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Q.
Density of nuclear matter is nearly
AMUAMU 2017
Solution:
As we know that,
density of the nucleus $ =\frac{\text{mass of nucleus}}{\text{volume of nucleus}}$
Mass of the nucleus $= A \times 1.66 \times 10^{-27}\, kg$,
where $A$ is the mass number of nucleus.
Volume of the nucleus $ = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi(R_0A^{1/3})^3$
$ = \frac{4}{3} \pi R_0^2 A$
Thus, density $ = \frac{A \times 1.66 \times 10^{-27}}{(\frac{4}{3} \pi R_0^3)A}$
$ = \frac{1.66\times 10^{-27}}{(\frac{4}{3} \pi R_0^3)}$
which shows that the density is independent of
mass number $A$ Using,
$R_0 = 1.2 \times 10^{-15}\, m$, we get
density $= 2.99 \times 10^{17}\, kg\, m^{-3}$