AtomsAlpha-particle Scattering and Rutherford's Nuclear Model of Atom
In 1911, at Rutherford's suggestion, Hans Geiger and his 20-year-old student Ernest Marsden fired α-particles at a thin gold foil (NCERT Figs. 12.1 and 12.2). A radioactive source, bismuth-214 in a lead block, sends out α-particles of 5.5 MeV. Two lead bricks let only a narrow beam through to a gold foil just 2.1 × 10⁻⁷ m thick. A detector, a zinc sulphide screen with a microscope behind it, can be swung round the foil. Each α-particle that hits the screen makes a tiny flash, a scintillation, and the flashes were counted by eye, in the dark, for hours. Everything sits in a vacuum. Swing the detector round and count.
Most α-particles go straight through the foil as if nothing were there. Only about 0.14 % are turned through more than 1°, and only about 1 in 8000 through more than 90°. Notice the scale of the graph (NCERT Fig. 12.3): each grid line is ten times the one below it. The purple curve is Rutherford's prediction for an atom with a small, dense, positive nucleus, and the counts follow it closely.
How Small Is the Positive Charge?
To turn an α-particle right round, something must push it back very hard. In Thomson's atom the positive charge is spread through a ball 10⁻¹⁰ m across, and the push it can give is feeble. Rutherford's idea was that the positive charge, and most of the mass, are squeezed into a tiny region at the centre. The graph shows the potential energy of an α-particle at a distance r from the centre of a gold atom whose charge 79e fills a ball of radius R. A head-on α-particle can only be turned back if this hill rises above its 5.5 MeV of kinetic energy.
Squeeze the positive charge into a smaller ball until a head-on α-particle is turned back.
Shrink the ball of positive charge by
While the charge is spread out, the hill flattens off inside the ball at a height of only a few keV, far below the α-particle's 5.5 MeV. Squeeze the same charge into about 10⁻¹⁴ m and the hill climbs a thousand times higher: an α-particle that happens to come close is thrown back. This is why Rutherford is credited with the discovery of the nucleus.
Rutherford's nuclear model: all the positive charge and nearly all the mass of an atom are concentrated in a tiny nucleus, about 10⁻¹⁵ m to 10⁻¹⁴ m across, with the electrons going round it at a distance of about 10⁻¹⁰ m. The atom is 10 000 to 100 000 times larger than its nucleus, so it is mostly empty space.
Empty space explains why most α-particles sail straight through. And the electrons of the atom are far too light, about 7300 times lighter than an α-particle, to turn it noticeably.
Example 12.1: An Atom the Size of the Solar System
In Rutherford's model the nucleus, radius about 10⁻¹⁵ m, is like the Sun, and the electron goes round it in an orbit of radius about 10⁻¹⁰ m, like the Earth. Suppose the solar system had the same proportions as the atom. Would the Earth be closer to the Sun than it really is, or farther away? The radius of the Sun is 7 × 10⁸ m and the Earth's orbit has a radius of about 1.5 × 10¹¹ m.
The electron's orbit is (10⁻¹⁰ m)/(10⁻¹⁵ m) =
Scaled the same way, the Earth's orbit would have a radius of
Solution
The Earth would be about 470 times farther from the Sun than it really is. An atom holds a far larger fraction of empty space than even our solar system.
Alpha-particle Trajectory
The foil is so thin that an α-particle meets at most one nucleus on its way through, so it is enough to follow one α-particle past one nucleus. An α-particle is a helium nucleus, with charge 2e; a gold nucleus has charge Ze with Z = 79 and is about 50 times heavier, so it hardly moves. The only force is the Coulomb repulsion between them:
It acts along the line joining the two and changes in size and direction all the time. Which way the α-particle leaves depends on the impact parameter b: the perpendicular distance of its starting line of motion from the centre of the nucleus (NCERT Fig. 12.4). The particles of a beam all have nearly the same energy but a spread of impact parameters. Drag the α-particle up and down.
The smaller the impact parameter b, the more the α-particle is turned; head-on, with b = 0, it stops and comes straight back, θ = 180°. Far from the nucleus it is hardly deflected. The purple curve of θ against b falls steeply: only α-particles aimed almost exactly at a nucleus come back.
So the tiny fraction that bounce back tells us that very few α-particles meet a nucleus nearly head-on, and that means the mass and positive charge of the atom sit in a very small volume. Rutherford scattering is a powerful way to find an upper limit to the size of a nucleus.
Distance of Closest Approach
Fire an α-particle straight at a nucleus. As it slows down, its kinetic energy K turns into electric potential energy, until it stops for an instant at the distance of closest approach d and is pushed back. The total mechanical energy stays the same, so at d all of K has become potential energy.
Set the α-particle's energy and complete all three missions.
Kinetic energy of the α-particle K =
The dashed line is the α-particle's kinetic energy, and it can only climb the potential-energy hill as far as the point where the two meet. A faster α-particle climbs higher and gets closer; a nucleus with less charge has a lower hill.
Example 12.2: How Close to a Gold Nucleus?
In a Geiger–Marsden experiment, what is the distance of closest approach to a gold nucleus of a 7.7 MeV α-particle, the most energetic found in nature?
At the stopping point all the kinetic energy has become potential energy:
With K = 7.7 MeV = 1.2 × 10⁻¹² J, e = 1.6 × 10⁻¹⁹ C and 1/4πε₀ = 9.0 × 10⁹ N m² C⁻²:
For gold, Z = 79, so d =
A gold nucleus actually has a radius of about 6 fm, so the α-particle turns back
Solution
This tells us only that the gold nucleus is smaller than 30 fm. The real radius, about 6 fm, is much smaller still: the α-particle stops far outside the nucleus, because d is much larger than the sum of the radii of the two nuclei. To measure the size of a nucleus this way you need faster particles.
Electron Orbits
Rutherford's atom is a classical picture: a tiny, heavy, positive nucleus with electrons going round it in stable orbits. In a hydrogen atom the Coulomb attraction Fe on the electron supplies the centripetal force Fc it needs:
So the kinetic energy and the potential energy of the electron are
and the total energy is
The graph plots all three against the orbit radius, using e²/4πε₀ = 14.4 eV Å.
Change the orbit radius and complete both missions.
Orbit radius r =
At every radius the kinetic energy K is exactly minus the total energy E, and the potential energy U is twice E. The total energy is always negative: the electron is bound to the nucleus, and it takes energy −E to pull it free. If E were positive, the electron would not follow a closed orbit at all.
Example 12.3: Size of the Hydrogen Atom
It takes 13.6 eV to separate a hydrogen atom into a proton and an electron. Find the orbital radius and the speed of the electron. Take m = 9.1 × 10⁻³¹ kg.
The total energy of the electron is E = −13.6 eV = −13.6 × 1.6 × 10⁻¹⁹ J = −2.2 × 10⁻¹⁸ J. From Eq. (12.4), r =
Then Eq. (12.3) gives the speed of the electron, v =
Solution
The hydrogen atom is about 10⁻¹⁰ m across, and its electron moves at less than 1 % of the speed of light.
Speed Round: Rutherford's Atom
Small impact parameter, big deflection; d = 2Ze²/(4πε₀K); for an orbiting electron K = −E and U = 2E. Get five right to move on.