A pion lives 26 ns in its rest frame. It travels 15 cm down a beam pipe in a particle accelerator before it decays into other particles. How far is this in the rest frame of the pion?

Answer :

Explanation:

It is given that,

Life time of a pion, [tex]t=26\ ns=26\times 10^{-9}\ s[/tex]

Distance covered by a pion, d = 15 cm = 0.15 m

We need to find distance from the rest frame of the pion. Firstly, we can calculate the speed of the pion as :

[tex]v=\dfrac{d}{t}[/tex]

[tex]v=\dfrac{0.15}{26\times 10^{-9}}[/tex]

v = 5769230.76 m/s

or

[tex]v=5.76\times 10^6\ m/s[/tex]

Its length in the rest frame is given by the formula as :

[tex]d_o=d\sqrt{1-\dfrac{v^2}{c^2}}[/tex]

[tex]d_o=0.15\sqrt{1-\dfrac{(5.76\times 10^6)^2}{(3\times 10^8)^2}}[/tex]

[tex]d_o=0.149\ m[/tex]

or

[tex]d_o=14.9\ cm[/tex]

So, the length of the pion in rest frame is 14.9 cm. Hence, this is the required solution.

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