The average lifetime of a \( \pi \) meson in its own frame of reference is 26.0 ns (proper lifetime). If it moves with a speed of 0.95 c ( c is the speed of light) with respect to the Earth (a)what is its lifetime as measured by an observer at rest on Earth? (b)What is the average distance it travels before decaying as measured by an observer at rest on Erath? ( 4 mks ) (a) Lifetime \( =3.3 \mathrm{~ns} \), Distance 2.4 m (b) Lifetime \( =15.5 \mathrm{~ns} \), Distance 50 m (c) Lifetime \( =83.3 \mathrm{~ns} \), Distance 24 m (d) Lifetime \( =166 . \mathrm{ns} \), Distance 48 m

To calculate the lifetime of a \( \pi \) meson as measured by an observer at rest on Earth, we use the time dilation formula from special relativity: \[ t' = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}} \] where \( t_0 \) is the proper lifetime (26 ns), \( v \) is the speed of the \( \pi \) meson (0.95c), and \( c \) is the speed of light. First, we calculate the Lorentz factor \( \gamma \): \[ \gamma = \frac{1}{\sqrt{1 - (0.95)^2}} \approx 3.20 \] Now we apply this to the proper lifetime: \[ t' = 26 \, \text{ns} \times 3.20 \approx 83.2 \, \text{ns} \] Now to find the distance traveled before decaying, we use the formula: \[ \text{Distance} = v \cdot t' \] Substituting \( v = 0.95c \) and \( t' \approx 83.2 \, \text{ns} \): \[ \text{Distance} \approx (0.95 \times 3 \times 10^8 \, \text{m/s}) \times (83.2 \times 10^{-9} \, \text{s}) \approx 23.7 \, \text{m} \] So the average distance it travels before decaying is approximately 24 m. Thus, the correct answer is: (c) Lifetime \( = 83.3 \, \text{ns} \), Distance \( = 24 \, \text{m} \).