A proton, a deuteron ion and an α-particle of equal kinetic energy perform circular motion normal to a uniform magnetic field B. If the radii of their paths are rₚ, r_d and r_α respectively then ...... (Here q_d=qₚ, m_d=2mₚ, q_α=2qₚ, m_α=4mₚ)

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(1) r_α=rₚ<r_d. Given: a proton, a deuteron and an α-particle of the same kinetic energy K in the same field B, with q_d=qₚ, m_d=2mₚ, q_α=2qₚ, m_α=4mₚ. Idea: write the radius in terms of the kinetic energy rather than the speed, because it is K that the three share. r=(mv)/(qB) and the momentum is mv=√2mK, so r=(√2mK)/(qB), that is r∝(√m)/q. Proton: (√mₚ)/(qₚ). Deuteron: (√2mₚ)/(qₚ)=√2 (√mₚ)/(qₚ). α-particle: (√4mₚ)/(2qₚ)=(√mₚ)/(qₚ). So the α-particle matches the proton while the deuteron is √2 times larger: r_α=rₚ<r_d.

A proton, a deuteron ion and an α-particle of equal kinetic energy perform circular motion normal to a uniform magnetic field B. If the radii of their paths are rₚ, r_d and r_α respectively then ...... (Here q_d=qₚ, m_d=2mₚ, q_α=2qₚ, m_α=4mₚ)

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