# Elastic collisions in Franck-Hertz experiment

Looking at a Franck-Hertz experimental setup, and given a potential difference such as $4.0\ V$ which is too small to excite out the first electron orbital, the electrons moving through the tube will have elastic collisions with the Hg atoms. I'm supposed to show that given some kinetic energy $E$, of a flowing electron, that the maximum recoil of a Hg atom that it hits is approximately $4Em/M$ (where $m$ is the mass of the electron, and $M$ is the mass of the Hg atom). I know to use conservation of energy, and given that the Hg atom was initially at rest, I know the electron is probably bouncing backward (so something like $2\times E-\text{sub - elec} = E\text{ - sub - hg}$). I can't seem to find where the 4 comes from. Is it the $4.0\ V$ potential difference?

I tried playing with conservation of energy formula but can't get $\frac{4Em}{M}$. (Taking lower case to be the electron and UPPER CASE TO BE THE ATOM:) [$mv^2 = MV^2 - mv^2$] so [$mv^2 = 0.5MV^2$] so [$2e = E$] but I'm supposed to find that [$E = 4Em/M$]. I know there is something I'm missing here.

P.S.: The original problem is #44 here (or as an image here) in case I'm misunderstanding something.

-
You know the kinetic energy, E, already. So you don't need to know the field. This is just a collision problem. Use conservation of momentum. In the limit m << M (what is m/M?) it simplifies because the energy transfer is very small and the recoil velocity of the electron doesn't change very much in magnitude (despite a huge change in direction). – Michael Brown Jan 1 '13 at 3:05
To see that the field is irrelevant just consider the length scale of the collision: roughly the size of an atom. How much does the potential change over that distance? Compare that to the kinetic energy. – Michael Brown Jan 1 '13 at 3:07
I'm afraid the link isn't working for me. The equation you have E=4Em/M is inconsistent unless M=4m. I think you mean something like dE = 4Em/M? Your other condition is momentum conservation. What does that tell you? – Michael Brown Jan 1 '13 at 3:44
Also, your energy conservation equation is incorrect. It should be energy before = energy after: m v1^2 = m v2^2 + M V^2 – Michael Brown Jan 1 '13 at 3:46
@MichaelBrown It's working for me. Here I've turned it into a imgur link: i.stack.imgur.com/Oo1VT.png and will suggest such an edit to the post. I think tapper means that kinetic energy of the atom = 4(m/M) * kinetic energy of the electron. Meaning I think he meant to write [E=4em/M]. – Double AA Jan 1 '13 at 3:56

The problem is probably simpler than you think. If the electron's initial velocity is $v_1$ and it's final (recoil) velocity is $v_2$, and the Mercury atom's velocity is $V$ then conservation of energy and momentum give us two equations:
$$\frac{1}{2} m v_1^2 = \frac{1}{2} m v_2^2 + \frac{1}{2} M V^2$$
$$m v_1 = -m v_2 + MV$$
Use the second equation to get $v_2 = something$ and substitute for $v_2$ in the first equation to get a quadratic equation for $V$. You'll find this turns out to be simpler than you probably expect. To get the approximate expression in the question note that $M + m \approx M$ but save this approximation until the end.