Proton, Deuteron and alpha particle of the same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, - Physics MCQ AIEEE 2012 2026 | ExamTest.live

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Proton, Deuteron and alpha particle of the same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively r_{p}, r_{d} and r_{α} . Which one of the following relations is correct?

Solution & Step-by-step Explanation

Radius of a circular path in a magnetic field is r = \frac{2 m K}{qB} .Since K (kinetic energy) and B are constant: r \propto \frac{m}{q} For Proton (p): m_{p} = 1, q_{p} = 1 ⟹ r_{p} \propto \frac{1}{1} = 1 .For Deuteron (d): m_{d} = 2, q_{d} = 1 ⟹ r_{d} \propto \frac{2}{1} \approx 1.414 .For Alpha (α): m_{α} = 4, q_{α} = 2 ⟹ r_{α} \propto \frac{4}{2} = 1 .Thus, r_{p} = r_{α} and r_{d} > r_{p} .Relationship: r_{α} = r_{p} < r_{d} .

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Proton, Deuteron and alpha particle of the same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively r_{p}, r_{d} and r_{α} . Which one of the following relations is correct?

r_\alpha = r_p = r_d

r_\alpha = r_p < r_d

r_d > r_p > r_\alpha

r_d = r_\alpha > r_p

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