Alpha scattering and Rutherford’s model
Firing α particles at gold
By 1900 we knew that atoms contain electrons, and that an atom as a whole is neutral. So where is the positive charge, and how is it arranged?
In 1898 J. J. Thomson suggested a simple picture: a sphere of positive charge with the electrons embedded in it, like seeds in a watermelon. This is the plum pudding model.
In 1911 Rutherford proposed a test. His assistants, Geiger and Marsden, fired a narrow beam of α particles at a very thin gold foil and watched where they went.
Here is the set-up. A radioactive source, bismuth 214, gives α particles of 5.5 MeV. Lead blocks with slits shape them into a narrow beam. The beam hits a gold foil only 2.1 × 10⁻⁷ metre thick.
Around the foil, a zinc sulphide screen and a microscope can be turned to any angle θ. Each α particle that strikes the screen makes a tiny flash, and the flashes are counted, angle by angle.
In a Thomson atom the positive charge is spread out, so the force on an α particle is weak everywhere. Every α particle should pass straight through, bent by a fraction of a degree at most.
Most did go straight through. Only about 0.14 percent were scattered by more than 1 degree. But about 1 in 8000 was turned back by more than 90 degrees!
Rutherford said it was as if you fired a heavy shell at a sheet of tissue paper and it came back and hit you. Something tiny, massive and strongly charged must be inside each atom.
His conclusion: all the positive charge and almost all the mass of an atom sit in a tiny central nucleus. The nucleus is 10⁻¹⁵ to 10⁻¹⁴ metre across, while the atom is about 10⁻¹⁰ metre. The atom is mostly empty space.
Impact parameter and closest approach
Zoom in on one gold nucleus. It has charge plus Z e, with Z equal to 79. The α particle has charge plus 2e. They repel with the Coulomb force F equals 1 over 4πε₀, times 2e times Ze, over r².
The gold nucleus is about 50 times heavier than the α particle, so we can treat it as fixed. The electrons are far too light to deflect an α particle much, so we ignore them.
The path of an α particle depends on its impact parameter, b. This is the perpendicular distance of its starting line of motion from the centre of the nucleus.
Watch several α particles with different b. With a large b, the force stays weak and the path bends only slightly. As b gets smaller, the deflection grows.
With b equal to zero, the α particle heads straight at the nucleus. It slows down, stops, and comes straight back: a scattering angle of 180 degrees.
Since the nucleus is so tiny, very few α particles have a small impact parameter. That is why large deflections are so rare.
In a head-on collision, how close does the α particle get? At the turning point all its kinetic energy K has become electric potential energy. So K equals 1 over 4πε₀, times 2e times Ze, over d.
This gives the distance of closest approach: d equals 2Ze² over 4πε₀ K.
Take a 7.7 MeV α particle. K equals 7.7 × 1.6 × 10⁻¹³, which is 1.2 × 10⁻¹² joule.
Then d equals 2 × 9 × 10⁹ × 1.6 × 10⁻¹⁹ squared × 79, divided by 1.2 × 10⁻¹². That is 3.0 × 10⁻¹⁴ metre, or 30 femtometre. The nucleus must be smaller than this.
Electrons in orbit
If the nucleus is tiny, where are the electrons? Rutherford pictured them moving in orbits round the nucleus, like planets round the Sun. This is the planetary model of the atom.
Take hydrogen: one proton, one electron in a circle of radius r. The electrostatic attraction provides the centripetal force: m v² over r equals e² over 4πε₀ r².
From this, the kinetic energy, ½ m v², equals e² over 8πε₀ r.
The potential energy of the electron and proton is U equals minus e² over 4πε₀ r. It is twice as big as the kinetic energy, and negative.
Adding them, the total energy is E equals minus e² over 8πε₀ r. It is negative: the electron is bound to the nucleus. To set it free, we must supply energy.
Experiments show that it takes 13.6 eV to separate a hydrogen atom into a proton and an electron. So E equals minus 13.6 eV, which is minus 2.2 × 10⁻¹⁸ joule.
Then r equals e² over 8πε₀ times 13.6 eV. With 1 over 4πε₀ equal to 9 × 10⁹, r equals 9 × 10⁹ × 1.6 × 10⁻¹⁹ squared, over 2 × 2.2 × 10⁻¹⁸. That is 5.3 × 10⁻¹¹ metre.
And from m v² equals e² over 4πε₀ r, the speed comes out at about 2.2 × 10⁶ metres per second.
So the hydrogen atom is about 10⁻¹⁰ metre across. That is roughly 10⁵ times the size of its nucleus, as Rutherford found.
Try it: scatter α particles
- Drag up or down in the picture to change the impact parameter b. Smaller b, bigger scattering angle θ.
- Set b = 0: the α particle stops at the blue circle, distance d = 2kZe²/K, and comes straight back.
- Raise the energy K: d shrinks, because d ∝ 1/K. Switch to aluminium (Z = 13): d shrinks too, because d ∝ Z.
- Choose “Beam + detector” and fire a million α particles. Only about 1 in 10 000 bounces back, and the counts follow 1/sin⁴(θ/2).
How close?
An α particle of kinetic energy 5.5 MeV is fired head-on at a gold nucleus (Z = 79). What is its distance of closest approach? (Take 1/4πε₀ = 9 × 10⁹ N m² C⁻², e = 1.6 × 10⁻¹⁹ C.)
- 2.1 × 10⁻¹⁴ m
- 4.1 × 10⁻¹⁴ m
- 8.3 × 10⁻¹⁴ m
- 5.2 × 10⁻¹⁶ m
Answer
B. K = 5.5 × 1.6 × 10⁻¹³ J = 8.8 × 10⁻¹³ J. At closest approach all of K is potential energy: K = (1/4πε₀)(2e)(Ze)/d, so d = 2 × 9 × 10⁹ × (1.6 × 10⁻¹⁹)² × 79 / 8.8 × 10⁻¹³ = 4.1 × 10⁻¹⁴ m (about 41 fm). Taking the α charge as e instead of 2e gives 2.1 × 10⁻¹⁴ m; leaving out Z gives 5.2 × 10⁻¹⁶ m.
Energy of the orbiting electron
In Rutherford’s model of hydrogen the electron’s total energy is −13.6 eV. What is its potential energy?
- −13.6 eV
- +13.6 eV
- −6.8 eV
- −27.2 eV
Answer
D. For a circular orbit, K = e²/8πε₀r and U = −e²/4πε₀r, so U = −2K and E = K + U = −K. With E = −13.6 eV, K = +13.6 eV and U = −27.2 eV. +13.6 eV is the kinetic energy, not the potential energy.
Summary
- Coulomb force on the α particle:
F = (1/4πε₀) (2e)(Ze)/r² - Distance of closest approach:
d = 2Ze² / (4πε₀ K) - Electron in a hydrogen orbit:
mv²/r = e²/(4πε₀ r²) - Total energy:
E = K + U = −e²/(8πε₀ r)
- In the Geiger–Marsden experiment most α particles went straight through a gold foil; about 1 in 8000 was scattered by more than 90°.
- Rutherford concluded that the positive charge and nearly all the mass are in a tiny nucleus (10⁻¹⁵–10⁻¹⁴ m), while the atom is about 10⁻¹⁰ m: mostly empty space.
- The smaller the impact parameter b, the larger the scattering angle; b = 0 gives a head-on rebound (θ = 180°).
- A 7.7 MeV α particle gets within about 30 fm of a gold nucleus, so the nucleus is smaller than that.
- In the planetary model the electron’s energy is negative (bound): for hydrogen E = −13.6 eV at r = 5.3 × 10⁻¹¹ m.