Question

In the Bohr model of the hydrogen atom, the electron moves around the one- proton nucleus on circular orbits of well-determined radii, given by \(r_{n}=n^{2} a_{\mathrm{B}}\), where \(n=1,2,3, \ldots\) is an integer that defines the orbit and \(a_{\mathrm{B}}=5.29 \cdot 10^{-11} \mathrm{~m}\) is the radius of the first (minimum) orbit, called the Bohr radius. Calculate the force of electrostatic interaction between the electron and the proton in the hydrogen atom for the first four orbits. Compare the strength of this interaction to the gravitational interaction between the proton and the electron.

Short answer

Based on the calculations above, we can conclude that the electrostatic forces between the proton and electron in the hydrogen atom are much stronger than the gravitational forces for all four orbits. For example, the electrostatic force for orbit 1 is \(8.238 \times 10^{-8}\,\text{N}\), while the gravitational force is only \(3.624 \times 10^{-47}\,\text{N}\), a significantly smaller magnitude. This pattern is consistent for all other orbits as well.

Step-by-step solution

Write down the known values and constants

For this exercise, we need the following values and constants: - Bohr radius (\(a_B\)): \(5.29 \times 10^{-11}\,\text{m}\) - Electron charge (\(e\)): \(1.60 \times 10^{-19}\,\text{C}\) - Proton charge (\(e_p\)): \(1.60 \times 10^{-19}\,\text{C}\) - Vacuum permittivity (\(\varepsilon_0\)): \(8.85 \times 10^{-12}\,\text{C}^2\text{/N·m}^2\) - Proton mass (\(m_p\)): \(1.67 \times 10^{-27}\,\text{kg}\) - Electron mass (\(m_e\)): \(9.11 \times 10^{-31}\,\text{kg}\) - Gravitational constant (\(G\)): \(6.67 \times 10^{-11}\,\text{N·m}^2\text{/kg}^2\)

Calculate the orbit radii

We need to find the radii for the first four Bohr orbits. Use the equation \(r_n = n^2 a_B\) for \(n = 1, 2, 3, 4\). \(r_1 = 1^2 \times 5.29 \times 10^{-11}\,\text{m} = 5.29 \times 10^{-11}\,\text{m}\) \(r_2 = 2^2 \times 5.29 \times 10^{-11}\,\text{m} = 4 \times 5.29 \times 10^{-11}\,\text{m} = 21.16 \times 10^{-11}\,\text{m}\) \(r_3 = 3^2 \times 5.29 \times 10^{-11}\,\text{m} = 9 \times 5.29 \times 10^{-11}\,\text{m} = 47.61 \times 10^{-11}\,\text{m}\) \(r_4 = 4^2 \times 5.29 \times 10^{-11}\,\text{m} = 16 \times 5.29 \times 10^{-11}\,\text{m} = 84.64 \times 10^{-11}\,\text{m}\)

Calculate the electrostatic forces

For the electrostatic forces between the proton and electron, we use Coulomb's law: \(F_e = \frac{e \cdot e_p}{4 \pi \varepsilon_0 r_n^2}\) Now calculate and fill in for each \(r_n\) found in Step 2: \(F_{e1} = \frac{1.6 \times 10^{-19}\,\text{C} \cdot 1.6 \times 10^{-19}\,\text{C}}{4 \pi \cdot 8.85 \times 10^{-12}\,\text{C}^2\text{/N·m}^2 \cdot (5.29 \times 10^{-11}\,\text{m})^2} = 8.238 \times 10^{-8}\,\text{N}\) \(F_{e2} = \frac{1.6 \times 10^{-19}\,\text{C} \cdot 1.6 \times 10^{-19}\,\text{C}}{4 \pi \cdot 8.85 \times 10^{-12}\,\text{C}^2\text{/N·m}^2 \cdot (21.16 \times 10^{-11}\,\text{m})^2} = 5.149 \times 10^{-9}\,\text{N}\) \(F_{e3} = \frac{1.6 \times 10^{-19}\,\text{C} \cdot 1.6 \times 10^{-19}\,\text{C}}{4 \pi \cdot 8.85 \times 10^{-12}\,\text{C}^2\text{/N·m}^2 \cdot (47.61 \times 10^{-11}\,\text{m})^2} = 1.533 \times 10^{-9}\,\text{N}\) \(F_{e4} = \frac{1.6 \times 10^{-19}\,\text{C} \cdot 1.6 \times 10^{-19}\,\text{C}}{4 \pi \cdot 8.85 \times 10^{-12}\,\text{C}^2\text{/N·m}^2 \cdot (84.64 \times 10^{-11}\,\text{m})^2} = 3.861 \times 10^{-10}\,\text{N}\)

Calculate the gravitational forces

To calculate the gravitational forces between the proton and electron for each orbit, use: \(F_g = \frac{G \cdot m_p \cdot m_e}{r_n^2}\) Fill in the values for each of the radii found in Step 2: \(F_{g1} = \frac{6.67 \times 10^{-11}\,\text{N·m}^2\text{/kg}^2 \cdot 1.67 \times 10^{-27}\,\text{kg} \cdot 9.11 \times 10^{-31}\,\text{kg}}{(5.29 \times 10^{-11}\,\text{m})^2} = 3.624 \times 10^{-47}\,\text{N}\) \(F_{g2} = \frac{6.67 \times 10^{-11}\,\text{N·m}^2\text{/kg}^2 \cdot 1.67 \times 10^{-27}\,\text{kg} \cdot 9.11 \times 10^{-31}\,\text{kg}}{(21.16 \times 10^{-11}\,\text{m})^2} = 5.659 \times 10^{-48}\,\text{N}\) \(F_{g3} = \frac{6.67 \times 10^{-11}\,\text{N·m}^2\text{/kg}^2 \cdot 1.67 \times 10^{-27}\,\text{kg} \cdot 9.11 \times 10^{-31}\,\text{kg}}{(47.61 \times 10^{-11}\,\text{m})^2} = 1.236 \times 10^{-48}\,\text{N}\) \(F_{g4} = \frac{6.67 \times 10^{-11}\,\text{N·m}^2\text{/kg}^2 \cdot 1.67 \times 10^{-27}\,\text{kg} \cdot 9.11 \times 10^{-31}\,\text{kg}}{(84.64 \times 10^{-11}\,\text{m})^2} = 3.147 \times 10^{-49}\,\text{N}\)

Compare the forces

Comparing the electrostatic forces to gravitational forces, we observe that the electrostatic forces are much stronger than the gravitational forces for all four orbits. The electrostatic force for orbit 1, for instance, is \(8.238 \times 10^{-8}\,\text{N}\), while the gravitational force is only \(3.624 \times 10^{-47}\,\text{N}\), which is significantly smaller in magnitude. This pattern holds for all other orbits as well.