# Homework Help: A quick uncertainty question

1. Jan 31, 2012

### StephenD420

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

2. Feb 1, 2012

### wukunlin

i'll give you a few clues:

$$k = \frac{2 \pi}{\lambda}$$
$$c = f \lambda$$

3. Feb 1, 2012

### StephenD420

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

4. Feb 4, 2012

5. Feb 4, 2012

### wukunlin

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.

6. Feb 4, 2012

### StephenD420

How??? I am not seeing it. Would you show me the first couple of steps....

Thanks.
Stephen

7. Feb 4, 2012

### wukunlin

since $k = \frac{2 \pi }{\lambda}$ and $c = f \lambda$
then $k = \frac{2 \pi f}{c}$

all of those are constants except for f, therefore

$$Δk = \frac{2 \pi}{c} Δf$$

8. Feb 4, 2012

### StephenD420

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

9. Feb 4, 2012

### wukunlin

i don't see what RMS has to do with this question? hbar is your uncertainty (or the product of two uncertainties...)

10. Feb 4, 2012

### StephenD420

ok am I right on the preceding work???

11. Feb 4, 2012

### wukunlin

looks good, just make sure you understand every step

12. Feb 5, 2012

### StephenD420

13. Feb 5, 2012

### wukunlin

also stated on the site that it is an approximation for extreme relativistic velocities

14. Feb 5, 2012

### StephenD420

which is needed for a photon which is why Ephoton=pc

15. Feb 5, 2012

### wukunlin

ah right, sorry

16. Feb 5, 2012

### StephenD420

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