Question

Particle track detectors are used to measure the speed of particles if the lifetime of the particle is known. Particle-X has a lifetime of 256.2 ps. These particles are created in an experiment inside the detector by a given reaction. The particles leave 10.7 cm long tracks on average before they decay into other particles not observable by the detector. What is the average speed of the particles in terms of the speed of light

Answer 1

Answer:

the average speed of the particles in terms of the speed of light is 0.81c

Explanation:

Given the data in the question;

$t_p$ = 256.2 ps = 256.2 × 10⁻¹² s

speed of light c = 2.99 × 10⁸ m/s

d = 10.7 cm = 0.107 m

we know that; Average speed v = d/t ------- let this be equation 1

Also, given that 256.2 ps is the lifetime of particle X frame, proper time will be;

t = Y$t_p$ = $t_p$ / √( 1 - $\frac{v^2}{c^2}$ ) --------- let this be equation 2

Next, we input equation 2 into equation 1'

v = d / [ $t_p$ / √( 1 - $\frac{v^2}{c^2}$ )  ]

v = d/$t_p$[ √( 1 - $\frac{v^2}{c^2}$ ) ]

v$t_p$/d = √( 1 - $\frac{v^2}{c^2}$ )

we square both sides

( v$t_p$/d )² = (√( 1 - $\frac{v^2}{c^2}$ ) )²

v²$t_p$²/d² = 1 - $\frac{v^2}{c^2}$

v²$t_p$²/d² +  $\frac{v^2}{c^2}$ = 1

v²( $t_p$²/d² +  $\frac{1}{c^2}$ ) = 1

v²( ($t_p$²c² + d²)/d²c² ) = 1

∴

v² = (d²c²) / ($t_p$²c² + d²)

v = √[ (d²c²) / ($t_p$²c² + d²) ]

v = (dc) / √($t_p$²c² + d²)

so we substitute

v = (0.107 m × c) / √( (256.2 × 10⁻¹² s)²(2.99 × 10⁸ m/s)² + (0.107 m )²)

v = 0.107c / √( 0.00586814 + 0.011449 )

v = 0.107c / √( 0.01731714 )

v = 0.107c / 0.1315946

v = 0.81c

Therefore, the average speed of the particles in terms of the speed of light is 0.81c