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I'm using Dr. Wei-Kan Chu's book on backscattering spectrometry for self-study. Early on in the book he describes how a projectile entering a material loses energy as it progresses through the material due to small-angle nuclear scattering and something akin to drag through Coulombic interaction with electrons. He presents the following simplified model that I am having some trouble understanding.

For this model we will consider particle interactions as occurring between two isolated particles. We will assume that the direction and speed of the incident particle is perturbed only slightly by the interaction and that the collision is elastic. Assume a projectile with mass $M_1$, a charge $Z_1e$ and a velocity $v_1$ and an initially stationary target particle with mass $M_2$ and charge $Z_2e$. The text states that the perpendicular momentum transferred to the target particle is

$$P_\bot = \frac{2Z_1Z_2e^2}{bv_1}$$

where b is the impact parameter, the distance of closest approach between the two particles if the mass $M_2$ were held fixed while the projectile flew past it along a straight trajectory. This is the first part where I get stuck. I don't really understand how the author calculates this. Do we simply set up an electric field associated with $M_2$ and find the closest the projectile comes to the origin (where I set my target)? That seems to me to be the most straightforward way to do this, but would that be the correct approach?

The book goes on to say that therefore he energy transferred to the target particle is

$$E_\bot = P_\bot^2/2M_2 = (2/M_2)(Z_1Z_2e^2/bv_1)^2.$$

Since the perturbation is only slight this is very close to the total energy lost by the projectile. At this point it is noted in the book that electrons have a much smaller mass than the nuclei of atoms and thus absorb more energy in a collision based on the formula above. He then indicates that he will direct the rest of the discussion in this section of his book towards the electron friction.

The next step is to calculate the total energy loss $\Delta E$ of the incident particle as it travels through the target penetrating through a depth of $\Delta x$. The book states that the probability of an encounter with the impact parameter between $b$ and $b+db$ is $2 \pi b db$ per unit area since the target particle (electron in our case) may lie anywhere on a circle of radius $2 \pi b$ around the projectile's track. If our target is homogeneous, consists of atoms with atomic number $Z_2$, and the density of atoms in our target is given by N then the number of electrons per unit area over the length $\Delta x$ of the track is $NZ_2 \Delta x$. Dr. Chu then states that the average number $d \Delta n(E_\bot)$ of encounters that will generate a quantum $E_\bot$ of energy loss is

$$d \Delta n(E_\bot) = NZ_2 \Delta x \cdot 2 \pi b db.$$

I don't really understand how these quantities are physically interpreted. Why is this the average number of encounters to generate an energy loss of $E_\bot$?

Finally, we are told that these losses contribute the average differential about d$\Delta E$ to the total energy loss $\Delta E$ across $\Delta X$, hence

$$d \Delta E = NZ_2 \Delta x \left[ 2(Z_1e^2)^2/m_ev_1^2 \right] 2 \pi (db/b).$$

The mathematical details to go from the previous equation to this equation is unclear to me. I think I'm a little confused with the notation, and a little confused about the physical interpretation of some of these things. Could anyone please help me understand? Thanks, and regards.

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  • $\begingroup$ Cross-sections, lovely cross-sections. Step back and look at the mechanical analog. The target is a sphere with $\pi r^{2}$ equal to the cross section. The incident ion is a point. A line through the center of the sphere is your reference. The incident point particle is a distance $b$ from the line, and traveling parallel to it. Sketch out the kinematics of the collision - what angle will the incident particle scatter at (this is related to $b$ and the cross section only). In an elastic collision, what are the resulting momenta of the two? Please edit to say which step is confusing. $\endgroup$ – Jon Custer Sep 24 '15 at 15:34

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