From the equation $R = {R_ \circ }{A^{1/3}}$ show that the nuclear matter density is nearly constant. [Where ${R_ \circ }$ is a constant and $A$ is the mass number.]
Answer
597.6k+ views
Hint:Here we have to use the formula for density of nuclear matter.Nuclear matter is an idealised nucleon-interacting system that operates in many phases that are not yet fully formed. It is a fictional material, not matter of a nucleus. Composed of a vast number of protons and neutrons interacting only with nuclear forces and not with the coulomb forces. The mass of an atom’s nucleus, averaging about $2.3 \times
1017\,kg/{m^3}$, is the nuclear energy. For conditions where equally high densities exist, such as inside of neutron stars, the descriptive term nuclear density is often added.
Complete step by step solution:
Given, $R = {R_ \circ }{A^{1/3}}$
From this equation we can deduce that $A = $ mass of the nucleus, $R = $Radius of the nucleus,${R_ \circ }$ is a constant. We know that density is used to define how much space an object or substance takes up in contrast to the amount of matter in the object or substance. The formula for density is given by –
Density =mass/volume
We shall apply this formula itself to show that the nuclear matter density is nearly constant.
Density of nuclear matter = mass of nucleus/volume of nucleus
$\Rightarrow$ Density of nuclear matter $= \dfrac{A}{{\dfrac{4}{3}\pi {R^3}}}$
$\Rightarrow$ Density of nuclear matter $ = \dfrac{A}{{\dfrac{4}{3}\pi {{({R_ \circ }{A^{1/3}})}^3}}} \\ $
$\therefore$ Density of nuclear matter $=\dfrac{1}{{\dfrac{4}{3}\pi {R_ \circ }^3}} = $ constant
The value of constant nuclear density is almost $2.3 \times {10^{17}}\,kg{m^{ - 3}}$. Nuclear density is of the order of $1017\,kg{m^{ - 3}}$.
Hence, it is proved that the nuclear matter density is nearly constant.
Note:Nuclear matter density is nearly constant implies that the nuclear matters are all approximately the same distance apart, and since protons and neutrons have virtually equal mass, regardless of how many nucleons there are, there would be the same mass for a given amount, thus the constant density. Also nuclear density is independent of mass numbers.
1017\,kg/{m^3}$, is the nuclear energy. For conditions where equally high densities exist, such as inside of neutron stars, the descriptive term nuclear density is often added.
Complete step by step solution:
Given, $R = {R_ \circ }{A^{1/3}}$
From this equation we can deduce that $A = $ mass of the nucleus, $R = $Radius of the nucleus,${R_ \circ }$ is a constant. We know that density is used to define how much space an object or substance takes up in contrast to the amount of matter in the object or substance. The formula for density is given by –
Density =mass/volume
We shall apply this formula itself to show that the nuclear matter density is nearly constant.
Density of nuclear matter = mass of nucleus/volume of nucleus
$\Rightarrow$ Density of nuclear matter $= \dfrac{A}{{\dfrac{4}{3}\pi {R^3}}}$
$\Rightarrow$ Density of nuclear matter $ = \dfrac{A}{{\dfrac{4}{3}\pi {{({R_ \circ }{A^{1/3}})}^3}}} \\ $
$\therefore$ Density of nuclear matter $=\dfrac{1}{{\dfrac{4}{3}\pi {R_ \circ }^3}} = $ constant
The value of constant nuclear density is almost $2.3 \times {10^{17}}\,kg{m^{ - 3}}$. Nuclear density is of the order of $1017\,kg{m^{ - 3}}$.
Hence, it is proved that the nuclear matter density is nearly constant.
Note:Nuclear matter density is nearly constant implies that the nuclear matters are all approximately the same distance apart, and since protons and neutrons have virtually equal mass, regardless of how many nucleons there are, there would be the same mass for a given amount, thus the constant density. Also nuclear density is independent of mass numbers.
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