In a hydrogen atom's ground state, the average separation between the proton and the electron is 5.3 10⁻¹¹ m . When the atom transitions into its first excited state, this average separation increases to four times its value in the ground state. Determine the Coulomb force in this excited state.

Step-by-step solution

The average separation between the proton and the electron in the first excited state is given as: r = 4 5.3 10⁻¹¹ m = 21.2 10⁻¹¹ m The Coulomb force between the proton and the electron is given by Coulomb's law: F = 1 4 ₀ e^2 r^2 Substituting the standard values e = 1.6 10⁻¹⁹ C and 1 4 ₀ = 9 10^9 N m ^2/ C ^2 : F = 9 10^9 (1.6 10⁻¹⁹)^2 (21.2 10⁻¹¹)^2 F = 9 10^9 2.56 10⁻³⁸ 449.44 10⁻²² F = 23.04 10⁻²⁹ 449.44 10⁻²² F 5.1 10⁻⁹ N