Final answer:
Using the time dilation formula from special relativity, Δt = γΔτ, with the proper time Δτ of 2.2 microseconds and the Lorentz factor γ for a velocity of 0.98c, we find that the observed half-life of the muon from Earth is 11.055 microseconds.
Step-by-step explanation:
Calculating the Time Dilation of a Muon
To calculate the time interval Δt that a fast-moving muon experiences before decaying according to an observer on Earth, we can use the time dilation formula from special relativity, Δt = γΔτ. Here, Δτ is the proper time or the half-life of the muon at rest, and γ is the Lorentz factor which is calculated using the formula γ = 1 / √(1 - v^2/c^2), where v is the velocity of the muon, and c is the speed of light in a vacuum.
To find Δt, we first need to determine γ. With v = 0.98c, we get:
γ = 1 / √(1 - (0.98)^2) = 5.025
With Δτ given as 2.2 microseconds, we can now calculate Δt:
Δt = (5.025)(2.2 µs) = 11.055 µs
The time interval Δt experienced by the fast-moving muon before decaying, as measured by an observer on Earth, is 11.055 microseconds.