How to Prove |\DeltaE/E| = |\Delta\lambda/\lambda|<<1?

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Homework Help Overview

The discussion revolves around proving the relationship |\DeltaE/E| = |\Delta\lambda/\lambda|<<1, focusing on the context of energy and wavelength in quantum mechanics. Participants are exploring the implications of this relationship and its mathematical underpinnings.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss deriving expressions for \Delta\lambda in terms of E and \DeltaE, with some attempting to find differentials and rates of change. Questions arise regarding the interpretation of notation and the implications of the relationship being much less than one.

Discussion Status

The discussion is active, with participants providing insights into the derivation process and questioning the assumptions behind the relationship. Some have expressed uncertainty about their understanding, while others have clarified key points regarding the mathematical expressions involved.

Contextual Notes

There is mention of specific values for \DeltaE and E, which are used to illustrate the relationship, but the implications of these values are still being explored. Participants are also clarifying the meaning of notation such as "<<".

ProPatto16
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Homework Statement



show that |\DeltaE/E| = |\Delta\lambda/\lambda|<<1

Homework Equations



\DeltaE>hbar/2pi\Deltat

\lambda=hc/E

The Attempt at a Solution



dunno where to start.
 
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Start by finding an expression for Δλ in terms of E and ΔE.
 
\Delta\lambda=(\DeltaE/E)\lambda

...
 
ProPatto16 said:
\Delta\lambda=(\DeltaE/E)\lambda

...
If λ = hc/E, what is dλ? Can you find the differential?
 
you mean the derivative of the LHS? so \lambda d\lambda ?? that'd just be 1?
 
Can you find dλ/dE? It is not 1.
 
ah okay just wasnt sure what you were asking.

d\lambda/dE would be \lambdad\lambda = hc/E dE

which would be 1 = -hc/E2 ?
 
no wait. d\lambda wouls be a rate of change so then \Delta\lambda = -hc/E2 ?
 
ProPatto16 said:
ah okay just wasnt sure what you were asking.

d\lambda/dE would be \lambdad\lambda = hc/E dE

which would be 1 = -hc/E2 ?
How do you figure?
One more time. What is

\frac{d}{dE}\left( \frac{hc}{E} \right)
 
  • #10
ProPatto16 said:
no wait. d\lambda wouls be a rate of change so then \Delta\lambda = -hc/E2 ?
Not quite. Δλ = -(hc/E2) ΔE. Now can you find Δλ/λ ?
 
  • #11
oh gee. i feel so dumb -.- this should be so easy.

the notation i use for this differential would be:

hc/E dE
= hcE-1 dE
= (-1)hcE-2
= (-hc)/E2

how else would i derive that with respect to e :/... and i just left h and c as is since theyre constant?
 
  • #12
ProPatto16 said:
oh gee. i feel so dumb -.- this should be so easy.

the notation i use for this differential would be:

hc/E dE
= hcE-1 dE
This is correct except you need E2 in the denominator.
= (-1)hcE-2
= (-hc)/E2
This is not. You just can't drop the dE.
 
  • #13
oh so i did do it right.

well then, \Delta\lambda/\lambda = [(-hc/E2)\DeltaE]/\lambda
 
  • #14
oh i think i got it!... that expands to leave (\DeltaE/E2)(-hc/\lambda) which is \LambdaE/E so then \Delta\lambda/\lambda=\LambdaE/E
 
  • #15
yes i meant to put in deltaE
 
  • #16
whats the relevance of |\Delta\lambda/\lambda|<<1?

does << mean less than? or something else?
 
  • #17
ProPatto16 said:
whats the relevance of |\Delta\lambda/\lambda|<<1?

does << mean less than? or something else?
"<" means "less than"; "<<" means "much less than"; "<<<" means "even less than that".
 
  • #18
well in this particular question deltaE = 4.1neV and E=2.58eV so deltaE/E would be much much much much less than one. which shows that the equality would only hold if deltalambda/lambda was equaly small. would that be sufficient reason?
 
  • #19
You have shown that Δλ/λ = ΔΕ/Ε. This equality holds no matter what. If ΔΕ/Ε << 1, then Δλ/λ << 1 no matter what.
 
  • #20
So it's all good then :)
 
  • #21
ProPatto16 said:
So it's all good then :)
Unless there is more.
 
  • #22
Nah the other parts to the question I could do. I found all the answers just wasn't sure how to actually show that proof. So thanks :)
 

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