All Problems

Optics of Moving Sources

Problem 5.244

Find the lowest values of the kinetic energy of an electron and a proton causing the emergence of Cherenkov's radiation in a medium with refractive index n=1.60.n=1.60 . For what particles is this minimum value of kinetic energy equal to Tmin⁡=29.6MeV?T_{\min }=29.6 \mathrm{MeV} ?

Reveal Answer
 5.244. Tmin⁡=(n/Vn2−1‾−1)mc2;0.14MeV and 0.26GeV respectively. For muons. \begin{aligned} &\text { 5.244. } T_{\min }=\left(n / V \overline{n^{2}-1}-1\right) m c^{2} ; 0.14 \mathrm{MeV} \text { and } 0.26 \mathrm{GeV}\\ &\text { respectively. For muons. } \end{aligned}
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