Question

Suppose a particle called a kaon is created by cosmic radiation striking the atmosphere. It moves by you at 0.980c0.980c, and it lives 1.24×10^-8when at rest relative to an observer. How long does it live as you observe it?

Short answer

The life of the kaon as viewed in the laboratory is 6.23×10^-8s.

Step-by-step solution

To find the velocity of the kaon

Here the velocity of the kaon relative to an observer at rest is V=0.980cV=

The life of the kaon when at rest relative to an observer i.e., Proper time is

∆t₀= 1.24 × 10^-8 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=\[{\rm{3}}{\rm{.00}}\] X \[{10^8}\] \[{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}\] 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)

Therefore, from equation (1), we get

\begin{aligned}\gamma=\frac{1}{{\sqrt {1 - {{\left( {\frac{v}{c}} \right)}^2}} }}\\ &= \frac{1}{{\sqrt{1-\left({\frac{{0.980c}}{c}}\right)}}}\\&=\frac{1}{{\sqrt{1- \left({\frac{{0.980c}}{c}} \right)} }}\\ &= 5.025\end{aligned}

The life of the kaon as viewed in the laboratory

Now, from the equation (2), the life of the kaon as viewed in the laboratory by us is

\begin{aligned}\Delta t=\gamma\Delta {t_0}\\ &=5.025\times1.24\times{10^{-8}}\;{\rm{s}}\;\\ &= 6.23 \times {10^{ - 8}}\;{\rm{s}}\end{aligned}

Hence the life of the kaon as viewed in the laboratory is 6.23 ×10^-8 s.