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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)

  • physics -

    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.

  • physics -

    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.

  • physics -

    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.

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