Discovery of the Nucleus Nuclear Physics Lesson 1.

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Presentation transcript:

Discovery of the Nucleus Nuclear Physics Lesson 1

Homework Research Rutherford’s experiment using books and the internet. Please reference properly. Use the level ladder provided to help focus your answer  Let’s all aim high and go for an A*!

Learning Objectives Describe the experimental setup used by Rutherford to probe the structure of the atom. Describe the experimental setup used by Rutherford to probe the structure of the atom. Explain how the results of this experiment provide evidence for the nuclear model of an atom. Explain how the results of this experiment provide evidence for the nuclear model of an atom. Explain how an estimate for the nuclear radius can be obtained from Rutherford’s Experiment. Explain how an estimate for the nuclear radius can be obtained from Rutherford’s Experiment.

Video In Search of Giants (3 of 15) The Discovery of the Atomic Nucleus In Search of Giants (3 of 15) The Discovery of the Atomic Nucleus

Rutherford’s Alpha Scattering At P, the point of closest approach, all of the initial kinetic energy of the alpha is converted to electrostatic potential energy.

Electrostatic Potential Energy The equation for Electrostatic Potential Energy is given by this equation:- Where:- E P is the Electrostatic Potential Energy r c is the distance of closest approach ε 0 is the permittivity of free space (constant - see data booklet) Q 1 Q 2 are the charges of the two particles involved.

Little bit of Maths… Solving to find r C, what do we get? Rearranging… Remember: Q 1 =2e = 2 × 1.60 × C (α is He nucleus) Q 2 =79e = 79 × 1.60 × C (79 protons in Gold nucleus) E P =768 MeV = 768 × 10 6 × 1.60 × J (K.E. of α particles fired at the foil.) ε 0 = 8.85 × Fm -1 (from the data booklet)  r c = 2.96 × m (a bit large)

A couple of points… The nucleus is treated as a point charge. At this level it is not. The alpha particles are stopped some distance away from the nucleus. It takes higher energy alpha particles to penetrate the nucleus. The values for the nuclear radius given by other particles such as protons, neutrons and electrons are slightly different. …so really it is only an upper limit on the nuclear radius.

Method 2 About 1 in 10,000 particles are deflected by more than 90º. About 1 in 10,000 particles are deflected by more than 90º. For a thin foil (so that only one scattering) with n layers of atoms, the probability of being deflected by a given atom about 1 in 10,000n. For a thin foil (so that only one scattering) with n layers of atoms, the probability of being deflected by a given atom about 1 in 10,000n. This probability depends on effective cross section of nucleus to the atom:- This probability depends on effective cross section of nucleus to the atom:- Typically n=10,000 so d = D/10,000 Typically n=10,000 so d = D/10,000

Questions A) For a metal foil which has layers of atoms, explain why the probability of an α particle being deflected by a given atom is therefore about 1 in 10,000n. (assume 1 in deflected by more than 90º) A) For a metal foil which has layers of atoms, explain why the probability of an α particle being deflected by a given atom is therefore about 1 in 10,000n. (assume 1 in deflected by more than 90º) B) Assuming this probability is equal to the ratio of cross sectional area of the nucleus to that of the atom, estimate the diameter of a nucleus for atoms of diameter 0.5 nm in a metal foil of thickness 10 μm. B) Assuming this probability is equal to the ratio of cross sectional area of the nucleus to that of the atom, estimate the diameter of a nucleus for atoms of diameter 0.5 nm in a metal foil of thickness 10 μm.