1. Nuclear Radius
Probing Matter

2. Homework
Why are spent fuel rods more radioactive after removal from the reactor?
What are the different types of radioactive waste?
How is each type stored/disposed?

3. Filling the Gaps
γ emitters can indeed be used as medical tracers as we thought.
Cell membranes can also be destroyed as well as affecting DNA.

4. Learning Objectives
Explain how an estimate for the nuclear radius can be obtained from Rutherford’s Experiment.

5. 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.

6. Electrostatic Potential Energy
The equation for Electrostatic Potential Energy is given by this equation:-
Where:-
EP is the Electrostatic Potential Energy
rc is the distance of closest approach
ε0 is the permittivity of free space (constant - see data booklet)
Q1Q2 are the charges of the two particles involved.

7. Little bit of Maths… Solving to find rC, what do we get? Rearranging…
Remember:
Q1=2e = 2 × 1.60 × C (α is He nucleus)
Q2=79e = 79 × 1.60 × C (79 protons in Gold nucleus)
EP=7.68 MeV = 768 × 106 × 1.60 × J (K.E. of α particles fired at the foil.)
ε0 = 8.85 × Fm-1 (from the data booklet)
rc= 2.96 × m (a bit large)

8. 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.

9. Method 2 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 is about 1 in 10,000n.
This probability depends on effective cross section of nucleus to the atom:-
Typically n=10,000 so d = D/10,000

10. 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)
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.