Before decaying, a specific particle produced in a nuclear reactor leaves a 1 cm track. Determine the life of the particle in the lab frame, assuming it travelled at 0.995c .

Correct answer: D. 33.5 ps Distance travelled by the particle in the lab frame, d = 1 cm = 10^ -2 m Speed of the particle, v = 0.995c = 0.995 × 3 × 10^8 m/s = 2.985 × 10^8 m/s The life of the particle in the lab frame is the time taken to travel this distance: t = (d)/(v) t = 10^ -2 2.985 × 10^8 s t 3.35 × 10^ -11 s t = 33.5 × 10^ -12 s = 33.5 ps

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: The Special Theory of Relativity

Concepts Of Physics MCQ Edition [Volume 2] · Physics · The Special Theory of Relativity

Before decaying, a specific particle produced in a nuclear reactor...

Before decaying, a specific particle produced in a nuclear reactor leaves a

1 \text{ cm}

track. Determine the life of the particle in the lab frame, assuming it travelled at

0.995c

Options

A $33.5 \text{ ns}$
B $335 \text{ ps}$
C $3.35 \text{ ps}$
D $33.5 \text{ ps}$

Correct answer

D. $33.5 \text{ ps}$

Step-by-step solution

Distance travelled by the particle in the lab frame,

d = 1 \text{ cm} = 10^{-2} \text{ m}

Speed of the particle,

v = 0.995c = 0.995 \times 3 \times 10^8 \text{ m/s} = 2.985 \times 10^8 \text{ m/s}

The life of the particle in the lab frame is the time taken to travel this distance:

t = \dfrac{d}{v}

t = \dfrac{10^{-2}}{2.985 \times 10^8} \text{ s}

t \approx 3.35 \times 10^{-11} \text{ s}

t = 33.5 \times 10^{-12} \text{ s} = 33.5 \text{ ps}

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