From the relation $R = R_0 A^{1/3}$, where $R_0$ is a constant and $A$ is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of $A$).

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> [!TLDR]
> Nuclear density is nearly constant because both the mass and the volume of a nucleus scale as $A$, so their ratio — density — becomes independent of $$A$$. The result is $ho \approx 2.3 	imes 10^{17} \ 	ext{kg/m}^3$.

The key insight is that the nuclear radius follows $R = R_0 A^{1/3}$, where $R_0 \approx 1.2 \ 	ext{fm}$ is a constant. This means the volume of a nucleus grows in proportion to its mass number $A$. Since the mass of the nucleus is also proportional to $A$ (each nucleon has roughly the same mass), the density — mass per unit volume — ends up being independent of $A$.

Let’s walk through the reasoning step by step.

1. **Mass of the nucleus**  
   The mass number $A$ tells us the total number of nucleons (protons + neutrons). Each nucleon has a mass approximately equal to $1.67 	imes 10^{-27} \ 	ext{kg}$ (the atomic mass unit $u$). So the nuclear mass is  
   $$ M \approx A \cdot m_p $$  
   where $m_p$ is the proton mass (we ignore the small neutron-proton mass difference and binding energy effects, which are negligible here).

2. **Volume of the nucleus**  
   The radius is given by $R = R_0 A^{1/3}$. Assuming the nucleus is roughly spherical, its volume is  
   $$ V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A $$  
   Notice that $A$ appears linearly — the volume is directly proportional to $A$.

3. **Density calculation**  
   Nuclear matter density $ho$ is mass divided by volume:  
   $$ ho = \frac{M}{V} = \frac{A m_p}{\frac{4}{3} \pi R_0^3 A} = \frac{m_p}{\frac{4}{3} \pi R_0^3} $$  
   The $A$ cancels out completely. This is the central result: density does not depend on $A$.

4. **Numerical value**  
   Using $R_0 = 1.2 \ 	ext{fm} = 1.2 	imes 10^{-15} \ 	ext{m}$ and $m_p = 1.67 	imes 10^{-27} \ 	ext{kg}$, …

Q.From the relation R=R0A1/3R = R_0 A^{1/3}R=R0​A1/3, where R0R_0R0​ is a constant and AAA is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of AAA).

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Q.From the relation $R = R_0 A^{1/3}$ , where $R_0$ is a constant and $A$ is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of $A$ ).

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