Physics Asked on April 1, 2021
I am learning high school modern physics. I came across a formula that is Radius of nucleus is directly proportional to cube root of atomic mass. However upon searching some of books( which I could actually afford to read at high school level) I could not found a satisfactory derivation for this result. Is this an experimental result or there is a possible derivation to this.
Ur answer would be of great help
The strong nuclear force between nucleons (protons or neutrons) becomes rapidly and extremely repulsive if you try and cram them close together. The force is also "short range" which means that only adjacent nucleons will exert a force on each other. This means that to first order they behave like "hard" balls when you pack them together in a nucleus.
Since the volume of such a nucleus would be proportional to (again to first order) the number of nucleons (i.e. the mass number), then the radius will be proportional to the cube root of the volume and hence the cube root of the number of nucleons.
Answered by ProfRob on April 1, 2021
From a variety of experiments, we know some general features of nuclear density. . Because the nuclear force is the strongest of the forces, we might expect that this strong force would cause the protons and neutrons to congregate at the center of the nucleus, giving an increased density in the central region. However, it turns out ( experimentally) that this is not the case—the density remains quite uniform.
Another interesting feature that experiments show is that the density of a nucleus seems not to depend on the mass number $A$; very light nuclei, such as $C^{12}$, have roughly the same central density as very heavy nuclei, such as $Bi^{209}$. Stated another way, the number of protons and neutrons per unit volume is approximately constant over the entire range of nuclei:
$$frac{mathrm{Number of neutrons and protons}}{mathrm{Volume of nucleus}}=frac{A}{frac{4}{3}pi R^3}approx mathrm{constant}$$ assuming the nucleus to be a sphere of radius $R$. Thus $Apropto R^3$, which suggests a proportionality between the nuclear radius $R$ and the cube root of the mass number; $Rpropto A^{1/3}$ or, defining a constant of proportionality $R_0$, $$R=R_0A^{1/3}$$ The constant $R_0$ must be determined by experiment, and a typical experiment might be to scatter charged particles from the nucleus and to infer the radius of the nucleus from the distribution of scattered particles.
~References :
Answered by Young Kindaichi on April 1, 2021
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