A proton and ` alpha - particle` are accelerated with same potential difference and they enter in the region of constant magnetic field `B` perpendicular to the velocity of particles. Find the ratio of radius of curvature of ` alpa - particle`.
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KEY CONCEPT : ` eV = (1)/(2) mv_(p)^(2) and eV = (1)/(2) mv_(alpha)^(2)` `V` is the potential, difference `v_(p) = velocity of proton` `v_(alpha) = velocity of alpha- particle` `m = mass of proton , mass of alpha - particle = 4m` rArr `v_(p) = sqrt((2ev)/(m)) , v_(alpha) = sqrt((2ev)/( 4m))` Now when the particles enter in magnetic field , the force on proton is `ev_(p)B = ( mv_(p)^(2))/( r_(p)) or r_(p) = ( mv_(p))/(eB) rArr r_(alpha) = (m)/(eB)` ` sqrt((2eV)/( m)) = (1)/(B) sqrt((2mV)/(e)) and r_(alpha) = (1)/(B)sqrt(( 4 mV)/(e))` :. ( r_(p))/(r_(alpha)) = (1)/(sqrt(2))`
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