Atoms & Nuclei

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Question 161 of 239

For a nucleus of mass number AA and radius RR, the mass density of nucleus can be represented as

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Answer: (D) Independent of A

The mass density ( ρ\rho) of a nucleus is defined as the ratio of its mass to its volume:
ρ=\rho= Mass of nucleus / Volume of nucleus
For a spherical nucleus with radius R , the volume is:
 Volume =(4/3)πR3\text { Volume }=(4 / 3) \pi R^3
The mass of the nucleus is related to the mass number A :
\text { Mass }-\mathrm{A} \times \mathrm{m} \text { _u }
(where m_u is the atomic mass unit, approximately 1.66×10−27 kg1.66 \times 10^{-27} \mathrm{~kg})
Therefore:
ρ=(A×m_u)/[(4/3)πR3]\rho=\left(\mathrm{A} \times \mathrm{m} \_\mathrm{u}\right) /\left[(4 / 3) \pi \mathrm{R}^3\right]
Now, there's an important relationship between the mass number A and nuclear radius R. Empirically, it has been found that:
R=Ro∗ A∧(1/3)\mathrm{R}=\mathrm{Ro} * \mathrm{~A}^{\wedge}(1 / 3)
Where Ro is a constant approximately equal to 1.2×10−15 m(1.21.2 \times 10^{-15} \mathrm{~m}(1.2 fermi).
Substituting this into the density equation:
$\begin{aligned}
& \rho=\left(\mathrm{A} \times \mathrm{m} \_\mathrm{u}\right) /\left[(4 / 3) \pi\left(\mathrm{Ro} \times \mathrm{A}^{\wedge}(1 / 3)\right)^3\right] \\
& \rho=(\mathrm{A} \times \mathrm{m} \text { _u }) /\left[(4 / 3) \pi \mathrm{Ro}^3 \times \mathrm{A}\right] \\
& \rho=\mathrm{m} \_\mathrm{u} /\left[(4 / 3) \pi \mathrm{Ro}^3\right]
\end{aligned}$
This shows that the nuclear density is approximately constant for all nuclei, regardless of their mass number. This is a fundamental property of nuclear matter known as nuclear saturation density.
The numerical value is:
ρ=2.3×1017 kg/m3\rho=2.3 \times 10^{17} \mathrm{~kg} / \mathrm{m}^3

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