Question

ionization energy is the energy required to remove an electron from an atom
explain how atomic radii affects ionization energy in terms of coulomb’s law (she does not specifically mention coulomb’s law-think about it):

Explanation:

Coulomb’s Law is \( F = k\frac{q_1q_2}{r^2} \), where \( F \) is force, \( k \) is a constant, \( q_1/q_2 \) are charges, and \( r \) is distance. For an atom, the nucleus (positive charge \( q_1 \)) and the outermost electron (negative charge \( q_2 \)) interact. Atomic radius is related to the distance \( r \) between them. As atomic radius (and thus \( r \)) increases, the force \( F \) between the nucleus and the electron decreases (since \( F \) is inversely proportional to \( r^2 \)). A weaker attractive force means less energy is needed to remove the electron, so ionization energy decreases with increasing atomic radius. Conversely, a smaller atomic radius (smaller \( r \)) means a stronger attractive force, requiring more energy to remove the electron (higher ionization energy).

Answer:

According to Coulomb’s Law (\( F = k\frac{q_1q_2}{r^2} \)), the force (\( F \)) between the nucleus (positive charge, \( q_1 \)) and the outermost electron (negative charge, \( q_2 \)) is inversely proportional to the square of the distance (\( r \)) between them (atomic radius is related to \( r \)). As atomic radius increases, \( r \) increases, so the attractive force \( F \) between the nucleus and the electron decreases. A weaker attractive force means less energy is required to remove the electron, so ionization energy decreases. When atomic radius decreases, \( r \) decreases, the attractive force \( F \) increases, and more energy (higher ionization energy) is needed to remove the electron.