Posted by Gracie on .
A tungsten target is struck by electrons that have been accelerated from rest through a 38.9kV potential difference. Find the shortest wavelength of the radiation emitted.

Physics please help! 
Elena,
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•3.89•10^4 = 3.19•10^11 m. 
Notation got messed up 
Damon,
Î» = 6.63â€¢10^34â€¢3â€¢10^8/1.6â€¢10^19â€¢3.89â€¢10^4 = 3.19â€¢10^11 m.