Animated Solution for Physics - Atoms and Nuclei: Order of magnitude of density of uranium nucleus is ($m_p=1.67\times 10^{-27}\text{ kg}$)

Have you ever wondered just how dense the matter inside an atom really is? We often hear that atoms are mostly empty space, but the tiny nucleus at the center holds almost all the mass. In this problem, we are going to calculate the order of magnitude of the density of a uranium nucleus.

To find the density of any object, we need two fundamental properties: its mass and its volume. Let's start by defining the mass number of our uranium nucleus as $A$. This number represents the total count of nucleons—that is, the protons and neutrons combined.

Since protons and neutrons have roughly the same mass, we can approximate the total mass of the nucleus $M$ by multiplying the mass number $A$ by the mass of a single proton $m_p$.

Next, we need the volume. Assuming the nucleus is a perfect sphere, its volume $V$ is given by the standard geometric formula:

But what is the radius $R$ of a nucleus? Experimental scattering data has shown that the radius of a nucleus scales with the cube root of its mass number. The empirical formula is:

Here, $R_0$ is a constant approximately equal to $1.25\times 10^{-15}\text{ m}$ (or $1.25\text{ fm}$).

Now that we have expressions for both mass and volume, we can set up our density equation. Density $\rho$ is simply mass divided by volume:

Let's substitute our expressions into this formula:

Watch what happens when we expand the denominator. Cubing the radius term gives us:

This is the magical moment! The mass number $A$ appears in both the numerator and the denominator, which means they perfectly cancel each other out.

This reveals a profound truth about the universe: the density of nuclear matter is a constant. It doesn't matter if you are looking at a light hydrogen nucleus or a massive uranium nucleus; the nuclear density remains exactly the same.

All that's left is to plug in the standard values and compute the final number. We know the mass of a proton $m_p=1.67\times 10^{-27}\text{ kg}$ and the constant $R_0=1.25\times 10^{-15}\text{ m}$.

First, let's cube the constant $R_0$:

Now, multiply by $\frac{4}{3}\pi$:

Finally, divide the proton mass by this volume:

The order of magnitude of the density is $10^{17}{\text{ kg/m}}^3$.

To put this staggering number into perspective, a single teaspoon of nuclear matter would weigh billions of tons! This is the exact same density found in the core of neutron stars, where gravity has crushed atoms so tightly that the electrons and protons have merged into a solid ball of neutrons.