Heisenberg Uncertainty Maximum Lifetime of Photon

[diggleM]

Homework Statement

In special conditions, it is possible to measure the energy of a gamma ray photon to 1 part in 10^15. For a photon energy of 50 keV, estimate the maximum lifetime that could be determined be a direct measurement of the spread of photon energy.

Homework Equations

ΔtΔE≥$\hbar$/2

The Attempt at a Solution

It seems simple enough using the given equation to solve for the maximum lifetime, however I'm not sure I'm understanding how the relative uncertainty of 1 part in 10^15 effects the problem. If someone would be able to explain how the relative uncertainty effects the calculation to me, it would be much appreciated.

 

[Simon Bridge]

I'm not sure I'm understanding how the relative uncertainty of 1 part in 10^15 effects the problem.

Take another look at Heisenbergs relation - what do each of the terms mean?

 

[diggleM]

The way I understand it, they can be thought of as terms of uncertainty for time and energy, which are inversely related. That is to say, if you measure the Energy to a high level of certainty, the level of uncertainty will be high as well.

Looking back in my text describing the relation, it mentions that if a particle has a definite energy, ΔE = 0. This leads me to thinking that to solve a value for ΔE, you could find the relative uncertainty where Relative Uncertainty = | uncertainty/measured quantity|, in this case the relative uncertainty would be 10^15 and measured quantity would be 50,000 eV, to solve for the uncertainty, ΔE, which could be used to solve for Δt.

These are still pretty new concepts to me so I'm not sure if that is an appropriate mode of thinking about this. Does that seems to makes sense or do you see flaws in my thought process?

 

[Simon Bridge]

If you know a measurement x to 1 part in 1000 then the relative uncertainty would be about 1/1000 and the uncertainty would be Δx=x/1000.

Use a value of planks constant in units of keV.(time) to avoid conversions.

 

[paranormal]

Dear student,

Thank you for your question. The Heisenberg uncertainty principle states that the product of the uncertainties in the position and momentum of a particle (or in this case, a photon) cannot be smaller than a certain value, which is equal to Planck's constant divided by 2π. This means that if we know the position of a particle very precisely, we cannot know its momentum very precisely, and vice versa. In the case of a photon, the uncertainty in its energy (ΔE) is related to the uncertainty in its lifetime (Δt) by the equation ΔtΔE≥\hbar/2. This means that the more precisely we measure the energy of a photon, the less precisely we can measure its lifetime, and vice versa.

In this problem, we are given that the uncertainty in the energy of the photon is 1 part in 10^15. This means that if the photon has an energy of 50 keV, the uncertainty in its energy is 50 keV/10^15 = 5x10^-11 keV. Using this value for ΔE in the equation above, we can solve for the maximum lifetime (Δt). Plugging in the value for Planck's constant (6.626x10^-34 J s) and converting the energy units to Joules, we get Δt≥ 1.3x10^-24 seconds. This is the maximum lifetime that can be determined by a direct measurement of the spread of photon energy, given the uncertainties in our measurement.

I hope this helps to clarify the problem for you. It is important to understand the effects of uncertainty in measurements, as it is a fundamental aspect of quantum mechanics. Keep up the good work in your studies!

Best regards,