Question Video: Determining the Minimum Lifetime of a Particle due to a Rest Energy Uncertainty

An unstable elementary particle has a rest energy of 80.41 GeV and an uncertainty in rest energy of 2.06 GeV. What is the minimum lifetime of this particle?


Video Transcript

An unstable elementary particle has a rest energy of 80.41 giga electron volts and an uncertainty in rest energy of 2.06 giga electron volts. What is the minimum lifetime of this particle?

Given this unstable elementary particle, we know the particle might decay or disappear at any point in time. Though we can’t know exactly when that might happen, based on the uncertainty and rest energy we’re given, we can solve for a minimum amount of time that the particle must exist. This minimum lifetime is based on the uncertainty relationship or uncertainty equation developed by Werner Heisenberg.

One form of this relationship says that the uncertainty in the energy of a particle multiplied by the uncertainty in the lifetime of that particle must be greater than or equal to Planck’s constant ℎ divided by four times 𝜋. So given either Δ𝐸 or Δ𝑡, this relationship sets a lower bound on which the other of the two parameters can be.

In our case, we’re given Δ𝐸. So we’ll solve for the least that Δ𝑡 can possibly be. And that lower bound will be the minimum lifetime of this particle.

To solve for the smallest that Δ𝑡 can possibly be, we’ll set Δ𝑡 times Δ𝐸 equal to ℎ over four 𝜋. ℎ, Planck’s constant, we’ll treat as exactly 6.262 times 10 to the negative 34th joule seconds. Our problem statement tells us Δ𝐸, that it’s 2.06 times 10 to the ninth electron volts.

As we solved then for Δ𝑡, the minimum lifetime of the particle, we see it’s Planck’s constant over four 𝜋 times Δ𝐸. And when we plug in for ℎ and for Δ𝐸, we see we’re ready to calculate Δ𝑡, except for one thing. The units we’re using for energy in our numerator and denominator don’t agree. In the one case they’re joules and in the other they’re electron volts. So we’ll want to convert, in this case, the denominator from electron volts to joules.

One joule is roughly equivalent to 6.242 times 10 to the 18th electron volts. When we multiply by this conversion factor in our denominator, we now have a consistent bases for the units in our expression.

Calculating this expression to three significant figures, we find it’s 1.60 times 10 to the negative 25th seconds. The unstable particle we’re considering must exist for at least this long, based on the limit provided by Heisenberg’s uncertainty equation.

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