physics
posted by andrej on .
A tungsten target is struck by electrons that have been accelerated from rest through a 24.5kV potential difference. Find the shortest wavelength of the radiation emitted. (in nm)

Lets look at energy levels in the Tungsten orbitals.
Ek= Z^2*13.5eV/1^2=74^2*13.6ev/1=74k eV
El=74^2*13.6ev/2^2=18.6k eV
Em=74^2*13.6ev/3^2=6.4k eV
So investigatin of what trasitions a 24.5keV electron could make, well, it cant go from k to m, but it can go from l to m.
Energy of transition: 18.66.4 =12.2kev
Using plancks equation;
E=hf=hc/lambda
lambda=hc/E=4.1E15 eV s *3E8m/s *1/12.2E3 eV
lambda= 1E10 meters=0.1 nm
check my work. 
For Bremsstrahlung radiation (or "braking X radiation" )
the lowwavelength cutoff may be determined from Duane–Hunt law :
λ =h•c/e•U,
where h is Planck's constant, c is the speed of light, V is the voltage that the electrons are accelerated through, e is the elementary charge
λ = 6.63•10^34•3•10^8/1.6•10^19•2.45•10^4 = 5.06•10^11 m. 
I agree with Elena on the brakding cuttoff. My answer ignores the continuous spectrum. So for the answer, consider this: What have you covered in your physics class: transitions from energy levels, or the "continuous" spectrum?
Good work, Elena.