Q3PE
Question
Particles called π-mesons are produced by accelerator beams. If these particles travel at \({\rm{2}}{\rm{.70 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{\;m/s }}\) and live \({\rm{2}}{\rm{.60 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}\) when at rest relative to an observer, how long do they live as viewed in the laboratory?
Step-by-Step Solution
VerifiedThe life of the \({\rm{\pi }}\)-mesons viewed in the laboratory is \(5.97 \times {10^{ - 8}}\;{\rm{s}}\).
Here the velocity of the \({\rm{\pi }}\)-mesons relative to an observer at rest is,
\(\begin{align}v &= 2.70 \times {10^8}\;{\rm{m/s}}\\ &= 2.70 \times {10^8}\;{\rm{m/s}}\left( {\frac{c}{{3.00 \times {{10}^8}\;{\rm{m/s}}}}} \right)\\ &= 0.900c\end{align}\)
The life of the \({\rm{\pi }}\)-mesons when at rest relative to an observer i.e.proper time is
\(\Delta {t_0} = 2.60 \times {10^{ - 8}}\;{\rm{s}}\)
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 to an observer and c= \({\rm{3}}{\rm{.00}}\) X \({10^8}\) \({\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}\)is the speed of light
As viewed in the laboratory
\(\begin{align}\Delta t &= \frac{{\Delta {t_0}}}{{\sqrt {1 - {{\left( {\frac{v}{c}} \right)}^2}} }}\\ &= \gamma \Delta {t_0}\end{align}\) ……. (2)
Therefore, from equation (1), we get
\(\begin{align}\gamma &= \frac{{\sqrt {1 - {{\left( {\frac{v}{c}} \right)}^2}} }}{{\sqrt {1 - {{\left( {\frac{{0.900c}}{c}} \right)}^2}} }}\\ &= \frac{1}{{\sqrt {1 - {{\left( {\frac{{0.900c}}{c}} \right)}^2}} }}\\ &= \frac{1}{{\sqrt {1 - {{(0.900)}^2}} }}\end{align}\)
Now from the equation (2), the life of the \({\rm{\pi }}\)-mesons as viewed in the laboratory
\(\begin{align}\Delta t &= \gamma \Delta {t_0}\\ &= 2.294.60 \times {10^{ - 8}}\;{\rm{s}}\\ &= 5.97 \times {10^{ - 8}}\;{\rm{s}}\end{align}\)
Hence the life of the \({\rm{\pi }}\)-mesons viewed in the laboratory is \(5.97 \times {10^{ - 8}}\;{\rm{s}}\).