# The equation of the wavelength of the bremsstrahlung

In my text book, I came across this question:

What is the equation used to determine the shortest wavelenght of the bremsstrahlung spectrum of x-rays?

λ = hc/eV. (Where: h is planck constant, c is the speed of light, e is the electron charge and V is the voltage across the cathod and the anode in Coolidge tube.

I thought the above equation is used to determine the shortest wavelength of the charecteristic spectrum of x-rays, so why it is used here? Could some ome clarify things for me, please?

That represents all the kinetic energy gained by an electron $eV$ becoming a photon of energy $\frac {hc}{\lambda}$.

If electrons do not have enough energy to produce a characteristic X-ray photon then none such photons will be produced.

If the cut off wavelength for the continuous spectrum was $1\,\rm Å$ (corresponding to an electron energy of approximately $12 \,\rm kV$) then the K-lines would be produced from the Copper target but not from the Molybdenum target. So the shortest wavelength of the continuous spectrum must be less than the wavelength of a characteristic X-ray line.
In the case of Molybdenum the minimum energy of the incoming electron has be be approximately $18 \,\rm kV$ (photon wavelength $0.7\,\rm Å$) to produce a K-line.