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QPage 458: Q16: HC Verma: Concepts of Physics - Volume 2 The Special Theory of Relativity Page 458: Q16 Solution: A particular particle created in a nuclear reactor leaves a 1…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse which frame observes the proper time, incorrectly applying the Lorentz factor to the lab frame or assuming the particle's lifetime is longer in its own rest frame. Remember that the proper time $\tau$ is measured in the frame where the particle is at rest (the particle frame), so it is the shortest time interval; the lab frame time $t$ is dilated and calculated as $t = \gamma \tau$.

A particular particle created in a nuclear reactor leaves a 1 cm track before decaying. Assuming that the particle moved at 0.995c, calculate the life of the particle a) in the lab frame and b) in the frame of the particle.

Step-by-Step Verified Solution

**Answer:** (a) Lab frame: t=3.35×10−11t = 3.35 \times 10^{-11} s (b) Particle frame: t′=3.35×10−12t' = 3.35 \times 10^{-12} s. **Step-by-Step Explanation:** v=0.995cv = 0.995c. γ=1/1−0.9952=1/1−0.990025=1/0.009975=10.01\gamma = 1/\sqrt{1 - 0.995^2} = 1/\sqrt{1 - 0.990025} = 1/\sqrt{0.009975} = 10.01. (a) Lab frame time: t=d/v=0.01/(0.995×3×108)=3.35×10−11t = d/v = 0.01/(0.995 \times 3 \times 10^8) = 3.35 \times 10^{-11} s. (b) Particle frame (proper time): t′=t/γ=3.35×10−11/10.01=3.35×10−12t' = t/\gamma = 3.35 \times 10^{-11}/10.01 = 3.35 \times 10^{-12} s.

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