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Need help with X-ray production

  1. Apr 21, 2005 #1
    Need help with "X-ray production"

    Hi everyone. i am in 12th grade now.
    im having problems with the production of X-ray...
    Can someone tell me how X-ray is generated? From the very begining...to the charateristic raditaoin and the Bremsstrahlung Radiation...to the intensity graph (the one with two peaks).

    I don't understand why the graph is like that...i mean, what is the relationship between wavelength and x-ray intensity?

    I also don't get the calculation: when do we use qV=1/2 mv^2, and when do we use qV=hf? (q is the charge of electrons, V is the potential difference, 1/2mv^2 is the kinetic energy, h is plank's constant, f is frequency.)

    And how can we find out the followings if only an x-ray intensity graph is given?
    1. the potential difference
    2. the difference of energy between the ground state and the first excited state?

    And...if a rough x-ray graph is given...what will the new graph be when the potential difference is half?

    I'm going to take the IB physics exam soon in May(similar to AP in US, A-level in UK) and i still dont get this X-ray stuff. Can somebody help me? thanks a lot!
  2. jcsd
  3. Apr 21, 2005 #2
    The only thing you need to worry about is at what wavelengths the intensity is higher than zero (the cutoff at short wavelengths).
    The kinetic energy of the electrons after they have been accelerated by the potential diffference V is qV. That is also the highest energy of the photons that can be produced. (Photon energy is hf.)[/QUOTE]
  4. Apr 21, 2005 #3
    X-rays are generated by using a potential difference to accelerate and collide electrons into a target. This also answer's when you're supposed to use qV=1/2 mv^2. You use qV=hf to find out what's the maximum energy of a photon that can be produced when the electrons collide to the target.

    Bremsstrahlung, the breaking radiation, is a product of the elctrons slowing down. It creates the continuous part of the spectrum. The characteristic radiation is a property of the target you collide the electrons into. The colliding electrons with suffcient energy can knock off electron's from one of the target atom species' inner shells causing an electron from one the outer shells to "fall in" and take it's place emitting an photon in the x-ray wavelength area.

    1: Weird question. Usually folks now what accelerating voltage they're using. From the spectrum you should see a point where the intensity starts to differ from 0. This point corresponds to the maximum energy a photon generated by the collision. eV=h/lambda_min, solve for V.

    2. These you can calculate from the positions of the characteristic peaks.

    The new graph is of similiar form but it may be missing the charcteristic peak if the electrons do not have enough energy to generate characteristic radiation via the collisions. And lambda_min is larger.
  5. Apr 21, 2005 #4
    so...should i find out the wavelengths corresponding to the two peaks, and use E=hf1-hf2 (c=frequency x wavelength) ? is this the energy difference b/t the ground state and the first excited state?

    Also...please see the graph attached...A is the continuous radiation, B is the charasteristic radiation...then how about C? is C also the continuous radiation? what's the difference between A and C?

    And when the potential is half (suppose that half V still has enough energy to remove the inner-shell electrons and produce the charactereistic radiation...), will the entire graph shift to the right (wavelength_min larger)? will the peaks and the continuous radiation be of smaller intensity?


    Attached Files:

  6. Apr 21, 2005 #5
    Both are bremsstrahlung.
    The graph is on a photon energy scale. It looks like a silver spectrum taken at 35 kV. At half the voltage, no holes can be created in the K-shell of silver.
  7. Apr 21, 2005 #6
    oh, another question....for the bremsstrahlung radiation part, the projectile eletrons avoid all outer-shell electrons of the target and interact with the nucleus of the target. because the nucleus is of positive charge, the projectile electrons will be attracted. How come the electrons slow down? i thought they should be accelerating toward the nucleus due to the stronger attraction force from the protons.
  8. Apr 21, 2005 #7
    In classical terms: when high-energy electrons get close to nuclei, they are deflected (think of a hyperbolic trajectory). This change of the direction of the electron is an acceleration (whatever the sign), and classically, an accelerated charge must radiate. So there will be a loss of kinetic energy.

    But when the spectrum is studied at very high resolution, one can see at the highest photon energies the unoccupied part of the density of states of the anode material, and (if this is a metal) the Fermi level cutoff. This is studied in BIS (Bremsstrahlung Isochromat Spectroscopy). And that requires a quantum-mechanical explanation.
    Last edited: Apr 21, 2005
  9. Apr 21, 2005 #8
    I see. Thanks Pieter Kuiper and Inha. I just bombed my physics mid-term today...i just hope that i can do better on the real IB exam.

    Oh, and by the way...do anyone know any useful website about particle physics (leptons, mesons, bosons...and all those)...cuz this is also my weak point. Thanks~
  10. Apr 21, 2005 #9
    Check out the sticky thread at the nuclear and particle physics forum here. IIRC it should go through the basics and it had some nice link to some particle physics page which doesn't require that much previous knowledge on the subject.
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