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Show change in de Broglie wavelength from change in speed

  1. Jul 14, 2015 #1
    1. The problem statement, all variables and given/known data

    Show that for a nonrelativistic particle, a small change in speed leads to a change in de Broglie wavelength given from

    cramster-equation-20091191039506336795839013912505186.gif



    3. The attempt at a solution

    I have tried to expand the left hand side of the equation, but found that it gave the answer of v0/delta v. My definition of delta lambda is the final wavelength minus the initial wavelength.
     
  2. jcsd
  3. Jul 14, 2015 #2

    Orodruin

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    You need to actually show us what you did. How else can we find out where and if you went wrong?
     
  4. Jul 14, 2015 #3
    Sorry, my attempt is as follows.

    In the book (Eisberg, Resnick - quantum physics of atoms, molecules, solids, nuclei and particles, pg. 82, question 10) it has the answer as that given above, however, my attachment proves that wrong. Is there anywhere I may have made a mistake?
     

    Attached Files:

  5. Jul 14, 2015 #4

    BvU

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    In your first step you write something that looks like ##{1\over a -b} = {1\over a} - {1\over b}## to me o0) ...
     
  6. Jul 14, 2015 #5
    I can't find where I have done this. How would you do this question?
     
  7. Jul 15, 2015 #6

    SammyS

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    What you did is equivalent to that.

    You had ##\displaystyle \ \Delta\lambda=\frac{h}{mv_f}-\frac{h}{mv_i} \ .##

    Then you did this:
    ##\displaystyle \ \frac1{\Delta\lambda}=\frac{mv_f}{h}-\frac{mv_i}{h} \ .##

    However, ##\displaystyle \ \frac1{\displaystyle\frac{h}{mv_f}-\frac{h}{mv_i}}\ne\frac{mv_f}{h}-\frac{mv_i}{h} \ .##
     

    Attached Files:

  8. Jul 15, 2015 #7
    Okay, I understand that! I was able to try and attempt to solve this with the new knowledge, however I got stuck. I derive an answer that is

    ##\displaystyle \ \frac{\Delta \lambda}{\lambda_0}=\frac{-\Delta v}{v_f}##
    -

    If my mathematics is correct, that would mean that

    ##\displaystyle \ \frac{-\Delta v}{v_f}=\frac{\Delta v}{v_0}##

    But I can't think of any relations that would make the above true?
     
  9. Jul 15, 2015 #8

    BvU

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    ##\displaystyle \ \frac{\Delta \lambda}{\lambda_0}=\frac{-\Delta v}{v_f}\ \ ## is correct. The book means to say ##
    \displaystyle \ \frac{\Delta \lambda}{\lambda}=\frac{|\Delta v|}{v}\ \ ## but finds the sign so trivial that it leaves out the ##|\ |##.
    And for a small change ##v = v_0 \approx v_f## in the denominator -- NOT, of course in the difference.

    This reminds me of differentiation and error propagation:

    With ##y = 1/x## you have ##dy = -1/x^2 \; dx## so ##dy/y = -dx/x ## !
     
  10. Jul 15, 2015 #9
    mindblow moment. Thankyou for your help!
     
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