Question

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.

Answer 1

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Given : v = 0.995c =0.995 \times 3\times 10^8 = 2.985 \times 10^8v=0.995c =0.995×3×10

8

=2.985×10

8

m/s

Distance covered in lab frame d = 1d=1 cm =10^{-2} =10

−2

m

\therefore∴ Life of particle in lab frame t = \dfrac{d}{v} = \dfrac{10^{-2}}{2.985\times 10^8} =33.5t=

v

d

=

2.985×10

8

10

−2

=33.5 ps

Life of particle in frame of particle \tau = t\sqrt{1-(v/c)^2}τ=t

1−(v/c)

2

\therefore∴ \tau = 33.5 \times \sqrt{1 - (0.995)^2} =3.35τ =33.5×

1−(0.995)

2

=3.35 ps

Answer 2

Explanation:

It is given:

Track has the length denoted as d = 1 cm

Particle’s velocity, v = 0.995c

(a) In the lab frame, the life of the particle is shown as

$\mathrm{t}=\mathrm{dv}=0.010 .995 \mathrm{c}=0.010 .995 \times 108 \times 3=33.5 \times 10-12 \mathrm{s}=33.5 \mathrm{ps}$

(b) In the particle’s frame, let the life of the particle be t‘. So,

$t^{\prime}=t 1-v 2 / c 2$

$\mathrm{t}^{\prime}=33.5 \times 10-121-0.9952 \mathrm{t}^{\prime}=3.3541 \mathrm{ps}$