Given below are two statements:one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: If oxygen ion ($$O^{-2}$$) and Hydrogen ion ($$H^+$$) enter normal to the magnetic field with equal momentum, then the path of $$O^{-2}$$ ion has a smaller curvature than that of $$H^+$$.
Reason R: A proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly.
In the light of the above statement,Choose the correct answer from the options given below.

When a charged particle of charge $$q$$ and linear momentum $$p$$ enters a uniform magnetic field $$\mathbf{B}$$ perpendicular to the field, the magnetic force $$|q|vB$$ provides the required centripetal force $$\dfrac{mv^{2}}{r}$$.

Substituting $$p = mv$$ and $$v = \dfrac{p}{m}$$, the radius (called radius of curvature) is obtained as
$$r = \dfrac{p}{|q|B}$$ $$-(1)$$

Case 1: Oxide ion $$O^{2-}$$ versus proton $$H^{+}$$
For $$O^{2-}: |q| = 2e \quad(\text{two electronic charges})$$
For $$H^{+}: |q| = e$$
Both ions are given to have the same momentum $$p$$ and the same field $$B$$, so using $$(1)$$:

$$r_{O^{2-}} = \dfrac{p}{2eB}, \qquad r_{H^{+}} = \dfrac{p}{eB} = 2\,r_{O^{2-}}$$

Thus $$r_{O^{2-}} \lt r_{H^{+}}$$, i.e. the oxide ion describes a circle of smaller radius (larger bending) than the proton. Consequently its curvature radius is smaller. Hence Assertion A is correct.

Case 2: Proton versus electron with the same linear momentum
For a proton $$|q_p| = e$$ and for an electron $$|q_e| = e$$. With equal momenta $$p$$, equation $$(1)$$ gives

$$r_p = \dfrac{p}{eB}, \qquad r_e = \dfrac{p}{eB}$$

The radii are identical; the proton does not form a path of smaller radius than the electron. Therefore Reason R is false.

Conclusion: Assertion A is true, Reason R is false. The correct choice is Option A.