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In the question number 34, the kinetic energy (in MeV) of the proton beam produced by the accelerator is (radius of dees =60 cm) |
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Answer» 5 `v=rxx2 pi upsilon=0.6xx2xx3.14xx12xx10^(6)=4.5xx10^(7)"m s"^(-1)` `therefore""K=(1)/(2)MV^(2)=(1)/(2)XX(1.67xx10^(-27)(4.5xx10^(7))^(2))/(1.6xx10^(-13))` `="10.6 MEV"` |
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