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Transition between excited states

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data

    An atom in an excited state has a lifetime of 1.2 x 10 -8 sec; in a second excited state the
    lifetime is 2.3 x 10 -8 sec. What is the uncertainty in energy for the photon emitted when
    an electron makes a transition between these two levels?

    2. Relevant equations

    [itex]\Delta[/itex]E[itex]\Delta[/itex]t[itex]\geq[/itex][itex]\frac{\hbar}{2}[/itex]

    3. The attempt at a solution

    So I just found the uncertainty in energies for the two excited states using the uncertainty principle, getting 2.74*10^-8 eV for the 1.2*10^-8 sec state, and 1.43*10^-8 eV for the 2.3*10^-8 sec state. And figured that the uncertainty in energy would just be the difference in the energies, giving 1.31*10^-8 eV. But the book gives an answer of 4.17*10^-8 eV, which I noticed is what you get if you add the energies instead.

    So is the book wrong, or is there some weird thing about the uncertainties combining such that I have to add them instead?
     
  2. jcsd
  3. Nov 4, 2013 #2
    Shouldn't it follow from

    [itex]\Delta E \Delta t \geq \frac{\hbar}{2}[/itex]

    that

    [itex]\Delta E \geq \frac{\hbar}{2 \Delta t}[/itex]

    ?
     
  4. Nov 4, 2013 #3

    gneill

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    Staff: Mentor

    Whether adding or subtracting two quantities, the uncertainties add.
     
  5. Nov 4, 2013 #4
    @Basic_Physics: Yes, that's how I got the energies.

    @gneill: That's a rule from statistics? Cause the book was a bit sparse on that point.
     
  6. Nov 4, 2013 #5

    gneill

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    Staff: Mentor

    Yup. Consider that subtraction is just adding the negative of one of the values. The uncertainty in the negative value is the same as for the positive value. So in terms of uncertainty, addition and subtraction are the same.
     
  7. Nov 4, 2013 #6
    Thanks gneill! At least now I know I'm not going crazy.
     
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