Question

A hydrogen atom contains one proton and one electron. It may be assumed that the electron revolves in a circle of radius 0.53 angstrom(1 angstrom=10ˉ¹⁰m and is abbreviated as A∙) with the proton at the centre. The hydrogen atom is said to be in the ground state in this case. Find the magnitude of the electric force between the proton and the electron of a hydrogen atom in its ground state.

Answer 1

The magnitude of the electric force between the proton and electron in a hydrogen atom in the ground state is found using Coulomb's Law, resulting in a force of approximately 8.2 × 10^-8 Newtons.

Step-by-step explanation:

To calculate the magnitude of the electric force between the proton and the electron in a hydrogen atom in its ground state, we use Coulomb's Law. The formula for the electric force (F) is F = k * |q1 * q2| / r^2, where k is Coulomb's constant (8.99 × 10^9 N·m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the separation distance.

For a hydrogen atom, this distance is 0.53 angstroms, which is 0.53 × 10^-10 meters. The charge of a proton (q1) is the same as the charge of an electron (q2), which is approximately 1.602 × 10^-19 Coulombs. Plugging these values into Coulomb's Law, we get F = (8.99 × 10^9 N·m^2/C^2)*(1.602 × 10^-19 C)^2 / (0.53 × 10^-10 m)^2.

By calculating this value, we find that the electric force between the electron and the proton in the hydrogen atom in its ground state is approximately 8.2 × 10^-8 Newtons.