Animated Solution for Physics - Atoms and Nuclei: If radius of the ${}_{13}^{27}\text{Al}$ nucleus is estimated to be $3.6\text{ fm}$, then the radius of ${}_{52}^{125}\text{Te}$ nucleus be nearly

Calculating Nuclear Radii

When we dive into the microscopic world of atoms, the nucleus sits at the very center, holding almost all the mass. But how big is it? Unlike everyday objects, we can't just take a ruler and measure a nucleus. Instead, physicists have discovered a beautiful, simple relationship between the number of nucleons (protons and neutrons) in a nucleus and its physical size.

Through scattering experiments, it was found that the volume of a nucleus is directly proportional to its mass number, $A$. Since volume scales with the cube of the radius ($V\propto R^3$), it naturally follows that the radius $R$ scales with the cube root of the mass number. This gives us the famous empirical formula:

Here, $R_0$ is a constant, roughly equal to $1.2\text{ fm}$ (femtometers, where $1\text{ fm}=10^{-15}\text{ m}$). This formula implies that nuclear matter has a constant density, regardless of whether it's a light element like Carbon or a heavy one like Uranium.

In our problem, we are given the radius of an Aluminum nucleus (${}_{13}^{27}\text{Al}$) as $3.6\text{ fm}$ and asked to find the radius of a Tellurium nucleus (${}_{52}^{125}\text{Te}$).

Instead of calculating $R_0$ from the Aluminum data and then plugging it into the Tellurium equation, we can use a much more elegant mathematical tool: ratios. By dividing the equation for Tellurium by the equation for Aluminum, the constant $R_0$ completely cancels out!

Now, we simply substitute the known values into our ratio equation. The mass number of Tellurium is $125$, and for Aluminum, it is $27$.

Notice how perfectly these numbers are chosen? Both $125$ and $27$ are perfect cubes! The cube root of $125$ is $5$, and the cube root of $27$ is $3$.

Cross-multiplying gives us our final answer:

The radius of the Tellurium nucleus is exactly $6\text{ fm}$. This method of using ratios is a powerful technique in physics, saving time and reducing the chance of calculation errors.