To solve this problem, we need to consider both the electrostatic and gravitational forces between two hydrogen atoms.$\\$Each hydrogen atom consists of a proton and an electron. Let's assume:$\\$• Charge of the proton $= e + \Delta e \\$• Charge of the electron $= -e \\$The net charge of a hydrogen atom is $(e + \Delta e) - e = \Delta e. \\$The electrostatic force between two hydrogen atoms separated by distance $d$ is given by:$\\ F_{elec} = \frac{k (\Delta e)^2}{d^2} \\$where $k$ is Coulomb's constant.$\\$The gravitational force between two hydrogen atoms is given by:$\\ F_{grav} = \frac{G (m_h)^2}{d^2} \\$where $G$ is the gravitational constant and $m_h$ is the mass of a hydrogen atom.$\\$According to the problem, the net force is zero, so:$\\ F_{elec} = F_{grav} \\ \frac{k (\Delta e)^2}{d^2} = \frac{G (m_h)^2}{d^2} \\$Cancel $d^2$ from both sides:$\\ k (\Delta e)^2 = G (m_h)^2 \\$Solve for $\Delta e: \\ (\Delta e)^2 = \frac{G (m_h)^2}{k} \\ \Delta e = \sqrt{\frac{G (m_h)^2}{k}} \\$Substitute the known values:$\\ G = 6.67 \times 10^{-11} \, \mathrm{N \cdot m^2/kg^2} \\ m_h = 1.67 \times 10^{-27} \, \mathrm{kg} \\ k = 8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \\ \Delta e = \sqrt{\frac{6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2}{8.99 \times 10^9}} \\ \Delta e = \sqrt{\frac{6.67 \times 10^{-11} \times 2.7889 \times 10^{-54}}{8.99 \times 10^9}} \\ \Delta e = \sqrt{\frac{1.860 \times 10^{-64}}{8.99 \times 10^9}} \\ \Delta e = \sqrt{2.068 \times 10^{-74}} \\ \Delta e \approx 1.44 \times 10^{-37} \, \mathrm{C} \\$Thus, $\Delta e$ is of the order of $10^{-37} \, \mathrm{C}. \\$This corresponds to Option 2.