A proton and a singly charged ion of mass 76 atomic mass units ( amu ) are accel

Question

A proton and a singly charged ion of mass 76 atomic mass units (amu) are accelerated through the same potential difference and enter a region of uniform magnetic field moving perpendicular to the magnetic field. What is the ratio of their kinetic energies?

If the radius of the proton's circular path in the magnetic field is 5.4 cm , what is the radius of the path of the singly charged ion of 76 amu?

By what factor must the magnetic field be multiplied by for the singly charge ion to have the same radius path as the proton?

Explanation / Answer

r=mv/qB

so r1/r2=(2mKE)/qB

or r1/r2=(76/1)^0.5

or r2=47.076 cm

b)the path ratio is independent of the manetic field, so we cannot make the two radii equal

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