188k views
2 votes
The proper lifetime of a certain particle is 161.0 ns (a) How long does it live in the laboratory if it moves at v- 0.945c? ns (b) How far does it travel in the laboratory during that time? (c) What is the distance traveled in the laboratory according to an observer moving with the particle?

1 Answer

2 votes

(a) 492.2 ns

The lifetime of the particle in the frame of reference of the laboratory is given by


T= \frac{T_0}{\sqrt{1-((v)/(c))^2}}

where


T_0 is the proper lifetime

v is the speed of the particle

c is the speed of the light

For this particle we have:


T_0 = 161 ns


v=0.945c

Substituting into the equation, we get:


T= (161)/(√(1-(0.945)^2))=492.2 ns

(b) 139.5 m

In the reference frame of the laboratory, the distance travelled by the particle is given by


L = v T

where

v is the speed of the particle

T is the lifetime of the particle in the laboratory frame of reference (found in part a)

The speed of the particle is


v=0.945c=0.945(3\cdot 10^8)=2.84\cdot 10^8 m/s

The lifetime of the particle in the laboratory's frame of reference is


T=492.2 ns = 492.2\cdot 10^(-9)s

And substituting into the equation, we find:


L=(2.84\cdot 10^8)(492.2\cdot 10^(-9))=139.5 m

(c) 426.5 m

Here we have to calculate the distance travelled by the particle in its frame of reference. This can be calculated by using the equation


L=L_0 \sqrt{1-((v)/(c))^2}

where


L_0 is the distance travelled measured by an observer moving with the particle

L is the distance travelled measured by an observer in the laboratory

We already know that

L = 139.5 m

So, solving the formula for
L_0,


L_0 = \frac{L}{\sqrt{1-((v)/(c))^2}}=(139.5)/(√(1-(0.945)^2))=426.5 m

User MarGin
by
6.5k points