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(A) For an electron and nucleus pair,the potential energy is given by $V(r) = \frac{K(-e)(+Ze)}{r} = -\frac{KZe^2}{r}$.Since the potential energy is n

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$(1, n), (2, p^{+}), (3, e^{-})$

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(A) For an electron and nucleus pair,the potential energy is given by $V(r) = \frac{K(-e)(+Ze)}{r} = -\frac{KZe^2}{r}$.Since the potential energy is negative and approaches zero as $r$ increases,graph $(3)$ represents the electron.For a neutron,there is no electrostatic force outside the nucleus $(r > r_0)$. Thus,the potential energy is zero for $r > r_0$. Graph $(1)$ represents the neutron.For a proton,the electrostatic force is repulsive for $r > r_0$ because both the nucleus and the proton are positively charged. Thus,the potential energy is positive and decreases as $r$ increases. Graph $(2)$ represents the proton.Therefore,the correct pairing is $(1, n), (2, p^{+}), (3, e^{-})$,which corresponds to option $A$.

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