Magnitudes of the atom

In a nutshell

It is possible to find an estimate for the size and density of any nucleus by starting with an equation that relates the radius of a nucleus and the nucleon number $$A$$. The strong force is one of the four fundamental forces and it overpowers the repulsive electrostatic force between protons in a nucleus.

Equations

Constants

Variable definitions

​The atomic mass unit

The masses of atoms and nucleons are often represented with the atomic mass unit $$u$$. $$1\ u$$ is a twelfth of the mass of a neutral carbon-$$12$$ atom and is roughly the mass of a nucleon.​

​$$1\ u=1.661\times10^{-27}\ kg$$​​

The rough mass in $$kg$$​ of the nucleus of any element can then be found by knowing the nucleon number:

​
$$m=A \times u$$​​

Size and density of the nucleus

The radius of the nucleus depends on $$A$$. It can be found using the equation:

$$R=r_0A^{\dfrac{1}{3}}$$​​

Where $$r_0=1.2\ fm=1.2\times10^{-15}\ m$$.​

Using the radius it is now possible to find the volume and density of the nucleus using:

​$$V=\dfrac{4}{3}\pi R^{3}\newline \rho=\dfrac{m}{V}$$​​

Nuclei usually have density of order $$10^{17}\ kgm^{-3}$$.​

Example

Calculate an estimate for the density of a lithium-$$7$$ nucleus to $$2$$ s.f.

Firstly, write down the known values:

$$A=7$$​    

Next, write down the equation needed to find the radius:

$$R=r_0A^{\dfrac{1}{3}}$$​​

Substitute the values:

$$R=1.2\times10^{-15}\times7^{\frac{1}{3}}=2.295...\times10^{-15}\ m$$​​

Now write the equation for volume:

$$\newline V=\dfrac{4}{3}\pi R^3$$

​

Substitute the radius you just found:

$$V=\frac{4}{3}\times \pi \times (2.295\times10^{-15})^3=5.063...\times10^{-44}\ m^3$$​​
​

Write the equation for density:

$$\newline \rho=\dfrac{m}{V}$$​​

Remember how to find mass in $$kg$$ from a nucleon number:

$$m=A\times u$$​​

Substitute mass into the equation for density:

$$\rho=\dfrac{A\times u}{V}$$​​

​

Substitute all the values to find the density:

$$\rho=\dfrac{7\times 1.661\times 10^{-27}}{5.063\times10^{-44}}$$​​

Finally calculate the value and make sure to write down the correct units:

$$\rho=2.30\times10^{17}\ kg\ m^{-3}\ (2\ s.f.)$$​​

The density of the nucleus of a lithium-$$7$$ atom is $$\underline{2.30\times10^{17}\ kg\ m^{-3}}$$.

The strong nuclear force

The strong nuclear force is one of the four fundamental forces of the Universe which keeps nucleons from coming apart.

Inside a nucleus, protons are in close proximity to one another and since they are positively charged they will experience a repulsive electrostatic force. This can be found with Coulomb's law:

​$$F=\dfrac{Qq}{4\pi\varepsilon_0r^2}$$​​

Since the protons are not flying apart it means that another force is keeping them together, this being the strong nuclear force. It has a very short range of only a few femtometres ($$10^{-15}\ m$$​), it is attractive up to $$3\ fm$$ and repulsive below $$0.5\ fm$$ and it acts on all nucleons. 

A graph showing the relationship between the force $$F$$ and the distance $$r$$ is shown below;