Relating Uncertainty in Time to Uncertainty in Wavelength

[Blanchdog]

Homework Statement

An unusually long lived unstable atomic state has a lifetime of 1 ms. Assuming that the photon emitted when this state decals is visible (λ ≈ 550 nm), what are the uncertainty and fractional uncertainty in its wavelength.

Relevant Equations

ω = 2πf (Definition of angular frequency)
ΔE Δt ≥ ħ/2 (Uncertainty Principle)
E = hf (De Broglie Relation)
c = λf

I actually have a solution available to me, but I don't understand what it's doing so I'll include my attempt at a solution and briefly describe the correct solution that I don't understand.

ΔE Δt ≥ ħ/2 (Uncertainty)
ΔE = hΔf (De Broglie)
ΔE = hc/Δλ

Substituting,
(hc/Δλ)Δt ≥ ħ/2
Rearranging,
Δλ =(2hc Δt)/ħ
Δλ = 4π c Δt

The factor of the speed of light makes this answer very large and obviously wrong, though why I'm not sure.

The correct solution that I don't understand is as follows:

ω = 2πf
ω = 2π(c/λ)

Δω = ∂ω/∂λ * Δλ
Δω = 2π(c/λ)*1/λ * Δλ
Δω = 2πc/λ² * Δλ

hΔf = ΔE
ħΔω = ΔE
Δω = ΔE/ħ ≈ 1/(2 Δt)

2πc/λ² * Δλ = 1/(2 Δt)
Δλ = λ²/4πcΔt

This is the correct solution, but I don't understand what's going on with the rate change equation with angular frequency and wavelength, and the approximation of Δω ≈ 1/(2 Δt) seems a little weird too.

Thanks in advance for the help!

 

[collinsmark]

Homework Statement: An unusually long lived unstable atomic state has a lifetime of 1 ms. Assuming that the photon emitted when this state decals is visible (λ ≈ 550 nm), what are the uncertainty and fractional uncertainty in its wavelength.
Homework Equations: ω = 2πf (Definition of angular frequency)
ΔE Δt ≥ ħ/2 (Uncertainty Principle)
E = hf (De Broglie Relation)
c = λf

I actually have a solution available to me, but I don't understand what it's doing so I'll include my attempt at a solution and briefly describe the correct solution that I don't understand.

ΔE Δt ≥ ħ/2 (Uncertainty)
ΔE = hΔf (De Broglie)
ΔE = hc/Δλ

Substituting,
(hc/Δλ)Δt ≥ ħ/2
Rearranging,
Δλ =(2hc Δt)/ħ
Δλ = 4π c Δt

The factor of the speed of light makes this answer very large and obviously wrong, though why I'm not sure.

The correct solution that I don't understand is as follows:

ω = 2πf
ω = 2π(c/λ)

Δω = ∂ω/∂λ * Δλ
Δω = 2π(c/λ)*1/λ * Δλ
Δω = 2πc/λ² * Δλ

hΔf = ΔE
ħΔω = ΔE
Δω = ΔE/ħ ≈ 1/(2 Δt)

2πc/λ² * Δλ = 1/(2 Δt)
Δλ = λ²/4πcΔt

This is the correct solution, but I don't understand what's going on with the rate change equation with angular frequency and wavelength, and the approximation of Δω ≈ 1/(2 Δt) seems a little weird too.

Thanks in advance for the help!

Just to keep things clear before I start, the given [correct] solution makes some substitutions between the differential dE and the uncertainty \Delta E; and similarly with d \lambda and \Delta \lambda; and d \omega and \Delta \omega.

That's a fine approach in my opinion. Just be aware that you might be substituting the uncertainties with the differentials. In other words, what this all means, using different variables, is that \frac{dy}{dx} \approx \frac{\Delta y}{\Delta x}.

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The major mistake [in your attempted solution] is assuming that just because

E = \frac{hc}{\lambda}, <---- (So far this is correct)

that,

\Delta E = \frac{hc}{\Delta \lambda}. <---- (This is incorrect.)

You can't just change from absolute variables to deltas all willy-nilly like that.

Instead, step back to differentials for a moment (don't worry, you can substitute the uncertainties in later). In order to find the relationship between dE and d \lambda, you need to take a derivative. I'll give you a hint to get you started:

Since

E = \frac{h c}{\lambda},

then,

\frac{dE}{d \lambda} = \frac{d}{d \lambda} \left\{ \frac{h c }{\lambda} \right\}.