Question

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

Answer 1

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: 18.6-6.4 =12.2kev

Using plancks equation;

E=hf=hc/lambda

lambda=hc/E=4.1E-15 eV s *3E8m/s *1/12.2E3 eV

lambda= 1E-10 meters=0.1 nm

check my work.

Answer 2

For Bremsstrahlung radiation (or "braking X- radiation" )
the low-wavelength 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.

Answer 3

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.