Q.Given the mass of iron nucleus as $55.85\ \text{u}$ and $A = 56$, find the nuclear density.

The key idea is that nuclear density is nearly constant for all nuclei because the nuclear volume scales linearly with mass number $A$.

Step 1 – Find the nuclear radius.

Using the empirical formula $R = R_0 A^{1/3}$, where $R_0 = 1.2 \times 10^{-15}\ \text{m}$:

Since $56^{1/3} \approx 3.83$,

Step 2 – Compute the nuclear volume.

Step 3 – Convert mass to kg and find density. …

The idea behind nuclear density is beautiful in its simplicity. Unlike ordinary matter, where density varies wildly from gas to solid, nuclear matter has an almost constant density. Why? Because a nucleus is a tightly packed sphere of protons and neutrons. If you add more nucleons, the volume increases proportionally — the radius follows $R = R_0 A^{1/3}$, so volume $\propto A$. Mass also $\propto A$ (since each nucleon has roughly 1 u). So density $\approx$ constant, independent of $A$.

We’ll now calculate it for iron, step by step.

1. Convert the mass to kilograms. The mass of the iron nucleus is given as $55.85\ \text{u}$. One atomic mass unit is $1\ \text{u} = 1.660539 \times 10^{-27}\ \text{kg}$. So:

2. Find the nuclear radius. The empirical formula for nuclear radius is:

where $R_0 \approx 1.2\ \text{fm}$ (1 femtometre = $10^{-15}\ \text{m}$).

For iron, $A = 56$, so:

Now $56^{1/3}$ is about $3.825$ (since $3.8^3 = 54.9$, close enough).

Thus:

3. Compute the volume. The nucleus is spherical, so:

First cube the radius:

Then:

4. Calculate density. Density $\rho = \frac{m}{V}$: …

Method: Direct Application of the Nuclear Density Formula

This problem uses the fact that nuclear density is nearly constant for all nuclei. The method is straightforward: find the nuclear volume from the radius formula, then divide mass by volume.

Step 1: Write the nuclear radius formula

The radius of a nucleus is given by:

where $R_0 = 1.2 \times 10^{-15}\ \text{m}$ (a constant) and $A$ is the mass number.

Step 2: Convert the given mass to kilograms

Mass of iron nucleus = $55.85\ \text{u}$.

Recall: $1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}$.

So:

Step 3: Calculate the nuclear radius

$56^{1/3} \approx 3.83$ (since $3.83^3 = 56.2$).

Step 4: Calculate the nuclear volume

The nucleus is spherical:

First find $R^3$:

Then:

$V = \frac{4}{3} \times 3.14 \times 9.73 \times 10^{-44}$ …

The most common mistake here is treating the mass number $A$ as the mass of the nucleus in kilograms. $A$ is just the number of nucleons — it has no units. The mass in kilograms must be calculated separately.

Mistake 1: Using $A$ directly as mass in kg

A student writes $\rho = \frac{56}{ \frac{4}{3}\pi R^3 }$ and gets a wildly wrong answer. The mass number $56$ is dimensionless, not a mass. You must convert the given mass from atomic mass units (u) to kg first: $1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}$.

Mistake 2: Forgetting the nuclear radius formula

The radius of a nucleus is $R = R_0 A^{1/3}$, where $R_0 \approx 1.2 \times 10^{-15}\ \text{m}$. Some students use the atomic radius (of the order $10^{-10}\ \text{m}$) instead, which makes the density off by a factor of $10^{15}$. The nucleus is tiny — always use the nuclear radius formula.

Mistake 3: Using the atomic mass instead of nuclear mass

The problem gives the mass of the iron nucleus as $55.85\ \text{u}$, so this is already correct. But if a question gives the atomic mass, remember that the mass of electrons is included. For iron ($Z = 26$), that’s about $26 \times 9.1 \times 10^{-31}\ \text{kg}$ — negligible for most exam purposes, but conceptually you should know the difference.

Mistake 4: Unit mismatch in the final answer

Nuclear density comes out around $2.3 \times 10^{17}\ \text{kg/m}^3$. A common slip is reporting it in $\text{g/cm}^3$ without converting, or forgetting that $1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}$ and $1\ \text{fm} = 10^{-15}\ \text{m}$.