Question

Find the mean path travelled by pions whose kinetic energy exceeds their rest energy `eta= 1.2 time s`. The mean lifetime of very slow pions is `tau_(0)= 25.5 ns`.

Answer 1

Energy of poins is `(1+eta)m_(0)c^(2)` so
`(1+eta)m_(0)c^(2) = (m_(0)c^(2))/(sqrt(1-beta^(2)))`
Hence `(1)/(sqrt(1-beta^(2)))=1+eta or beta =sqrt(eta(2+eta))/(1+eta)`
Here `beta=(v)/(c )` of pion. Hence time dilation factor is `1+eta` and the distance traversed by the pion in its lifetime will be
`(cbetatau_(0))/(sqrt(1-beta^(2)))=ctao_(0)sqrt(eta(2+eta))=15.0` metres
on substituting the values of various quantities (Note The factor `(1)/(sqrt(1-beta^(2)))` can be looked at as a time dilation effect in the laboratory frame or as length contraction factor brought to outside in the proper frame of the pion.