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X-ray production

  1. Oct 9, 2005 #1
    in my modern physics textbook 6/E by arthur beiser,
    the formula for x-ray production is written as
    (namdamin)=1.24*10^(-6)/V V*m
    is that a typo? how can namda have units of V*m?
     
  2. jcsd
  3. Oct 9, 2005 #2

    HallsofIvy

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    Perhaps it would help if you explained exactly what you are talking about!

    As far as I know "x-ray production" doesn't have units! What exactly does
    "nambdamin" mean? (Was that supposed to be "lambda"?) Oh, and what is V? volts?
     
  4. Oct 9, 2005 #3
    x-ray production doesn't have units???
    jeepers~
    i must be misunderstanding modern physics...
    :P
    i don't know how to type the symbols, but yes, "nambda min" means the "lambda" minimun...
    V stands for volts...
     
  5. Oct 9, 2005 #4

    lightgrav

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    the second V (that's multiplied by meters) is a UNIT, Volts.
    But the first V (that you divide by) is a VARIABLE ... Voltage.

    These are NOT the same thing, but the units cancel.

    "the lamda minimum ..."
    lamda of WHAT? V of WHAT?
     
  6. Oct 9, 2005 #5

    Gokul43201

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    You are talking about the minimum wavength of x-rays produced by accelerating electrons through a voltage V and slamming them against a target. Naturally, the maximum energy of an x-ray will be the final KE of the electron before it hits the target.

    That gives

    [tex]KE(electron) = e*V = \frac{hc}{\lambda_{min}} [/tex]
    [tex]\implies~ \lambda_{min} = \frac{hc}{eV} = \frac {hc}{e} \cdot \frac{1}{V} [/tex]

    The quantity [itex]hc/e[/itex] has a value of [itex]1.24 *10^{-6}~Vm [/itex]

    When you divide by the units of the applied voltage V (ie: volts), you are left with units of meters, the correct unit for a wavelength !
     
    Last edited: Oct 9, 2005
  7. Oct 10, 2005 #6
    thank you very much!!! :)
     
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