I would assume that the beam profile in RBS is circular and uniform motivated by the fact that I expect some beam appertures in the beamline to create a beam with a very small angular spread. I also assume that you use singly charged He.

Let A be the beam diameter and q be the electron charge and C be the totol accumulated charge than the fluence F is given by:

F = (C/q) * (1/A)

The first factor is just the number of He ions per unit charge.

Normally the incident beam is made of ions with the same state of ionization (either protons or He2+) which makes things rather simple. The charge incident on a given area is simply the sum of all the charges of single ions. Divide the total charge to the charge of a single ion and you will have the number of ions incident on that area, hence the fluence. For instance, if your beam is made out of He2+ ions, than the charge of a single ion will be 3.2E-13 microC. This means that for a 1 microC crossing a 1 sqcm you will have a fluence of 312.5E6 particles per square meter.

Of course, if you do not have a clean beam and you have ions with different states of ionisation, then you will need to know the ion charge distribution across the beam, but the principle remains.

Thanks guys (Radu and Dood) am greatful.

Hi Radu:  the last report of a dirty beam I can remember is Hemment et al,  "Effect of poor vacuum conditions on the analysis of thin films by RBS", Thin Solid Films, 28, 1975, L1-L4.  This is not usually a problem!

I get rather different numbers from you.  The charge on the electron is 1.60218e-19 Coulombs,  so 1 microC is 6.242e12 charges.  We only (usually) use He++ for energies>4MeV (since we have a 2MV tandem machine),  so with a He+ (or H+) beam 1microC is 6.242e12 particles.  If the beam is spread (rather incredibly) over 1 square meter this will give 6.242e8 particles/cm^2.  If the beam is He++ there will be half that number.

Curiously,  the only time one usually calculates actual fluence in RBS is when one is worried about beam damage (for which see arXiv:1303.3171 [cond-mat.mtrl-sci] for a very useful Handbook).  The RBS formula using fluence is in terms of collected charge Q:

A = Q.N.s.o

(see Eq.3 of the 2012 paper in Analytical Chemistry linked below) where the number of counts A (area) in a signal is given by the product of Q, the areal density N of atoms generating the counts,  the differential cross-section s ("sigma": this is in area per unit solid angle) and the solid angle of the detector o ("omega").  You see,  the fluence of the incident particles is not needed,  only the number.  But N is in number per unit area because the probability of scattering (s for sigma) is given by the scattering cross-sectioin,  which involves an area of course.

http://dx.doi.org/10.1021/ac300904c

You are right, I must have made some arithmetic mistake. It should have been e+12, not e+6. My logic was very similar to yours:

- the charge of the electron is 1.6e-19 C i.e. it has 1.6e-13 microC

- the He++ has two charges i.e. it has 3.2e-13 microC

- 1 microC will then have 3.12E12 He++ particles

- if the beam has a 1 sqcm area, then it means that the He++ fluence will be 3.12E12 per 1 sqcm or 3.12E16 particles per square meter.

Here was the mistake, I think, I transformed a E-4 in a E+4 at the denominator and instead of 3.12E16 I got 3.12E8. Thank you for drawing it to my attention.