Question

A neutral π-meson is a particle that can be created by accelerator beams. If one such particle lives 1.40×10⁴⁸sas measured in the laboratory, and0.840×10^-18s when at rest relative to an observer, what is its velocity relative to the laboratory?

Short answer

The velocity to the laboratory is v = 0.800 c.

Step-by-step solution

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^-16s

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

∆t = 1.40 × 10^-16s

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

$$\begin{array}{r}\mathrm{\Delta }t=\frac{\mathrm{\Delta }{t}_0}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\\ =\gamma \mathrm{\Delta }{t}_0\end{array}$$ … (2 )

To solve the equation

From the above equation (2), we get

_{$$\begin{array}{r}v=c\sqrt{1-{\left(\frac{\mathrm{\Delta }{t}_0}{\mathrm{\Delta }t}\right)}^2}\\ =c\sqrt{1-{\left(\frac{0.840\times {10}^{-16}\mathrm{s}}{1.40\times {10}^{-16}\mathrm{s}}\right)}^2}\\ =0.800c\end{array}$$}

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