I am so surprised that this topic is not in this blog.  But I am not always posting what I did. Anyway…

The Stopping power is energy loss per length, in unit of MeV/cm.

\displaystyle S = -\frac{dE}{dx}

Since the stopping power is proportional to density, it is often normalized with the density in literature and the unit becomes MeV/(ug/cm^2).

The Bethe-Bloch formula based on classical argument.

In order to calculate the range, we can integrate the stopping power. Consider the particle moved by \Delta x distance, the energy loss is

E(x+\Delta x) = E(x) - S(E(x)) \Delta x

rearrange, gives

\displaystyle  \frac{dE(x)}{dx} = - S(E)

\displaystyle \int_{E_0}^{E} \frac{-1}{S(E)} dE = x(E ; E_0)

This is the relation between particle energy and range.

We can plot S(E(x)) to get the Bragg peak.

In here, I use the physical calculator in LISE++, SRIM, and Bethe-Bloch formula to calculate the stopping power for proton in CD2 target.

The atomic mass of deuteron is 2.014102 u. The density of CD2 target is 0.913 g/cm3.


For the Bethe curve, I adjusted the density to be 0.9 mg/cm3, and simple use the carbon charge. And the excitation potential to be 10^{-5} z, where z = 6 , so that the Bethe curve agree with the others.

It is well known that the Bethe curve fails at low energy.

Assume the proton is 5 MeV. The energy loss against distance isEvsX.PNG

and the stopping power against distance is


We can see the Bragg peak is very sharp and different models gives different stopping range. It is due to the rapid decrease of energy at small energy. In reality, at small energy, the microscopic effect becomes very important and statistical. thus, the curve will be smoothed and the Bragg peak will be broaden.

In very short range, the stopping power increase almost linearly, Because the incident energy is large, so that the stopping power is almost constant around that energy range.

For S(E_0) = h ,

\displaystyle  x = \frac{-1}{h}(E - E_0 )

\displaystyle E = E_0 - hx,  hx << E_0

\displaystyle S(E) = S(E_0 - hx) \approx S(E_0) - \frac{dS(E_0)}{dE}(hx)