An unstable high-energy particle is created in the laboratory, and it moves at a

Question

An unstable high-energy particle is created in the laboratory, and it moves at a speed of 0.987c. Relative to a stationary reference frame fixed to the laboratory, the particle travels a distance of 1.00 × 10-3 m before disintegrating. What is (a) the proper distance and (b) the distance measured by a hypothetical person traveling with the particle? Determine the particle's (c) proper lifetime and (d) its dilated lifetime.
An unstable high-energy particle is created in the laboratory, and it moves at a speed of 0.987c. Relative to a stationary reference frame fixed to the laboratory, the particle travels a distance of 1.00 × 10-3 m before disintegrating. What is (a) the proper distance and (b) the distance measured by a hypothetical person traveling with the particle? Determine the particle's (c) proper lifetime and (d) its dilated lifetime.
An unstable high-energy particle is created in the laboratory, and it moves at a speed of 0.987c. Relative to a stationary reference frame fixed to the laboratory, the particle travels a distance of 1.00 × 10-3 m before disintegrating. What is (a) the proper distance and (b) the distance measured by a hypothetical person traveling with the particle? Determine the particle's (c) proper lifetime and (d) its dilated lifetime.

Explanation / Answer

Given,

v = 0.987 c

a)The proper distance is;

d = 1 x 10^-3 m

b)we know that

L = L0 sqrt (1 - v^2/c^2)

L = 1 x 10^-3 sqrt (1 - 0.987^2 c^2/c^2) = 0.1607 x 10^-3 m

Hence, 1.607 x 10^-4 m

c)t = d/v

t = 1.607 x 10^-4/0.987 x 3 x 10^8 = 5.43 x 10^-13 s

Hence, t = 5.43 x 10^-13

d)t = 1 x 10^-3/0.987 x 3 x 10^8 = 3.38 x 10^-12

Hence, t = 3.38 x 10^-12

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