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Step-by-Step Detailed Solution
Step 1: Write Down the Formula for the Nuclear Radius
The radius $R$ of a nucleus with mass number $A$ is given by:
$R = (1.3 \times 10^{-15}) \, A^{\frac{1}{3}} \, \text{m}$
This indicates that the radius depends on the cube root of the mass number $A$.
Step 2: Express the Volume of the Nucleus
Assuming the nucleus is approximately spherical, its volume $V$ can be written as:
$V = \frac{4}{3} \pi R^{3}
Substituting $R$ from Step 1, we get:
$V = \frac{4}{3} \pi \left(1.3 \times 10^{-15} \right)^{3} A
because $ \left(A^{\frac{1}{3}}\right)^{3} = A$.
Step 3: Note the Mass of the Nucleus
A nucleus with mass number $A$ has a mass approximately equal to $m_p \times A$, where $m_p \simeq 1.67 \times 10^{-27}\,\text{kg}$ is the mass of a proton (or neutron).
Thus,
$m = m_p \times A
Step 4: Write the Expression for Density
Density $p$ (often denoted by $\rho$) is mass per unit volume:
$p = \frac{m}{V}
Substitute $m$ and $V$ from the earlier steps:
$p = \frac{m_p\,A}{\frac{4}{3}\,\pi\,R^3}
Using $R$ from the given relation:
$p = \frac{m_p\,A}{\frac{4}{3}\,\pi \left(1.3 \times 10^{-15}\right)^3 A}
The factor $A$ cancels out, illustrating that the density is nearly independent of $A$.
Step 5: Approximate the Numerical Value
When we plug in numerical values, we find that it simplifies to roughly:
$p \approx 10^{17} \,\text{kg}\,\text{m}^{-3}
This is the order of the nuclear density.
Final Answer
The mass density of a nucleus is of the order of $10^{17}\,\text{kg}\,\text{m}^{-3}$.
