Question

Assume that in a hydrogen atom the electron circles the nucleus in a circle of radius r = 5.83 * 10- ⁻¹¹ m with an angular momentum L= 1.10 *10 ⁻³⁴ s. Determine the value of the magnetic field at the nucleus due to the current of the electron's orbital motion. T

Answer 1

The magnetic field at the nucleus of a hydrogen atom due to the electron's orbital motion can be calculated using the formula B = (μ₀ * I) / (2 * π * r), where B is the magnetic field, μ₀ is the permeability of free space, I is the current, and r is the radius of the electron's orbit. Substituting the given values into the formula, the magnetic field is approximately 1.763 × 10⁻⁶ T (Tesla).

Step-by-step explanation:

According to Ampere's law, the magnetic field created by a current-carrying wire is directly proportional to the current and inversely proportional to the radius of the circular path. In this case, the electron in a hydrogen atom can be considered as a current-carrying wire due to its orbital motion around the nucleus. Therefore, the magnetic field at the nucleus due to the electron's orbital motion can be calculated using the formula:

B = (μ₀ * I) / (2 * π * r)

Where B is the magnetic field, μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A), I is the current, and r is the radius of the electron's orbit.

Substituting the given values into the formula:

B = (4π × 10⁻⁷ T·m/A * 1.6 × 10⁻¹⁹ C * 2.18 × 10⁶ m/s) / (2 * π * 5.83 × 10⁻¹¹ m)

Simplifying the expression:

B ≈ 1.763 × 10⁻⁶ T (Tesla)