A proton and an alpha particle are accelerated from rest through the same electrical potential difference. They then enter a region of uniform magnetic field, moving perpendicular to the field lines. The ratio of the radius of the proton's path to the radius of the alpha particle's path is:

Step-by-step solution

When a particle of charge q is accelerated from rest through a potential difference V , the kinetic energy acquired is K = qV . The radius of the circular path in a uniform magnetic field B is given by r = 2mK qB . Substituting K = qV , we get: r = 2mqV qB = 1 B 2mV q Since V and B are the same for both particles, r m q . For a proton, mass is m and charge is e . For an alpha particle, mass is 4m and charge is 2e . Taking the ratio of their radii: r_p r_ = m_p m_ q_ q_p = m 4m 2e e = 2 4 = 1 2 Thus, the required ra