# Collision of a heavy charged particle with an electron at rest - Deduction

When I was studying Nuclear Physics I saw this formula about interaction of radiation with matter. The book mentioned that when considering a elastic collision of a heavy charged particle of mass $M$ with an electron (initially at rest) of mass $m$ the kinetic energy loss for the charged particle is given by:

$$∆T = T \frac{4m}{M}$$

I tried to understand where this expression came from using simple knowledge of collision mechanics (kinetic energy and linear momentum conservation), but I can't. Can someone help me finding where this expression came from?

Note: I found this formula on "Introductory Nuclear Physics" by Kenneth S. Krane

• I believe that precise form rises from 1D collision mechanics. The "first power of $m/M$" can be guessed just by considering the limits in which $m/M \to 0$ (where there is no loss) and $m/M = 1$ (where the transfer of kinetic energy is complete (in 1D)). – dmckee Jun 20 '14 at 14:47

Suppose the initial velocity of $M$ is $v_0$, and after the collision the velocity of $M$ is $v_1$ and the velocity of the electron is $v_2$. Before the collision you have:

\begin{align} E &= \tfrac{1}{2}Mv_0^2 \\ p &= Mv_0 \end{align}

and after the collision you have:

\begin{align} E &= \tfrac{1}{2}Mv_1^2 + \tfrac{1}{2}mv_2^2 \\ p &= Mv_1 + mv_2 \end{align}

Since energy and momentum are conserved this given us the pair of equations:

\begin{align} \tfrac{1}{2}Mv_0^2 &= \tfrac{1}{2}Mv_1^2 + \tfrac{1}{2}mv_2^2 \\ Mv_0 &= Mv_1 + mv_2 \end{align}

and some frenzied scribbling and tugging of hair later we eliminate $v_1$ to get:

$$v_2 = \frac{2v_0}{1 + m/M}$$

Now note that $\Delta T$ is just $\tfrac{1}{2}Mv_0^2 - \tfrac{1}{2}Mv_1^2$ so it's equal to $\tfrac{1}{2}mv_2^2$ (i.e. the energy loss of $M$ is the energy gain of the electron). So:

\begin{align} \Delta T &= \tfrac{1}{2}mv_2^2 \\ &= \frac{2mv_0^2}{(1 + m/M)^2} \\ &= \tfrac{1}{2}Mv_0^2 \frac{4m}{M} \frac{1}{(1 + m/M)^2} \end{align}

It just remains to note that $\tfrac{1}{2}Mv_0^2$ is the original kinetic energy $T$, and that because $m \ll M$ that means $1 + m/M \approx 1$, and we end up with:

$$\Delta T \approx T \frac{4m}{M}$$