Chapter 37: Q5P (page 1145)

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?

Short Answer

Expert verified

The proper life time for particle is $0.446 p s$ .

Step by step solution

Identification of given data

The speed of particle relative to detector is $v = 0.992 c$

The length of track is $L = 1.05 m m$

The time dilation is used to find the duration for particle before decay from detector.

Determination of proper life time before decay of particle

The proper life time for particle is given as:

$t_{0} = \frac{L}{V} \sqrt{1 - {(\frac{v}{c})}^{2}}$

Here, is the speed of light and its value is $3 \times 10^{8} m / s$ .

Substitute all the values in equation.

$t_{0} = (\frac{(1.05 m m) (\frac{1 m}{10^{3} m m})}{(0.992 c) (\frac{3 \times 10^{8} m / s}{c})}) \sqrt{1 - {(\frac{0.992 c}{c})}^{2}} t_{0} = 0.0446 \times 10^{- 11} s t_{0} = (0.0446 \times 10^{- 11} s) (\frac{1 p s}{10^{- 12} s}) t_{0} = 0.446 p s$