- #1

Blanchdog

- 57

- 22

- 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/λ

hΔf = ΔE

ħΔω = ΔE

Δω = ΔE/ħ ≈ 1/(2 Δt)

2πc/λ

Δλ = λ

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!

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/λ

^{2}* ΔλhΔf = ΔE

ħΔω = ΔE

Δω = ΔE/ħ ≈ 1/(2 Δt)

2πc/λ

^{2}* Δλ = 1/(2 Δt)Δλ = λ

^{2}/4πcΔtThis 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!