For a hydrogen atom of radius 10^-10 m how does the ratio of the gravitational force between the electron and the proton to the electrostatic force F_g/F_e change if the atomic radius is doubled? A)The ratio decreases by a factor of four. B)The ratio decreases by a factor of two. C)The ratio remains unchanged. D)The ratio doubles. E)The ratio increases by a factor of four.

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Quizplus · Accepted Answer

The gravitational force ($F_g$) between two masses is given by Newton's law of universal gravitation, $F_g = G \frac{m_1 m_2}{r^2}$, where $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the distance between their centers. The electrostatic force ($F_e$) between two charges is given by Coulomb's law, $F_e = k \frac{q_1 q_2}{r^2}$, where $k$ is Coulomb's constant, $q_1$ and $q_2$ are the charges of the two objects, and $r$ is the distance between their centers. When the atomic radius is doubled, the distance $r$ in both formulas is doubled, but since both forces depend on $1/r^2$, the ratio $F_g/F_e$ remains unchanged because the change in distance affects both forces equally.

For a Hydrogen Atom of Radius 10-10 M How Does

For a Hydrogen Atom of Radius 10^-10 M How Does