Theory of
Einstein, some of the
mass is converted into
energy and emitted so
that more stable atoms
are formed. This is why in
nuclear reactions energy
and mass are conserved
together.
ΔE=Δmc2
Where Δ E is binding energy
Δm is mass defect:
(m n0 + m p+)- m element itself
c is speed of light in vacuum which
is 2.99×108 m×s-1
When units are kg for mass defect
and m.s-1 for speed of light,
then we get kg×m2×s-2 which is
Joule.
To make comparison between different nuclides easily, we
express binding energies on a per-nucleon.
Example.1b will give information about BE/nucleon
Example.1:
Calculate
a) the change in energy if 1 mol 168 O nuclei were
formed from neutrons and protons.
b) Binding Energy per nucleon.
Mass of 10 n= 1.67493×10-24 g
Mass of
Mass of
1
1 p=
16
8O
1.67262×10-24 g
nucleus is 2.65535×10-23g
For Δm we ought to know the equation
of formation
16
1
1
8 0n + 8 1 p
8O
In nucleus the difference is=
(Mass of O) – (Mass of 8p and 8n)
= -2.269×10-25g
The difference in 1 mole is
= (6.022×1023nuclei/mol) × -2.269×10-25g/nucleus
Δm = -0.1366 g/mol
a)
ΔE= Δmc2
ΔE= -0.1366 g/mol × (3.00×108m/s)2 × 1 kg/1000g
ΔE= -1.23×1013 J/mol
b)
In order to find ΔE per nucleon, we
need to know ΔE per nucleus first
ΔE per O nucleus
-1.23×1013 J/mol
=
6.022×1023nuclei/mol
=-2.04×10-11 J/nucleus
In terms of a more convenient
energy unit, a million electron volts
where; 1 MeV = 1.60 × 10-13 J
-2.04×10-11 J/nucleus
=-1.28 × 102 MeV/nucleus
At last we can calculate ΔE per
nucleon by dividing by the sum of
neutrons and protons; which is 16
16
for 8 O
-1.28 × 102 MeV/nucleus
=
16 nucleons/nucleus
ΔE per nucleon for
=
16
8
O is
-7.98 MeV/nucleon
Example.2:
In the radioactive
decomposition of radium-226,
the equation for the nuclear
process is
226
88 Ra
222
86
Rn + 42 He
How much mass is converted
into energy during this
radioactive decay process?
Ra: 226.025360 amu
Rn: 222.017530 amu
He: 4.00260361 amu
Δm= m produced-m reacted
Δm=( 222.017530amu + 4.00260361amu ) - 226.025360amu
Δm= -5.226×10-3 amu is lost in this process.
This means that amount of mass is converted
into energy
ΔE=Δmc2
ΔE= (-5.226×10-3 g/mol) × 1 kg/1000g × (3.00×108 m/s)2
ΔE= -4.70×1011 J/mol
Negative sign indicates that the process is exothermic
The binding energy curve is
obtained by dividing the total
nuclear binding energy by the
number of nucleons. The fact that
there is a peak in the binding
energy curve in the region of
stability near iron means that either
the breakup of heavier nuclei
(fission) or the combining of lighter
nuclei (fusion) will yield nuclei
which are more tightly bound (less
mass per nucleon).