A Proton and an Alpha particle are accelerated through the same potential difference enter in a region of uniform magnetic field with their velocities perpendicular to the field they also have same Kinetic energy. compare the radii of circular path followed by them

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Solution

we know that

mv2/r = qvB

The radius of the cirxular path is given as

r = mv/qB

now as E = (1/2)mv2

we have, mv = (2mE)1/2

so, the radius will be

r = √(2mE)/qB

or

the radius of proton

rp = √(2mpE) / qpB

the radius of an alpha particle

ra = √(2maE) / qaB

we also know that

mp = 4ma

and

qp = 2qa

so, the ratios would be

rp : ra = 1 : 1

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