Question

A proton, a deuteron and an alpha particle, are accelerated through the same potential difference and then subjected to a unifrom magnetic field `vecB`, perpendicular to the direction of their motions. Compare (i) their kinetic energies, and (ii) if the radius of the ciruclar path described by proton is 5 cm, determine the radii of the paths described by deuteron and alpha partcile.

Answer 1

Kinetic Energy : `k=qV`
as V is constant and same for all particles
`k prop q_(1)`
`k_(p):k_(d):k_(alpha)-q_(p):q_(d):q_(alpha)-1:1:2`
Kinetic energy of a charge particle in uniform magnetic field is given by
`k=(1)/(2))(q^(2)B^(2)r^(2))/(M)`
`(k_(p))/(k_(d))=(q_(p)r_(p)^(2))/(M_(p))(M_(d))/(q_(d)r_(d)^(2))`
`(1)/(1)=(1^(2)(5)^(2))/((1))(2)/((1)^(2)r_(d)^(2))`
`r_(d)=5sqrt(2)=7.071`
Hence, radius of deutron will be 7.071 cm
`(k_(p))/(k_(alpha))=(q_(p)^(2)r_(p)^(2))/(M_(p))xx(M_(alpha))/(q_(alpha)^(2)r_(alpha)^(2))`
`(1)/(2)=((1)^(2)(5)^(2))/(1)xx(4)/(2^(2)r_(alpha)^(2))`
`r_(alpha)=5sqrt(2)=7.071 cm`