### Alpha Particle Scattering and Rutherford's Nuclear Model of an Atom The experimental set up used by Rutherford and his collaborators, Geiger and Marsden, is shown in the above figure.

Observations − A graph is plotted between the scattering angle θ and the number of α-particles N (θ), scattered at ∠θ for a very large number of α-particles. Conclusions:

• Most of the alpha particles pass straight through the gold foil.

• Only about 0.14% of incident α-particles scatter by more than 1.

• About one α-particle in every 8000 α-particles deflects by more than 90°.

Explanation

• In Rutherford’s model, the entire positive charge and most of the mass of the atom are concentrated in the nucleus with the electrons some distance away.

• The electrons would be moving in orbitals about the nucleus just as the planets do around the sun.

• The size of the nucleus comes out to be 10−15 m to 10−14 m. From kinetic theory, the size of an atom was known to be 10−10 m, about 10000 to 100,000 times larger than the size of the nucleus. Thus, most of an atom is empty space.

• The trajectory of an alpha particle can be computed employing Newton’s second law of motion andCoulomb’s law for electrostatic force of repulsion between the alpha particle and the positively charged nucleus.

The magnitude of this force is Where,

Ze − Charge of gold nucleus

2e − Charge on alpha particle

r − Distance between α-particle and the nucleus

Alpha particle trajectory

Trajectory traced by an α-particle depends on the impact parameter b of collision. The impact parameter is the perpendicular distance of the initial velocity vector of the α-particle from the centre of the nucleus. • For large impact parameters, force experienced by the alpha particle is weak because . Hence, the alpha particle will deviate through a much smaller angle.

When impact parameter is small, force experienced is large and hence, the alpha particle will scatter through a large angle.

• Electron orbits

Let

Fc − Centripetal force required to keep a revolving electron in orbit

Fe − Electrostatic force of attraction between the revolving electron and the nucleus

Then, for a dynamically stable orbit in a hydrogen atom,

Fc = Fe K.E. of electron in the orbit, From equation (i), Potential energy of electron in orbit, Negative sign indicates that revolving electron is bound to the positive nucleus.

∴ Total energy of electron in hydrogen atom 