Question

5 fill in the blank 1 point as the distance between a proton and valence electron decreases, the force of attraction choose your answer... previous next

Explanation:

Step1: Recall Coulomb's law

The force between two charged - particles (a proton and an electron) 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.

Step2: Analyze the relationship

Since $q_1$ (charge of proton) and $q_2$ (charge of electron) are constant, when the distance $r$ decreases, the value of $\frac{1}{r^2}$ increases. So, the force of attraction $F$ increases.

Answer:

increases