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(C) The radius of curvature $r$ of a charged particle moving perpendicular to a magnetic field $B$ is given by $r = \frac{mv}{qB} = \frac{p}{qB}$,wher

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$A$ is false but $R$ is true

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(C) The radius of curvature $r$ of a charged particle moving perpendicular to a magnetic field $B$ is given by $r = \frac{mv}{qB} = \frac{p}{qB}$,where $p$ is the momentum and $q$ is the magnitude of the charge.For Assertion $A$: Both ions have the same momentum $p$. Thus,$r \propto \frac{1}{q}$. Since the charge of $O^{-2}$ is $2e$ and $H^{+}$ is $1e$,the radius of curvature for $O^{-2}$ is $r_{O} = \frac{p}{2eB}$ and for $H^{+}$ is $r_{H} = \frac{p}{eB}$. Since $r_{O} For Reason $R$: For a proton and an electron with the same momentum $p$,the radius of curvature is $r = \frac{p}{qB}$. Since both have the same magnitude of charge $e$,their radii of curvature will be identical $(r_{p} = r_{e} = \frac{p}{eB})$. Thus,Reason $R$ is false.

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