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,

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