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

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

Vedclass · Accepted Answer

$1: \sqrt{2}: 1$

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(A) The radius $R$ of the circular path of a charged particle in a uniform magnetic field is given by $R = \frac{mv}{qB}$.Since kinetic energy $E = \frac{1}{2}mv^2$, we have $mv = \sqrt{2mE}$.Thus, $R = \frac{\sqrt{2mE}}{qB}$.For a proton $(p)$: $m_p = m$, $q_p = e$, so $R_p = \frac{\sqrt{2mE}}{eB}$.For a deuteron $(d)$: $m_d = 2m$, $q_d = e$, so $R_d = \frac{\sqrt{2(2m)E}}{eB} = \sqrt{2} \frac{\sqrt{2mE}}{eB}$.For an $\alpha$-particle $(\alpha)$: $m_{\alpha} = 4m$, $q_{\alpha} = 2e$, so $R_{\alpha} = \frac{\sqrt{2(4m)E}}{2eB} = \frac{2\sqrt{2mE}}{2eB} = \frac{\sqrt{2mE}}{eB}$.Comparing the radii: $R_p : R_d : R_{\alpha} = 1 : \sqrt{2} : 1$.

$A$ proton, a deuteron, and an $\alpha$-particle are projected perpendicular to the direction of a uniform magnetic field with the same kinetic energy. The ratio of the radii of the circular paths described by them is

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