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Quantum Theory

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data
    Electrons are accelerated by 2450 V in an electron microscope. What is the maximum possible resolution?
    Me=9.11 x 10^-31kg and e= 1.60 x 10^-19 C


    2. Relevant equations
    I'm not so sure which equations to use for this question.. maybe the planck's formula?
    λ=h/p

    3. The attempt at a solution
     
  2. jcsd
  3. Oct 30, 2011 #2

    rude man

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    Hint #1: deBroglie
    Hint #2: Rayleigh
     
  4. Oct 30, 2011 #3
    Thanks, Can you help me with the problem though? I'm confused..
     
  5. Oct 30, 2011 #4
    So by using deBroglie equation lambda=h/p.. i'd have to do
    6.63x10^-34Js / ((9.11x10^-31)(1.60x10^-19)) ??
     
  6. Oct 30, 2011 #5
    when i find the solution, do i multiply the answer by 2450? to get the maximum resolution?
     
  7. Oct 31, 2011 #6

    vela

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    No. Why would you multiply m and e? It doesn't make any sense.
     
  8. Oct 31, 2011 #7

    Redbelly98

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    Moderator's note: I have moved this thread to Introductory Physics. In the future, please post questions of this level in Introductory Physics.

    The information given for V can be used to find p.
     
  9. Oct 31, 2011 #8
    Lol, I'm not sure.. I'm confused on solving the question..

    So, to find p, it's p=mv right? so (9.11 x 10^-31kg)(2450 v) ?

    When i find the solution of p, plug it into lambda=h/p, so (6.63x10^-34 Js) / p , will give me the maximum resolution?
     
  10. Oct 31, 2011 #9
    Thank you. Sorry, I'm new to the forum
     
  11. Oct 31, 2011 #10

    vela

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    No. The v in p=mv represents the velocity of the particle. The V in 2450 V stands for volt; it's a unit of electric potential. They're not the same thing at all.

    Read about electric potential in your textbook so you understand what the problem means when it says "Electrons are accelerated by 2450 V."

    That's the basic idea, but you need to find the electron's momentum first.
     
  12. Oct 31, 2011 #11
    Okay. I read up on it, but it didn't say much on "electrons accelerating"
    I did see that 1 Volt = 1 J/C
    So 2450 V = 2450 J/C
    (1.60x10^-19 C)(2450 J/C) = ??

    Man.. i am so confused
     
  13. Oct 31, 2011 #12
    Okay. This is what I got, tell me if i did it correctly.

    E= (1.60x10^19C)(2450J/C) = 3.92x10-16 J
    V= Squareroot of √(2x3.92x10^-16) / (9.11x10^-31) = 2.93x10^7

    So using de Broglie equation

    lambda= (6.63x10^-34Js)/(9.11x10^-31)(2.93x10^7) = 2.48x10^-11 ?
     
  14. Oct 31, 2011 #13

    vela

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    10^11? That's way off!

    (Right method, though.)
     
  15. Oct 31, 2011 #14
    Lol, aw man.. I thought I had it in the bag. What did I do wrong?
     
  16. Oct 31, 2011 #15
    Wait.. I wrote, 10^-11, not 10^11. Am I still wrong?
     
    Last edited: Oct 31, 2011
  17. Oct 31, 2011 #16

    vela

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    I see you edited your original answer so it says 10^-11 now instead of 10^11. That would be the right answer if you included the units.
     
  18. Oct 31, 2011 #17
    Thank you for all your help!

    Would the units be in meters?
     
  19. Oct 31, 2011 #18

    vela

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    You tell me. :)
     
  20. Oct 31, 2011 #19
    I guess so! :) Thanks for your help!
     
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