Question

The unit of length convenient on the nuclear scale is Fermi: 1 f = 10⁻¹⁵ m. Nuclear sizes obey roughly the following empirical relation: r = r₀ A⁽¹÷³⁾ ,where r is the radius of the nucleus. A, its mass number and r₀ is a constant equal to about 1.2 f. Show that the rule applies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of the sodium nucleus. Compare it with the average mass density of sodium atom of size 2.5 Å.

Answer 1

Hii dear,

# Step-by-step Explaination-
# Given-
r0 = 1.2 f = 1.2×10^-15 m
M = A amu = A×1.66×10^–27 kg

# Solution-
Radius of nucleus r is given by the relation,
r = r0.A^(1/3)

Volume of nucleus,
V = (4/3)πr^3
V = (4/3)π[r0.A^(1/3)]^3
V = (4/3)πr0^3.A ...(i)

Density of nucleus,
ρ = Mass of nucleus / Volume of nucleus
ρ = AX1.66×10^-27 / (4/3)πr0^3.A
ρ = 3X1.66×10^-27 / 4πr^03 Kg m-3
ρ = 4×10^−28 / r0^3

This relation shows that nuclear mass depends only on constant r0. Hence, the nuclear mass densities of all nuclei are nearly the same.

Density of sodium nucleus is given by,
ρSodium = 4×10^-28 /(1.2×10^-15)^3
ρSodium = 2.29×10^17 Kg/m^-3

Hence density of sodium is
2.29×10^17 Kg/m^-3.

Hope this is useful...

Answer 2

$\huge\underline{Explanation:-}$

Volume of a nucleus of mass number A is

$v \: = \frac{4}{3} \pi {r}^{3} = \frac{4}{3} \pi {(r _{o}A \frac{1}{3})}^{3} = \frac{4}{3} \pi {r}^{3} _{o}A$

Mass of a nucleus of mass number A m

= Avg mass of a nucleus × Avg number of nucleus

$= m _{o}A \: (m _{o} = 1.66 \times 10 ^{ - 27} )$

Avg density of nucleus:-

$d = \frac{m}{v} \\ \\ \: \: \: \: \: \: \: \:= \frac{m_{o}A} { \frac{4}{3} \pi {r}^{3}_{o} } \\ \\ = \frac{3m _{o} }{4\pi {r}^{3} _{o} }$

Independent of A.

$\implies$ Constant for all nuclie.

For, Na or any nuclie :-

$d = \frac{3 \times 1.66 \times 10 ^{ - 27} }{ ({4 \times 3.14 \times 1.2 \times 10^{ - 15}) }^{3} } \\ = 0. 3\times {10}^{18} kg {m}^{3}$

Comparing with the density of atoms = d nuclie/d atoms

$= \frac{0.3 \times 10 ^{18} }{584} \\ = {10}^{15} times$