Ask Question
28 August, 05:46

An unstable high-energy particle enters a detector and leaves a track of length 1.05 mm before it decays. Its speed relative to the detector was 0.992c. What is its proper lifetime? That is, how long would the particle have lasted before decay had it been at rest with respect to the detector?

+3
Answers (1)
  1. 28 August, 06:36
    0
    You have to calculate, first, the speed under the normal frame, the observer one. Second, you have to use the formula of time dilation derived for speeds close to the speed of light.

    1) Observer frame

    V = d / t = > t = d / v = [1.05 mm * 1m/10^3mm] / (0.992c)

    c = 3.0*10^8m/s

    = > t = 1.05*10^ - 3 m / (0.992*3*10^8 m/s) = 3.328 * 10^ - 12 s

    2) Particle frame

    t observer = t particle / √[1 - (v/c) ^2] equation from special relativity theory

    => t particle = t observer * √[1 - (v/c) ^2 ]

    substituting the values

    t particle = 3.328 * 10^ - 12 s * √ [ 1 - (0.992) ^2 ] = 0.42 * 10^ - 12 s =

    = 4.2 * 10^-13 s

    Answer: The lifetime of the particle on its own frame is 4.2 * 10^-13 s
Know the Answer?