The density of nuclear matter is the ratio of the mass of a nucleus to its volume. As the volume of a nucleus is directly proportional to its mass number A, so the density of nuclear matter is independent of the size of the nucleus. Thus, the nuclear matter behaves like a liquid of constant density. Different nuclei are like drops of this liquid, of different sizes but of same density. Let A be the mass number and R be the radius of a nucleus. If m is the average mass of a nucleon, then. Mass of nucleus = mA
$\text{Volume of nucleus }=\frac{4}{3}\pi\text{R}^3=\frac{4}{3}\pi\big(\text{R}_0\text{A}\frac{1}{3}\big)=\frac{4}{3}\pi\text{R}^3_0\text{A}$
$\therefore\text{Nuclear density},\rho\text{nu}=\frac{\text{Mass of nucleus}}{\text{Volume of nucleus}}\text{or}\ \rho\text{nu}$
$=\frac{\text{MA}}{\frac{4}{3}\pi\text{R}_0^3\text{A}}=\frac{3\text{m}}{4\pi\text{R}_0^3}$
Clearly, nuclear density is independent of mass number A or the size of the nucleus. The nuclear mass density is of the order 1017kg m-3 This density is very large as compared to the density of ordinary matter, say water, for which
$\rho$ = 1.0 × 103kg m-3 - The nuclear radius of
$_8^{16}\text{O}$ is 3 × 10-15m. The density of nuclear matter is. - 2.9 × 1034kg m-3
- 1.2 × 1017kg m-3
- 16 × 1027kg m-3
- 2.4 × 1017kg m-3
- What is the density of hydrogen nucleus in SI units? Given Ro= 1.1 fermi and mP = 1.007825 amu.
- 1.2 × 1017kg m-3
- 3.0 × 1034kg m-3
- 1.99 × 1011kg m-3
- 7.85 × 1017kg m-3
- Density of a nucleus is.
- More for lighter elements and less for heavier elements.
- More for heavier elements and less for lighter elements.
- Very less compared to ordinary matter.
- A constant.
- The nuclear mass of
$_{23}^{56}\text{Fe}$ is 55.85 amu. The its nuclear density is. - 5.0 × 1019kg m-3
- 1.5 × 1019kg m-3
- 2.9 × 1017kg m-3
- 9.2 × 1026kg m-3
- If the nucleus of
$_{13}^{27}\text{Al}$ has s a nuclear radius of about 3.6 fm, then $_{52}^{125}\text{Te}$ would have its radius approximately as. - 9.6fm
- 12fm
- 4.8fm
- 6fm