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Two nuclei have their mass numbers in the ratio of 1 : 3. The ratio of their nuclear densities would be -
A
$(3)^{1/3} : 1$
B
1 : 1
C
1 : 3
D
3 : 1
Detailed Solution
Radius of a nucleus: $R = R_0 A^{1/3}$, where $A$ is the mass number.
Mass of the nucleus $\approx A\,m$, where $m$ is the mass of each nucleon.
Volume of the nucleus $= \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi R_0^3 A$
Nuclear density $\rho = \dfrac{\text{mass}}{\text{volume}} = \dfrac{A\,m}{\frac{4}{3}\pi R_0^3 A} = \dfrac{3m}{4\pi R_0^3}$
The mass number $A$ cancels, so the nuclear density is independent of $A$ and is the same for all nuclei.
Hence the ratio of the nuclear densities is 1 : 1.
Mass of the nucleus $\approx A\,m$, where $m$ is the mass of each nucleon.
Volume of the nucleus $= \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi R_0^3 A$
Nuclear density $\rho = \dfrac{\text{mass}}{\text{volume}} = \dfrac{A\,m}{\frac{4}{3}\pi R_0^3 A} = \dfrac{3m}{4\pi R_0^3}$
The mass number $A$ cancels, so the nuclear density is independent of $A$ and is the same for all nuclei.
Hence the ratio of the nuclear densities is 1 : 1.