Environmental Engineering Reference
In-Depth Information
Table 2.3
Various Particle and Nuclide Masses (1 u = 1.660565·10
-27
kg)
Particle or nuclide
Mass
Particle or nuclide
Mass
Electron (e
-
)
0.00054858 u
Hydrogen (
1
H)
1.007825032 u
Proton (
1
p)
1.00727647 u
Helium (
4
He)
4.002603250 u
Neutron (
1
n)
1.008664923 u
Alpha particle (
4
α
)
4.0015060883 u
For this reaction the result is:
∆
m
= 4
•
1.00727647 u - 4.0015060883 u - 2
•
0.00054858 u = 0.02650263 u
Thus, the total mass of all particles after the fusion is less than that before.
The mass difference is converted into energy
∆
E
, with the relationship:
E
=
∆
∆
m
•
c
2
(2.3)
With the speed of light c = 2.99792458
•
10
8
m/s, this equation determines the
E
= 3.955
•
10
-12
J = 24.687 MeV. The
binding energy
E
b
of a nucleus explains the different masses after the fusion
as well as the energy difference. An atomic nucleus consists of N neutrons
1
n
and Z protons
1
p. To maintain equilibrium, this binding energy has to be
released during the assembly of a nucleus with protons and neutrons. The mass
difference of the alpha particle and the two neutrons together with the two
protons determines the binding energy of a helium nucleus.
So far, only the atomic nuclei have been considered; the electrons in the
atomic shell have not been taken into account. There is one electron in the
atomic shell of a hydrogen atom
1
H, while there are two electrons in the
helium atom
4
He. During the nuclear fusion process, two of the four electrons
of the hydrogen atoms become the atomic shell electrons of the helium atom.
The two other electrons and the positrons convert directly into energy. This
radiative energy is four times the equivalent mass of an electron of 2.044 MeV.
The total energy released during the reaction is thus 26.731 MeV. This very
small amount of the energy does not appear to be significant at first glance;
however, the enormous number of fusing nuclei results in the release of vast
quantities of energy.
The sun loses 4.3 million metric tonnes of mass per second (
energy released by the fusion as
∆
m
4.3
•
10
9
∆
kg/s). This results in the solar radiant power
φ
e,S
of:
m
•
c
2
= 3.845
•
10
26
W
φ
e,S
=
∆
(2.4)
This value divided by the sun's surface area,
A
S
, provides the
specific emission
of the sun
:
