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Particles of mass 1.6xx10^(-27) kg and charge 1.6xx10^(-19)C are accelerated in a cyclotron of dee radius 40cm. It employs a magnetic field 0.4T. Find the kinetic energy (in MeV) of the particle beam imparted by the accelerator. |
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Answer» Solution :Here mass of charged PARTICLE `m=1.6xx10^(-27)kg`, charge `q=1.6xx10^(-19)C`, RADIUS of dee `R=40cm=0.4m` and magnetic field B = 0.4T. `therefore` KINETIC energy of the particle beam imparted by the cyclotron accelerator `K_("max")=(B^(2)q^(2)R^(2))/(2m)=((0.4)^(2)xx(1.6xx10^(-19))^(2)xx(0.4)^(2))/(2xx(1.6xx10^(-27)))J` `=((0.4)^(2)xx(1.6xx10^(-19))^(2)xx(0.4)^(2))/(2xx(1.6xx10^(-27))xx(1.6xx10^(-13)))=128MeV` |
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