Question

Prove that nuclear mass density is constant
assuming that mass of proton and neutron is
equal.​

Answer 1

Answer:

Jay here

Nuclear density is the density of the nucleus of an atom, averaging about 2.3×1017 kg/m3. The descriptive term nuclear density is also applied to situations where similarly high densities occur, such as within neutron stars.

The nuclear density of an 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 {\displaystyle R=A^{1/3}R_{0}}{\displaystyle R=A^{1/3}R_{0}} where {\displaystyle A}A is the mass number and {\displaystyle R_{0}}R_{0} is 1.25 fm, with typical deviations of up to 0.2 fm from this value. The density of the nucleus is thus:

{\displaystyle n={\frac {A}{{4 \over 3}\pi R^{3}}}}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:

{\displaystyle n={\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\cdot 10^{44}\ \mathrm {m} ^{-3}}{\displaystyle n={\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\cdot 10^{44}\ \mathrm {m} ^{-3}}

The experimentally determined value for n is 0.16 fm−3, that is 1.6·1044 m−3.

The mass density is the product of n by the nuclear mass. The calculated mass density, using a nucleon mass of 1.67×10−27 kg, is thus:

{\displaystyle (1.67\cdot 10^{-27}\ \mathrm {kg} )(1.22\cdot 10^{44}\ \mathrm {m} ^{-3})=2.04\cdot 10^{17}\ \mathrm {kg} \cdot \mathrm {m} ^{-3}}(1.67\cdot 10^{{-27}}\ {\mathrm {kg}})(1.22\cdot 10^{{44}}\ {\mathrm {m}}^{{-3}})=2.04\cdot 10^{{17}}\ {\mathrm {kg}}\cdot {\mathrm {m}}^{{-3}}

(For nucleon density {\displaystyle d=M/V}{\displaystyle d=M/V}. Where {\displaystyle M=mA}{\displaystyle M=mA} and {\displaystyle V={\frac {4}{3}}\pi R^{3}}{\displaystyle V={\frac {4}{3}}\pi R^{3}} {\displaystyle ={\frac {4}{3}}\pi (R_{0}.A^{1/3})^{3}}{\displaystyle ={\frac {4}{3}}\pi (R_{0}.A^{1/3})^{3}} {\displaystyle ={\frac {4}{3}}\pi A.R_{0}^{3}}{\displaystyle ={\frac {4}{3}}\pi A.R_{0}^{3}}.So {\displaystyle d={\frac {3m}{4\pi R_{0}^{3}}}}{\displaystyle d={\frac {3m}{4\pi R_{0}^{3}}}})