Maximum kinetic energy of positive ion in cyclotron is _____ (Where `r_(0)` = radius of cyclotron)

Options

A. `(q^(2)Br_(0))/(2m)`
B. `(qB^(2)r_(0))/(2m)`
C. `(q^(2)B^(2)r_(0)^(2))/(2m)`
D. `(qBr_(0))/(2m)`

Answer

C

Explanation

Kinetic energy of charged particle in a cyclotron,
`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)`