Find the lowest values o fthe kinetic energy of an electron and a proton causing the emergence of Cherenkov's radiation in a medium with refractive index `n = 1.60`. For what particles is this minimum value of kinetic enegry equal to `T_(min) = 29.6MeV`?

Explanation

We must have
`Vge (c )/(n) = (3)/(1.6) xx 10^(8) m//s` or `(V)/(c ) ge (1)/(1.6)`
For electrons this means a `K.E.` greater than
`T_(e) ge (m_(e)c^(2))/(sqrt(1-((1)/(1.6))^(2))) - m_(e)c^(2) = m_(e)c^(2) [(n)/(sqrt(n^(2) - 1))-1]`
`= 0.511 [(1)/(sqrt(1((1)/(1.6))^(2)))-1]` using `m_(e)c^(2) = 0.511 MeV = 0.144 MeV`
For protons with `m_(P)c^(2) = 938MeV`
`T_(P) ge 938 [(1)/(sqrt(1-((1)/(1.6))^(2)))-1] = 264MeV = 0.264 GeV`
Also `T_(min) = 29.6 MeV = mc^(2) [(1)/(sqrt(1-((1)/(1.6))^(2)))-1]`
Then `mc^(2) = 105.3MeV`. This is very nearly the mass of means.