A proton,a deuteron and an alpha-particle hav | vedclass

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Question

(A) Given that the kinetic energy $K$ is the same for all particles: ${K_p} = {K_d} = {K_\alpha} = K$.We know the mass and charge relations: $m_p = m$

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${r_\alpha} = {r_p} < {r_d}$

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(A) Given that the kinetic energy $K$ is the same for all particles: ${K_p} = {K_d} = {K_\alpha} = K$.We know the mass and charge relations: $m_p = m$,$m_d = 2m$,${m_\alpha} = 4m$ and $q_p = e$,$q_d = e$,${q_\alpha} = 2e$.The radius of a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.For the proton: ${r_p} = \frac{\sqrt{2mK}}{eB}$.For the deuteron: ${r_d} = \frac{\sqrt{2(2m)K}}{eB} = \sqrt{2} \left( \frac{\sqrt{2mK}}{eB} \right) = \sqrt{2} {r_p}$.For the $\alpha$-particle: ${r_\alpha} = \frac{\sqrt{2(4m)K}}{(2e)B} = \frac{2\sqrt{2mK}}{2eB} = \frac{\sqrt{2mK}}{eB} = {r_p}$.Comparing these,we get ${r_\alpha} = {r_p} < {r_d}$.

$A$ proton,a deuteron and an $\alpha$-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If ${r_p}$,${r_d}$ and ${r_\alpha}$ denote respectively the radii of the trajectories of these particles,then:

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