Calculating Emitted Photon Properties for Excited Atoms | Homework Help

In summary, the atom emits a photon with an energy of 4.35*10^14 Hz and a wavelength of 6.9*10^-7 meters.
  • #1
oldspice1212
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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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  • #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.
 
  • #3
Start from ##d\lambda/dE##, and then replace the differentials by deltas.
 
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  • #4
Is that for the last question, by the way does the rest look right?
 
  • #5
oldspice1212 said:
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
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
oldspice1212 said:
Ah ok, I just did number 3 again and got 1.66*10^-31J.
That looks right.

oldspice1212 said:
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
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
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
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
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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1. What is an atom?

An atom is the basic unit of matter, consisting of a nucleus made up of protons and neutrons, surrounded by electrons. It is the smallest particle of an element that retains its properties.

2. What is the uncertainty principle?

The uncertainty principle, also known as Heisenberg's uncertainty principle, is a fundamental principle in quantum mechanics that states that it is impossible to know both the exact position and momentum of a subatomic particle at the same time. This is due to the wave-like nature of particles at the atomic level.

3. How does uncertainty affect our understanding of atoms?

Uncertainty affects our understanding of atoms because it introduces a level of unpredictability in the behavior of particles at the atomic level. This means that we can never have a complete understanding of the behavior of individual particles and can only make predictions based on probabilities.

4. Can we measure the uncertainty of an atom?

No, we cannot measure the uncertainty of an individual atom. The uncertainty principle applies to the behavior of subatomic particles, not entire atoms. However, we can use mathematical equations and models to calculate the uncertainty of a particle's position and momentum within an atom.

5. How does the concept of uncertainty impact other areas of science?

The concept of uncertainty is not limited to the study of atoms and quantum mechanics. It also applies to other areas of science, such as thermodynamics and chaos theory. In these fields, uncertainty refers to the unpredictability and sensitivity of complex systems, and it plays a crucial role in shaping our understanding of the natural world.

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