Question

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`.

Answer 1

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))`