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

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(A) The radius of the circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $E = \frac{1}{2}mv^

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$r_{\alpha} = r_p < r_d$

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(A) The radius of the circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $E = \frac{1}{2}mv^2$, we have $mv = \sqrt{2mE}$.Substituting this, $r = \frac{\sqrt{2mE}}{qB}$.Given that $E$ and $B$ are the same for all particles, we have $r \propto \frac{\sqrt{m}}{q}$.For a proton $(p)$: mass $m_p = m$, charge $q_p = e$.For a deuteron $(d)$: mass $m_d = 2m$, charge $q_d = e$.For an $\alpha -$ particle $(\alpha)$: mass $m_{\alpha} = 4m$, charge $q_{\alpha} = 2e$.Calculating the ratio $r_p : r_d : r_{\alpha} = \frac{\sqrt{m}}{e} : \frac{\sqrt{2m}}{e} : \frac{\sqrt{4m}}{2e}$.Simplifying the ratio: $1 : \sqrt{2} : \frac{2}{2} = 1 : \sqrt{2} : 1$.Thus, $r_p = r_{\alpha} < r_d$.

$A$ proton, a deuteron, and an $\alpha -$ particle enter a region of perpendicular magnetic field (to their velocities) with the same kinetic energy. If $r_p, r_d,$ and $r_{\alpha}$ are the radii of the circular paths of these particles, then:

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