Uncertainty in Energy-Time Relation for Atomic Transitions

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StephenD420
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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
 

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i'll give you a few clues:

[tex]k = \frac{2 \pi}{\lambda}[/tex]
[tex]c = f \lambda[/tex]
 
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
 
bump...please any help
 
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.
 
How? I am not seeing it. Would you show me the first couple of steps...

Thanks.
Stephen
 
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]
 
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
 
i don't see what RMS has to do with this question? hbar is your uncertainty (or the product of two uncertainties...)
 
ok am I right on the preceding work?
 
looks good, just make sure you understand every step
 
also stated on the site that it is an approximation for extreme relativistic velocities
 
which is needed for a photon which is why Ephoton=pc
 
ah right, sorry
 
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