Question

Chapter 37, Problem 001 The mean lifetime of certain subatomic particles is measured to be 2.7196 μs when they are stationary. However, the mean lifetime of fast-moving particles of the same kind observed in a burst of cosmic rays is measured to be 17.258 μs. What is the speed parameter β of these cosmic-ray particles relative to Earth?

Answer 1

v = 0.9875 c

Step-by-step explanation:

This is an exercise of time dilation in special relativity, which is described by the equation

t =
`t_(p)` γ

γ = 1 / √ (1 — v² / c²)

The most important is to find respect for which system the measurement is performed.

Since the particle is still, the measured time is the proper time
`t_(p)` = 2.7196 10-6 s, in the system where the particles move t = 17,258 10-6 s

Let's clear dam

γ = t /
`t_(p)`

γ = 17,258 10-6 / 2.7196 10-6

γ = 6.345

γ = 1 / √ (1-β²)

1-β² = 1 /γ²

β = √ (1- 1 / γ²)

β = √ (1 - 1 / 6,345²)

β = 0.9875

v / c = 0.9875

v = 0.9875 c