Atoms, Uncertainty

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Homework Statement



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?

Homework 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.

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?
 
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Answers and Replies

  • #2
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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.
 
  • #3
DrClaude
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Start from ##d\lambda/dE##, and then replace the differentials by deltas.
 
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  • #4
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Is that for the last question, by the way does the rest look right?
 
  • #5
DrClaude
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Is that for the last question, by the way does the rest look right?
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.
 
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  • #6
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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?
 
  • #7
DrClaude
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Ah ok, I just did number 3 again and got 1.66*10^-31J.
That looks right.

4. [itex]\frac{ d \lambda }{ d E } => \frac{ \Delta \lambda }{ \Delta E }=\frac{ \lambda }{ E } => \frac{ \Delta \lambda }{ \lambda } = \frac{\Delta E }{ E } [/itex]
The first equality there is not correct. Start by writing ##\lambda## as a function of ##E##, then calculate its derivative.
 
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  • #8
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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?
 
  • #9
DrClaude
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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##.
 
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  • #10
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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]
 
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  • #11
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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
 
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