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(A, C) The radius of a nucleus is given by the formula $r = R_0 A^{1/3}$, where $R_0$ is a constant and $A$ is the mass number.Since $A_A = 64$ and $A

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$E_{bnA} > E_{bnB}$

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(A, C) The radius of a nucleus is given by the formula $r = R_0 A^{1/3}$, where $R_0$ is a constant and $A$ is the mass number.Since $A_A = 64$ and $A_B = 125$, we have $r_A = R_0 (64)^{1/3} = 4R_0$ and $r_B = R_0 (125)^{1/3} = 5R_0$.Thus, $r_A Regarding binding energy per nucleon $(E_{bn})$, the value of $E_{bn}$ increases with mass number for light nuclei and reaches a maximum near $A = 56$ (Iron). For nuclei with $A > 60$, $E_{bn}$ gradually decreases.Since $A_A = 64$ is closer to $56$ than $A_B = 125$, the binding energy per nucleon for nucleus $A$ is greater than that for nucleus $B$, i.e.,$E_{bnA} > E_{bnB}$.Therefore, both options $A$ and $C$ are correct.

Let the binding energy per nucleon of a nucleus be denoted by $E_{bn}$ and the radius of the nucleus be denoted by $r$. If the mass numbers of nuclei $A$ and $B$ are $64$ and $125$ respectively, then:

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