Q5PE - College Physics (Urone)

Question

A neutral π-meson is a particle that can be created by accelerator beams. If one such particle lives 1.40×10⁴⁸s  as measured in the laboratory, and  0.840×10^-18 s when at rest relative to an observer, what is its velocity relative to the laboratory?

 

Step-by-Step Solution

Answer

The velocity to the laboratory is v = 0.800 c.

Step 1: To find the velocity relative to the laboratory

The life of the π -meson when at rest relative to an observer i.e., proper time is

   ∆t₀ = 0.840 × 10^-16 s

The life of the π - meson when measured in the laboratory

  ∆t = 1.40 × 10^-16 s

As we know the relativistic factor is given by 

  \[\gamma=\frac{1}{{\sqrt{1-{{\left({\frac{v}{c}}\right)}^2}}}}\]                         ... (1 )

 

Where v is its velocity relative to an observer and c =  3 × 10 ⁸ m/s  is the speed of the light we have

 \begin{aligned}\Delta t = \frac{{\Delta {t_0}}}{{\sqrt {1 - {{\left( {\frac{v}{c}} \right)}^2}} }}\\= \gamma \Delta {t_0}\end{aligned}       … (2 )

                                                                

Step 2: To solve the equation

From the above equation (2), we get

_{\begin{aligned}v = c\sqrt {1 - {{\left( {\frac{{\Delta {t_0}}}{{\Delta t}}} \right)}^2}} \\= c\sqrt {1 - {{\left( {\frac{{0.840\times{{10}^{-16}}\;{\rm{s}}}}{{1.40\times{{10}^{ -16}}\;{\rm{s}}}}} \right)}^2}} \\= 0.800c\end{aligned}}

 

Hence the velocity to the laboratory is v = 0.800 c.