- Source: Nuclear density
Nuclear density is the density of the nucleus of an atom. For heavy nuclei, it is close to the nuclear saturation density
n
0
=
0.15
±
0.01
{\displaystyle n_{0}=0.15\pm 0.01}
nucleons/fm3, which minimizes the energy density of an infinite nuclear matter. The nuclear saturation mass density is thus
ρ
0
=
n
0
m
u
≈
2.5
×
10
17
{\displaystyle \rho _{0}=n_{0}m_{\rm {u}}\approx 2.5\times 10^{17}}
kg/m3, where mu is the atomic mass constant. The descriptive term nuclear density is also applied to situations where similarly high densities occur, such as within neutron stars.
Evaluation
The nuclear density of a typical nucleus can be approximately calculated from the size of the nucleus, which itself can be approximated based on the number of protons and neutrons in it. The radius of a typical nucleus, in terms of number of nucleons, is
R
=
A
1
/
3
R
0
{\displaystyle R=A^{1/3}R_{0}}
where
A
{\displaystyle A}
is the mass number and
R
0
{\displaystyle R_{0}}
is 1.25 fm, with typical deviations of up to 0.2 fm from this value. The number density of the nucleus is thus:
n
=
A
4
3
π
R
3
{\displaystyle n={\frac {A}{{4 \over 3}\pi R^{3}}}}
The density for any typical nucleus, in terms of mass number, is thus constant, not dependent on A or R, theoretically:
n
0
t
h
e
o
r
=
A
4
3
π
(
A
1
/
3
R
0
)
3
=
3
4
π
(
1.25
f
m
)
3
=
0.122
f
m
−
3
=
1.22
×
10
44
m
−
3
{\displaystyle n_{0}^{\mathrm {theor} }={\frac {A}{{4 \over 3}\pi (A^{1/3}R_{0})^{3}}}={\frac {3}{4\pi (1.25\ \mathrm {fm} )^{3}}}=0.122\ \mathrm {fm} ^{-3}=1.22\times 10^{44}\ \mathrm {m} ^{-3}}
The experimentally determined value for the nuclear saturation density is
n
0
e
x
p
=
0.15
±
0.01
f
m
−
3
=
(
1.5
±
0.1
)
×
10
44
m
−
3
.
{\displaystyle n_{0}^{\mathrm {exp} }=0.15\pm 0.01\ \mathrm {fm} ^{-3}=(1.5\pm 0.1)\times 10^{44}\ \mathrm {m} ^{-3}.}
The mass density ρ is the product of the number density n by the particle's mass. The calculated mass density, using a nucleon mass of mn=1.67×10−27 kg, is thus:
ρ
0
t
h
e
o
r
=
m
n
n
0
t
h
e
o
r
≈
2
×
10
17
k
g
m
−
3
{\displaystyle \rho _{0}^{\mathrm {theor} }=m_{\mathrm {n} }\,n_{0}^{\mathrm {theor} }\approx 2\times 10^{17}\ \mathrm {kg} \ \mathrm {m} ^{-3}}
(using the theoretical estimate)
or
ρ
0
e
x
p
=
m
n
n
0
e
x
p
≈
2.5
×
10
17
k
g
m
−
3
{\displaystyle \rho _{0}^{\mathrm {exp} }=m_{\mathrm {n} }\,n_{0}^{\mathrm {exp} }\approx 2.5\times 10^{17}\ \mathrm {kg} \ \mathrm {m} ^{-3}}
(using the experimental value).
Applications and extensions
The components of an atom and of a nucleus have varying densities. The proton is not a fundamental particle, being composed of quark–gluon matter. Its size is approximately 10−15 meters and its density 1018 kg/m3. The descriptive term nuclear density is also applied to situations where similarly high densities occur, such as within neutron stars.
Using deep inelastic scattering, it has been estimated that the "size" of an electron, if it is not a point particle, must be less than 10−17 meters. This would correspond to a density of roughly 1021 kg/m3.
There are possibilities for still-higher densities when it comes to quark matter. In the near future, the highest experimentally measurable densities will likely be limited to leptons and quarks.
See also
Electron degeneracy pressure
Nuclear matter
Quark–gluon plasma
References
External links
"The Atomic Nucleus". Retrieved 2014-11-18. (derivation of equations and other mathematical descriptions)
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