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 $A$).

Concept: Nuclear Density Calculation

The key idea is that nuclear volume scales linearly with mass number $A$, so density becomes independent of $A$.

Reasoning

1. Nuclear radius: $R = R_0 A^{1/3}$, where $R_0 \approx 1.2 \times 10^{-15} \, \text{m}$.
2. Volume of a spherical nucleus: $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3 A$.
3. Mass of nucleus: $m \approx A \times u$, where $u = 1.66 \times 10^{-27} \, \text{kg}$ (atomic mass unit). …

The key insight is that the nuclear radius follows $R = R_0 A^{1/3}$, where $R_0 \approx 1.2 \ \text{fm}$ 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 $1.67 \times 10^{-27} \ \text{kg}$ (the atomic mass unit $u$). So the nuclear mass is

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

Notice that $A$ appears linearly — the volume is directly proportional to $A$.

3. Density calculation Nuclear matter density $\rho$ is mass divided by volume:

The $A$ cancels out completely. This is the central result: density does not depend on $A$.

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

Method: Volume–Mass Scaling from the Empirical Radius Law

The idea is simple: if the radius of a nucleus scales as $A^{1/3}$, then its volume scales as $A$, and since the mass also scales as $A$, the density becomes independent of $A$.

Step 1 – Write the nuclear volume in terms of $A$

The nucleus is treated as a sphere of radius $R = R_0 A^{1/3}$.

Volume of a sphere:

Step 2 – Simplify the cube

So:

The volume is directly proportional to $A$.

Step 3 – Write the nuclear mass

The mass of the nucleus is approximately:

where $m_p$ is the proton mass (neutron mass is nearly the same; the small difference doesn't affect the conclusion).

Step 4 – Compute the density

Nuclear matter density:

The $A$ cancels:

Step 5 – Interpret the result …

Students often lose marks on this derivation not because the math is hard, but because they skip steps or confuse mass with mass number. Here are the most common mistakes and how to fix each.

Mistake 1: Using mass number $A$ directly as the mass of the nucleus

Many students write the nuclear mass as just $A$ (e.g., $M = A$). That is wrong — $A$ is a count of nucleons, not a mass. The actual mass is $M \approx A \times m$, where $m$ is the average mass of one nucleon (roughly $1.67 \times 10^{-27}$ kg). If you treat $A$ as the mass, your density expression will be off by a factor of $m$, and you won't get a constant — you'll get something that still depends on $A$.

How to avoid: Always write $M = A \cdot m$ explicitly. Keep $m$ as a symbol; it will cancel out later.

Mistake 2: Using the nuclear radius formula incorrectly

The relation $R = R_0 A^{1/3}$ is correct, but $R_0$ is a constant (about $1.2 \times 10^{-15}$ m). Some students mistakenly treat $R_0$ as the radius of a single nucleon, or they forget the $A^{1/3}$ factor entirely. Others write the volume as $\frac{4}{3}\pi R^3$ but then substitute $R = R_0 A$ (missing the cube root).

How to avoid: Write the volume step carefully:

The $A$ inside the cube root becomes $A$ when cubed — that's the whole point.

Mistake 3: Forgetting to cube $R_0$ in the volume

Even if they substitute correctly, some students write $V = \frac{4}{3}\pi R_0 A$ instead of $\frac{4}{3}\pi R_0^3 A$. This leads to a density that still depends on $A$.

How to avoid: When you cube $(R_0 A^{1/3})$, cube both factors: $(R_0)^3 \cdot (A^{1/3})^3 = R_0^3 A$. Write it out explicitly.

Mistake 4: Not simplifying the density expression fully

After substituting $M = A m$ and $V = \frac{4}{3}\pi R_0^3 A$, the density is:

$\rho = \frac{M}{V} = \frac{A m}{\frac{4}{3}\pi R_0^3 A} = \frac{m}{\frac{4}{3}\pi R_0^3}$ …