Question

the table below shows the distance between the atoms in an atom pair. all four pairs of atoms have the same amount of electric charge.

pair of atoms	distance between the atoms
c & d	9 units
p & q	25 units
x & y	16 units

which pair of atoms are repelled from each other with the most force?
a. pair x & y
b. pair p & q
c. pair a & b
d. pair c & d

Explanation:

Step1: Recall Coulomb's law

The electrostatic force $F$ between two charged - particles is given by $F = k\frac{q_1q_2}{r^2}$, where $k$ is a constant, $q_1$ and $q_2$ are the charges of the two particles, and $r$ is the distance between them. Since all four pairs of atoms have the same amount of electric charge ($q_1 = q_2$ is constant for all pairs), the force is inversely proportional to the square of the distance $r$ between the atoms, i.e., $F\propto\frac{1}{r^2}$.

Step2: Compare the distances

We have the distances: $r_{AB}=4$ units, $r_{CD}=9$ units, $r_{PQ}=25$ units, and $r_{XY}=16$ units. The smaller the value of $r$, the larger the value of $\frac{1}{r^2}$, and thus the larger the force $F$.

Step3: Identify the pair with the smallest distance

Among the distances $4$, $9$, $16$, and $25$ units, the smallest distance is $4$ units, which corresponds to the pair A & B.

Answer:

C. pair A & B