Question

Before the development of quantum theory, Ernest Rutherford's experiments with gold atoms led him to propose the so-called Rutherford Model of atomic structure. The basic idea is that the nucleus of the atom is a very dense concentration of positive charge, and that negatively charged electrons orbit the nucleus in much the same manner as planets orbit a star. His experiments appeared to show that the average radius of an electron orbit around the gold nucleus must be about 10−1010−10 m. Stable gold has 79 protons and 118 neutrons in its nucleus.

What is the strength of the nucleus' electric field at the orbital radius of the electrons?
What is the kinetic energy of an electron in a circular orbit around the gold nucleus?

Answer 1

a) F = -1.82 10⁻¹⁵ N, b) K = 9.1 10⁻¹⁶ J

Step-by-step explanation:

a) To calculate the force between the nucleus and the electrons, let's use the Coulomb equation

F = k q Q / r²

as the nucleus occupies a very small volume compared to electrons, we can suppose it as punctual

let's calculate

F = 9 10⁹ (-1.6 10⁻¹⁹) (79 1.6 10⁻¹⁹) / (10⁻¹⁰)²

F = -1.82 10⁻¹⁵ N

b) they ask us for kinetic energy

let's use Newton's second law

F = m a

acceleration is centripetal

a = v² / r

we substitute

F = m v² / r

v = √ (F r / m)

v = √ (1.82 10⁻¹⁵ 10⁻¹⁰ / 9.1 10⁻³¹)

v = √ (0.2 10⁻¹⁶)

v = 0.447 10⁸ m / s

kinetic energy is

K = ½ m v²

K = ½ 9.1 10⁻³¹ (0.447 10⁸)²

K = 0.91 10⁻¹⁵ J

K = 9.1 10⁻¹⁶ J