Understanding cut off wavelength at edge of Bremsstrahlung region

[Jimbo]

Hello,

I am trying to understand an equation stated in the book 'Principles of Protein X-Ray Crystallography' by Jan Drenth (The exact page I am referring to can be found here: http://books.google.co.uk/books?id=...y crystallography&pg=PA24#v=onepage&q&f=true")

The page discusses electrons being accelerated between a cathode and an anode, and the x-rays emitted as part of the process. It mentions a cut off wavelength where "photons obtain their full energy from the electrons when they reach the anode", and states that electron energy = electron charge * accelerating voltage and that photon energy is planks constant * (speed of light / wavelength) and that the minimum wavelength is therefore 12.4 / V where V is in kilovolts.

My working is as follows:

When the photons obtain their full energy from the electrons, eV = hc / λ

So, λ = hc / eV (as stated in the book)

hc = 1.24 * 10^-6 eVm

So just looking at the units I get:

eVm / eV which makes sense as I am left with m for the wavelength.

But my question is where does the V come from in the books equation? (They state that λ = 12.4 / V)

Also why is this necessarily the minimum wavelength? (it is stated as λ_min) Why not just wavelength?

Sorry for the long introduction but I wanted to give my question a bit of context.

Thanks very much for any guidance,

Jimbo

 

[Bill_K]

The process that creates the X-rays is bremsstrahlung radiation, in which a fraction of the electron's energy goes into the photon. The maximum energy the photon can have is 100 percent of the electron energy, and this corresponds to the shortest wavelength the photon can have, λ_min. The electron energy when it arrives at the cathode is e*V, charge e times voltage V.