# Relating Uncertainty in Time to Uncertainty in Wavelength

• Blanchdog
In summary: All of this said, you still need to solve for d \lambda , which you can do by taking the natural logarithm of both sides: \ln{d \lambda} = hc\ln\lambda .
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!

Blanchdog said:
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 $\frac{dy}{dx} \approx \frac{\Delta y}{\Delta x}$.

------

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,

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

## 1. What is the relationship between uncertainty in time and uncertainty in wavelength?

The relationship between uncertainty in time and uncertainty in wavelength is described by the Heisenberg uncertainty principle. This principle states that the more precisely we know the position of a particle in time, the less precisely we can know its momentum or wavelength, and vice versa.

## 2. How does the concept of time affect the measurement of wavelength?

The concept of time affects the measurement of wavelength through the uncertainty principle. As time and wavelength are inversely related, the more precisely we measure time, the less precisely we can measure wavelength. This means that as our measurement of time becomes more accurate, the range of possible wavelengths becomes larger.

## 3. What factors contribute to the uncertainty in time and wavelength measurements?

There are several factors that contribute to the uncertainty in time and wavelength measurements. These include the limitations of measurement instruments, the inherent randomness of quantum particles, and the effects of external disturbances on the measurement process.

## 4. How does the uncertainty principle impact scientific experiments?

The uncertainty principle has a significant impact on scientific experiments, particularly in the field of quantum mechanics. It means that there will always be a degree of uncertainty in our measurements of time and wavelength, and this must be taken into account when designing and interpreting experiments.

## 5. Can the uncertainty in time and wavelength ever be completely eliminated?

No, according to the uncertainty principle, the uncertainty in time and wavelength can never be completely eliminated. This is a fundamental limitation of our ability to measure and understand the behavior of quantum particles. However, we can minimize this uncertainty through precise and careful measurement techniques.

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