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Maximum kinetic energy of positive ion in cyclotron is _____ (Where r_(0) = radius of cyclotron) |
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Answer» `(Q^(2)Br_(0))/(2m)` `K=1/2mv^(2)` `thereforeK_(max)=1/2mv_(max)^(2)""...(1)` We have, `r=(mv)/(Bq)` `thereforev=(Bqr)/m` `thereforev^(2)=(B^(2)q^(2)r^(2))/m^(2)` `thereforev_(max)^(2)=(B^(2)q^(2))/m^(2)r_(max)^(2)""...(2)` `thereforev_(max)^(2)=(B^(2)q^(2))/m^(2)r_(0)^(2)""(becauser_(max)=r_(0))` From equation (1) and (2), `K_(max)=1/2mxx(B^(2)q^(2)r_(0)^(2))/m^(2)=(B^(2)q^(2)r_(0)^(2))/(2m)` |
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