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A quick uncertainty question

  1. Jan 31, 2012 #1
    If an atom transitions from an initially excited state emitting a photon E=hf and going to the ground state, and since ΔxΔk=1 show that ΔEΔt = hbar.

    So I know Ephoton=hf=hbar*sqrt(k/m) = hbar * sqrt(1/Δx*m)
    and since E=pc and p=hbar*k=hbar/Δx

    but I am not sure what to do from here???
    Any help would be greatly appreciated.
    Thank you.
    Stephen
     
  2. jcsd
  3. Feb 1, 2012 #2

    wukunlin

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    i'll give you a few clues:

    [tex]k = \frac{2 \pi}{\lambda} [/tex]
    [tex]c = f \lambda[/tex]
     
  4. Feb 1, 2012 #3
    ok so...
    E = pc=hbar*kc=hbar*2pi/λ*fλ=hf=Ephoton
    if E= hf=pc=hbar*sqrt(k/m)
    hbar^2/(4pi^2) = p^2*c^2*m/k
    and since ΔxΔy=1
    so h=p^2*c^2*m*Δx=E*p^2*Δx
    so would this lead to h=ΔE*p^2*Δx
    or ΔE*Δt=h
    and the uncertainty principle for RMS values be ΔEΔt=h/4pi

    I am not sure what to do to get the answer from here.
    Any help would be appreciated.
    Thanks.
    Stephen
     
  5. Feb 4, 2012 #4
    bump...please any help
     
  6. Feb 4, 2012 #5

    wukunlin

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    I think you are just over complicating the problem, don't use E=pc (is that even true?). Just use the identities I gave you and rearrange ΔxΔk=1 into ΔEΔt = hbar.
     
  7. Feb 4, 2012 #6
    How??? I am not seeing it. Would you show me the first couple of steps....

    Thanks.
    Stephen
     
  8. Feb 4, 2012 #7

    wukunlin

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    okay, how about:

    since [itex]k = \frac{2 \pi }{\lambda} [/itex] and [itex]c = f \lambda[/itex]
    then [itex] k = \frac{2 \pi f}{c} [/itex]

    all of those are constants except for f, therefore

    [tex] Δk = \frac{2 \pi}{c} Δf[/tex]
     
  9. Feb 4, 2012 #8
    ok so
    k=2*pi/λ and c=f*λ and k=2*pi*f/c
    so Δk= 2*pi*Δf/c
    since ΔxΔk=1
    1=2*pi/c * ΔxΔf
    h=2*pi/c *Δx*h*Δf
    hbar=ΔEΔt

    and since the RMS value would be the average so 1/2*the uncertainty value so h/4*pi = ΔEΔt

    is this it????

    Thanks for the help.
    Stephen
     
  10. Feb 4, 2012 #9

    wukunlin

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    i don't see what RMS has to do with this question? hbar is your uncertainty (or the product of two uncertainties...)
     
  11. Feb 4, 2012 #10
    ok am I right on the preceding work???
     
  12. Feb 4, 2012 #11

    wukunlin

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    looks good, just make sure you understand every step
     
  13. Feb 5, 2012 #12
  14. Feb 5, 2012 #13

    wukunlin

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    also stated on the site that it is an approximation for extreme relativistic velocities
     
  15. Feb 5, 2012 #14
    which is needed for a photon which is why Ephoton=pc
     
  16. Feb 5, 2012 #15

    wukunlin

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    ah right, sorry
     
  17. Feb 5, 2012 #16
    no problem at all and thanks so much for the help....I am really just repeating what my prof said as I am trying to understand all this..

    Thanks again.
    Stephen
     
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