# Dependence of atomic mass number in the Bragg Kleeman rule

I have just staring learning about radioactivity, more or less about alpha particles and how there range can be approximated by the Bragg Kleeman rule, $\displaystyle{R \varpropto \frac{\sqrt{A}}{\rho}}$, where $A$ is the atomic mass number and $\rho$ is the density of the material. I understand why increasing the density of the material will decrease the range, more atoms in a smaller volume thus greater chance for iterations to occur, but I don't understand how increasing the atomic mass number would increase the range.

My only assumption would be that as the atom size increases, the electron cloud surrounding the nucleus expands and hence the alpha particle is less likely to interact with them and have a larger range. An explanation would be greatly appreciated as I am really interested to know the reasoning behind this. Thanks in advance

• Higher atomic mass means more mass per atom, which means fewer atoms to achieve a given density, which means fewer targets to hit in a given volume. Admittedly, the targets are slightly bigger. – user3386109 May 22 '16 at 5:19
• There may be some clue in the original article " W. H. Bragg and R. Kleeman, Philosophical Magazine 10, 358 (1905). "? – jim May 22 '16 at 10:06

For detailed understanding please consult the book "Radiation detection and measurement" by Glenn Knoll.

The qualitative answer to your question is, that $\alpha$ particle losses its energy mainly by its interaction with electrons (not nucleus). Hence the electron density is the key factor which decides the range of $\alpha$ particle.

if there are N atoms in unit volume then there will be $N_e=NZ$ electrons per unit volume, where Z is atomic number, if density of the material is $\rho$ and atomic number is A then

N $\propto$ $\rho$/A

Hence range should be

R $\propto 1/N_e$

R $\propto \frac{A}{\rho Z}$

A/Z goes approximately as $A^{2/3}$

This relation is not exactly same to the Bragg Kleeman rule but can at least give a qualitative explanation of why the range has some specific form (A in numerator).

I hope this will help

with best regards