Lifetime of an Excited State

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Homework Statement



According to the energy-time uncertainty principle, the lifetime t of a state is inversely proportional to the uncertainty in the energy E.

We consider the line λ= 656nm resulting from a transition in a hydrogen atom,
from an excited state of lifetime 10-8s.

(a) What is the uncertainty in the energy of the emitted photon?
(b) What is the corresponding uncertainty in the wavelength, Δλ?

My question involves b, but let's do a first.

Homework Equations



[tex]\Delta E \Delta t \approx \hbar[/tex]

The Attempt at a Solution



So for a I just solved the above equation for ΔE which gave me.

[tex]\Delta E \approx \frac{\hbar}{\Delta t}[/tex]

Now part a is done. (the value doesn't matter for my actual question)

Next up is b, where my question comes in.

This is how I solved it, which gave me a huge wavelength (in meters) and was wrong.

I had....

[tex]E=hf = \frac{hc}{\lambda}[/tex]

Solving for lambda gave:

[tex]\lambda=\frac{hc}{E}[/tex]

I actually had it right up to this point.

The correct method was to take the derivative of what I had for E with respect to λ.

Which gives:

[tex]\frac{dE}{d\lambda}=-\frac{hc}{\lambda^2}[/tex]

Or something like that. I took notes on the paper I handed in.:frown:

The actual answer after the derivative was...which I don't really get either. I'm not sure how we have λ2 in the denominator and then the square goes away and a Δλ jumps up to the top.

[tex]E=\left|-\frac{hc}{\lambda}\Delta \lambda\right|[/tex]

This is then obviously solved for Δλ.

Finally...the question.

Why does one take the derivative in order to get the uncertainty for E? We didn't set it to zero or anything. So I don't think it was min/max type stuff.
 
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Answers and Replies

  • #2
ehild
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The energy of the photon is hf=E(exited)-E(stable), the stable state having infinite lifetime. So the uncertainty of the photon energy is the same as that of the exited state. You get a small change of a function by multiplying its derivative with the change of the independent variable:

ΔE=(dE/dλ)Δλ .

dE/dλ=-hc/λ2,

ΔE =-(hc/λ2) Δλ.

Solving for Δλ:

Δλ=-(λ2 ΔE)/(hc)

and take the absolute value as uncertainty of the wavelength.

ehild
 
  • #3
collinsmark
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The correct method was to take the derivative of what I had for E with respect to λ.

Which gives:

[tex]\frac{dE}{d\lambda}=-\frac{hc}{\lambda^2}[/tex]

Or something like that.
Okay, that looks about right to me.

You started with E = hc/λ, But you're not looking for the relationship between E and λ in-particular. Instead you're looking for the relationship between ΔE and Δλ. You already know what ΔE is -- you've just calculated that in the last part. But you don't know what Δλis yet. That's where the derivative comes in. You're trying to figure out how much λ "wiggles", if you wiggle E by a small amount. You don't have that information yet until you take the derivative.

Let me give a few derivative examples, if you don't understand where the square came from.

If y = x, .......... dy = dx

If y = 5x2, .......... dy = 5(2)xdx = 10xdx

If y = axn, .......... dy = anxn-1dx

If y = a/x = ax-1, .......... dy = -ax-2dx = -(a/x2)dx

So, if both h and c are constants (which of course they are), and

if E = hc/λ = hcλ-1, .......... dE = (-1)hcλ-2 = -(hc/λ2)

Note that you also could have done it the other way around if you wanted to,

If λ = hc/E, .......... dλ = -(hc/E2)dE

And since you know that E = hc/λ,

dλ = -(hc)(λ2/(h2c2)dE = -(λ2/(hc))dE
I took notes on the paper I handed in.:frown:

The actual answer after the derivative was...which I don't really get either. I'm not sure how we have λ2 in the denominator and then the square goes away and a Δλ jumps up to the top.

[tex]E=\left|-\frac{hc}{\lambda}\Delta \lambda\right|[/tex]

This is then obviously solved for Δλ.
There seems to be a couple of errors in your equation. The λ in the denominator still needs to be squared. The square shouldn't have gone away. It must have been a typo or mistake. Also the left hand side of the equation is ΔE, not just E by itself.

The absolute value is fine though. You're not interested in the direction of how λ wiggles with respect to a small wiggle on E, just the magnitude.
Finally...the question.

Why does one take the derivative in order to get the uncertainty for E? We didn't set it to zero or anything. So I don't think it was min/max type stuff.
You do it to find a relationship between Δλ and ΔE. Before taking the derivative, you haven't found that relationship yet.


[Edit: Doh! ehild beat me to the response again. Note to self: Must learn to type quicker. :smile:]
 
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  • #4
ehild
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Collinsmark: I was too lazy to explain how to get the derivative of 1/x, so it was easy to beat you :)

ehild
 
  • #5
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First thanks to both of you. I appreciate the responses.

Okay, that looks about right to me.

You started with E = hc/λ, But you're not looking for the relationship between E and λ in-particular. Instead you're looking for the relationship between ΔE and Δλ. You already know what ΔE is -- you've just calculated that in the last part. But you don't know what Δλis yet. That's where the derivative comes in. You're trying to figure out how much λ "wiggles", if you wiggle E by a small amount. You don't have that information yet until you take the derivative.
Oh! I get it now. I was confused because that is how it was presented on the board. I understand the derivative part (with λ[/I]2 on the bottom - thanks for derivative explanation though) but could not figure out why the square went away and the Δλ showed up in the numerator. That made no sense at all. However now that I see the square doesn't go away it makes sense.


You do it to find a relationship between Δλ and ΔE. Before taking the derivative, you haven't found that relationship yet.
So what does this say about E? I am sure this is a stupid question but I've never encountered it before and want to understand it fully because I am sure it will come up again. I know the end goal is to to find a relationship between Δλ and ΔE. We start with a ΔE and are taking the derivative of it with respect to λ. Doesn't that modify the original ΔE some how? If started with E and λ alone it would make more sense.
 
  • #6
ehild
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A light ray never has an absolutely defined wavelength. When we say that a 656 nm light is emitted by the hydrogen atom it can be any wavelength between 656-Δλ and 656+Δλ (in nm-s). A certain wavelength in this range corresponds to the energy E=hc/λ. 656 nm, the middle of the spectral line corresponds to 3.028 E-19 J. The energy has the uncertainty of about ΔE=1 E-26 J. It would be possible to find the the wavelength range if you calculated the wavelengths that correspond to (3.028 E-19)-ΔE and (3.028 E-19)+ΔE. Try to do it. You have to get about the same deviations from 656 nm as with the differential method. The essence of that method is that a function y(x) is approximated by a straight line in the vicinity of a certain point (a,b). The slope of this line is dy/dx, calculated at x=a. The deviation of the dependent variable from b is proportional to the deviation of the independent variable from a if we stay near a: Δy=(dy/dx|x=a) Δx.

ehild
 

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