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Question : 28 of 33
Marks: +1, -0
The unit of length convenient on the nuclear scale is a fermi : 1 f = 10−15^{-15} m. Nuclear sizes obey roughly the following empirical relation :
r=r0A1/3r = r_0 A^{1/3}
where r is the radius of the nucleus, A its mass number, and r0r_0 is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in question 27.
Solution:  
Let m be the average mass of a nucleon (neutron or proton).As the nucleus contains A nucleons,
∴  \therefore\; mass of nucleus M=mA,M = m A, radius of nucleus r=r0A1/3r = r_0 A^{1/3}
Nuclear density, ρ=massvolume\rho = \frac{\text{mass}}{\text{volume}}
=M43πr3= \frac{M}{\frac{4}{3} \pi r^3}
=3mA4π(r0A1/3)3= \frac{3 m A}{4 \pi (r_0 A^{1/3})^3}
=3m4πr03= \frac{3 m}{4 \pi r_0^3}
As m and r0r_0 are constant, therefore, nuclear density is constant for allnuclei.
Using m=1.66×10−27kgm = 1.66 \times 10^{-27} \text{kg} and r0=1.2f=1.2×10−15mr_0 = 1.2 \text{f} = 1.2 \times 10^{-15} \text{m}
we get ,ρ=3m4πr03,\rho = \frac{3 m}{4 \pi r_0^3}
=3×1.66×10−274×3.14(1.2×10−15)3= \frac{3 \times 1.66 \times 10^{-27}}{4 \times 3.14 (1.2 \times 10^{-15})^3}
=2.29×1017kgm−3= 2.29 \times 10^{17} \text{kg} \text{m}^{-3}.
As ρ is constant for all nuclei, this must be the density of sodium nucleusalso.
Density of sodium atom, ρ′=0.58×103kgm−3\rho' = 0.58 \times 10^{3} \text{kg} \text{m}^{-3}
ρρ′=2.29×10170.58×103\frac{\rho}{\rho'} = \frac{2.29 \times 10^{17}}{0.58 \times 10^{3}}
=4×1014= 4 \times 10^{14}
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