Calculation of the diameter of a nucleus

Exploitation of Rutherford's experiment

Remember the interpretation of Rutherford's experiment:

If we count the number of impacts of $\alpha$ particles on the photographic paper, we notice that only one particle in $10000$ is deviated by the nucleus. The gold leaflet had $10000$ gold atoms in thickness. Let's calculate the diameter of a gold nucleus !

How many particles would be deviated, if the gold leaflet had only a thickness of 1 atom ?

$10000$ times less, therefore one particle in $100 000 000$ !

Imagine the section of a gold atom where the size of the nucleus is exagerated !
What is the ratio of the atom's section area and the nucleus's section area?

$\frac{A_a}{A_n}$ $=$ $100000000$

Imagine the two sections as discs, then calculate !

$\frac{A_a}{A_n}$ $=$ $\frac{4 \pi R^2}{4 \pi r^2}$ $=$ $\frac{R^2}{r^2}$ $=$ $100 000 000$

Calculate the ratio $\frac{R}{r}$

$\frac{R}{r}$ $=$ $\sqrt{100 000 000}$ $=$ $10000$

The radius of an atom is $10 000$ times greater than the radius of its nucleus!

→  The experiment

→  Interpretation