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Atoms, Uncertainty

  1. Nov 11, 2014 #1
    1. The problem statement, all variables and given/known data

    Atoms of a certain material are in an excited state 1.8eV above the ground state and remains in that excited state for 2.0ms before to the ground state. Find
    1) The frequency of the emitted photon
    2) The wavelength of the emitted photon
    3)The uncertainty in the energy of the emitted photon
    4) The relative width of the special line [itex]\frac{ \Delta \lambda }{ \lambda }[/itex] and [itex]\frac{ \Delta f }{ f }[/itex] produced by these atoms.
    I'm not really sure what 2.0 ms means, is that 2 milliseconds?

    2. Relevant equations
    Well, I'm not entirely sure, this what I mostly need help with but I'll try my best in finding what equations I think should be used.
    1.[itex]f = \frac{ E }{ h }[/itex]

    2.[itex]\lambda=\frac{ c }{ f }[/itex]

    3. [itex]\Delta t \Delta E \ge \frac{ h}{ 2 }[/itex]

    4.[itex]\Delta \lambda = \lambda \frac{ \Delta E }{ E }[/itex] not exactly sure for this one.

    3. The attempt at a solution
    Ok so using equation 1. [itex]f = \frac{ E }{ h }[/itex] = [itex]\frac{ 2.88*10^{-19}J }{ 6.626*10^{-34} Js } = 4.35*10^14 Hz[/itex]

    2. [itex]\lambda=\frac{ c }{ f }[/itex] = [itex]\frac{ 3*10^8m/s }{ 4.35*10^14 Hz } = 6.9*10^-7m[/itex]

    3. [itex]\Delta t \Delta E \ge \frac{ h}{ 2 }[/itex] = [itex]\Delta E = \frac{ h }{ \Delta t 2 } = \frac{ 6.626*10^{-34}Js }{ 0.002s(2) } = 2.66 *10^-31 J[/itex]

    4. I'm not sure if my formula I gave for this one is correct, but I'll try it anyways and not sure for frequency?
    Last edited: Nov 11, 2014
  2. jcsd
  3. Nov 11, 2014 #2
    So, does the stuff I did above look right? I'm still having trouble with #4, as I'm still not sure what to do for it.
  4. Nov 11, 2014 #3


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    Staff: Mentor

    Start from ##d\lambda/dE##, and then replace the differentials by deltas.
  5. Nov 11, 2014 #4
    Is that for the last question, by the way does the rest look right?
  6. Nov 11, 2014 #5


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    Staff: Mentor

    Yes, it's for the uncertainty. Your answers 1 and 2 are correct, but for 3 the approach is correct by the numerical result is not.
  7. Nov 11, 2014 #6
    Ah ok, I just did number 3 again and got 1.66*10^-31J.

    4. [itex]\frac{ d \lambda }{ d E } => \frac{ \Delta \lambda }{ \Delta E }=\frac{ \lambda }{ E } => \frac{ \Delta \lambda }{ \lambda } = \frac{\Delta E }{ E } [/itex]

    But what exactly would delta lambda, and E be?
  8. Nov 12, 2014 #7


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    Staff: Mentor

    That looks right.

    The first equality there is not correct. Start by writing ##\lambda## as a function of ##E##, then calculate its derivative.
  9. Nov 12, 2014 #8
    Oh so would it be

    [itex]\lambda= c/f=> \frac{ hc }{ E }[/itex]

    [itex]\frac{ d \lambda}{ d E} \frac{ hc }{ E} =>-\frac{ hc }{ E^2 }[/itex]

    But I don't understand how this helps me get delta lambda/lambda and delta frequency/frequency?
  10. Nov 12, 2014 #9


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    Staff: Mentor

    Now you have
    \frac{d \lambda}{dE} = - \frac{hc}{E^2} \Rightarrow \frac{\Delta \lambda}{\Delta E} = \frac{hc}{E^2}
    You simply need to use that to find ##\Delta \lambda/\lambda## in terms of ##\Delta E## and ##E##.
  11. Nov 12, 2014 #10
    Yes, I remember you said this earlier and it hit me before bed haha :p
    Just wondering why is it positive now?

    To find [itex]\frac{ \Delta \lambda }{ \lambda }[/itex] I first solve for delta lambda

    [itex]\Delta \lambda = \frac{ hc \Delta E }{ E^2 }[/itex]

    For [itex]\frac{ \Delta f }{ f }[/itex] if I find it in terms of lambda I get the inverse so [itex]\frac{ f }{ \Delta f }[/itex]
    Last edited: Nov 13, 2014
  12. Nov 13, 2014 #11
    After I solve for delta lambda I can get [itex]\frac{ \Delta \lambda }{ \lambda }[/itex] I wouldn't really need the [itex]\frac{ \Delta E }{ E }[/itex] right? I can just use what numbers I find for the lambdas.
    But I still have the frequency question as above, if I solve it in terms of lambda I get f/ delta f.

    I just realized 3 is wrong it should be h bar, a little difference, where then h bar is 1.055*10^-34 Js and I get a weird answer of 2.6*10^-32 J
    Last edited: Nov 13, 2014
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