Building the atom

                                       Richard belongs to Jayne
                                       And Jayne belongs to yesterday.
                                       How can I go on
                                       When every alpha particle hides a neon nucleus?

                                       Billy Bragg, Richard

Rutherford's planetary model of the atom

Ernst Rutherford was interested in exploring the basic structure of the atom. He performed a special experiment known as scattering. In a scattering experiment, one takes something with an unknown shape or structure. That is the target. Then one throws something at the target. That is the scatterer. Think of taking a target, which might be a plate. You are asked to determine the size of the plate without feeling it or measuring it directly. One way to do that is to use a scattering experiment. You hang the plate in a hallway. You know the area of the hallway; you want to know the area of the plate. The plate is your target and at your target you throw tennis balls. You throw them randomly. Some will pass the plate and roll on. Others will hit the plate and bounce back. How does this help you figure out the size of the plate?

Easy! You know that the ratio of the area of the plate to the area of the hall must be equal to the number of balls scattered back at you. That is

Aplate / Ahall = # balls scattered / # balls thrown

Since we know Ahall, the number of balls thrown, and the number of balls scattered back, we can figure out Aplate - the area of the plate. So with that sort of scattering experiment, you can learn the size of the plate, its area, without ever measuring it, or even seeing it.

We can take this a bit further. What if we replace the plate with a ball. Now the target is a sphere. The tennis balls we throw won't just scatter back or pass by. They might also be scattered at different angles, this way and that, depending on where they hit the ball. If we predict all that - all the possible ways the balls can be scattered by the spherical target - we can build a model for the scattering pattern that we'd expect for a ball of a certain size. Given an observed scattering pattern, we would fit that pattern to our model to determine the size of the ball. So you can see that scattering is a very powerful method for learning the basic size and shape of objects that you can't see or directly measure.

For Rutherford, the target was a thin gold foil made of a layer or sheet of gold atoms. Since he knew the mass of the gold foil, and the molar mass of gold, he could calculate the number of gold atoms that the foil was made of. For Rutherford, the scatterer was an α particle - which we now know to be the nucleus of a helium atom, made of two neutrons and two protons. He would throw the α particles at the foil and look to see if the particles were scattered or if they passed through the foil unhindered.

What Rutherford found was remarkable. When he cast the α particles at the foil, most of the particles passed through the foil unhindered! How could that be? To our senses, the foil seems to be a continuous surface. Rutherford's result implied that the foil was, at the microscopic level, quite porous. In fact, it was mostly empty space! Interpreting his result, he developed the basic model of the atom that we still use today - the planetary model.

Now there were other possibilities. The atom might have been constructed like the positive and negative charges of a salt crystal. But Rutherford didn't think that was the way it worked. He proposed there is a nucleus where most of the mass of the atom is concentrated. The nucleus is like the Sun. Around that nucleus orbit the electrons that are much lighter. Those are the planets. Rutherford didn't know about the strong force that acts between nucleons and holds them close together in the nucleus of the atom, on the order of 1 Fermi (= 1 x 10-15m).

Rutherford knew that atoms (his "Solar System") were roughly of diameter

datom = 3 x 10-10 m

He wanted to known the size of the nucleus (his "Sun"). Applying our "balls in the hall" model to Rutherford's data, the ratio to consider is

Anucleus / Afoil = # α particles scattered / # α particles thrown

Since he knew Afoil and the number of α particles thrown and scattered back, he could compute Anucleus. He found that Anucleus = 10-28 m2 so that the diameter of the nucleus was

dnucleus = Anucleus1/2 < 1 x 10-14 m

That's shocking! The difference between the diameter of the nucleus and the diameter of the atom is nearly a factor of 100,000. While a billion Earths can fit in a volume the size of our Sun, a million billion nuclei can fit inside the volume of a single atom.

Isotopes

The tour de force experiment of Rutherford led to the general acceptance of the planetary model of the atom. The protons are concentrated in a small nucleus of heavy mass and positive charge. About that nucleus orbit the relatively light and negatively charged electrons.

Now that we have the subatomic building blocks of the atom, the protons and electrons, we can construct atoms and compare the mass of what we build to what is observed in Nature. Let's start with the fifth element in the Periodic Table which is boron. If we go to a mine and dig up some boron, refine it, then weigh a mole of it, we will find that the molar mass is MB = 10.81 g/mol. We can compute the mass of a mole of each of the subatomic particles. Here's what we find. 

                 Particle mass and molar mass

        particle name          mass             molar mass
          electron      0.000911 X 10-27 kg      0.000549 g
          proton        1.672623 X 10-27 kg      1.007276 g 
          neutron       1.674929 X 10-27 kg      1.008665 g

So let's build a boron atom. We know that boron has an atomic number of five. The atomic number tells us the number of protons in the nucleus

atomic number = # of protons

and defines that atom's chemical identity. The number of protons in the nucleus - the atomic number - determines the chemical identity of an atom. For the neutral atom, the number of negatively charged electrons will equal the number of positively charged protons. Let's add up the mass of five protons and five electrons

5 mp + 5 me = 5.039 g/mol < mB

The electron mass is only 1/1837 of a proton mass. It can almost be neglected relative to the mass of the proton. In any case, our result is far from the observed molar mass of boron, which is more than twice our mass of 5 protons and 5 electrons. What's missing?

Of course, it's the neutrons. It took many year to discover the neutron. Thomson discovered the electron in 1897 (after it was predicted to exist, and even named, by Stoney in 1891). Rutherford discovered the proton in 1919. But it was not until 1932 that James Chadwick discovered the neutron. Particles that are electrically neutral are difficult to work with. They are difficult to trap, to direct, and to detect.

Neutrons are nucleons, just like protons. They can form stable arrangements with protons, but only in certain magic numbers. Some grouping are unstable and decay rapidly, making the nuclei radioactive. Other groupings are stable and decay very slowly. For an atom of boron, we might guess from its atomic molar mass, which is approximately 11 g/mol, that there are six neutrons in the nucleus of each boron atom. We can call that 11B where the 11 indicates that there are 11 total nucleons in the nucleus. We call such an atom ``boron-11.'' (Open the image in its own window to view the detail, a distribution of all isotopes as a function of atomic number!)

Physicists often use the notation 5B where the subscripted prefix indicates the number of protons. To a chemist, that notation is redundant. Saying it is boron with a capital B already says that there are five protons in the nucleus. That's what makes it boron!

For a mole of boron-11 we find

5 mp + 5 me + 6 mn = 11.091 g/mol > mB

Bloody! Now the mass is too large. So what is wrong with this idea? It turns out that we can build a variety of relatively stable boron nuclei, by combining five protons with a varying number of neutrons. So a nucleus with five protons and one neutron is not stable. However, a nucleus of five protons and five neutrons forming boron-10 is relatively stable

5 mp + 5 me + 5 mn = 10.082 g/mol < m<B

We call these different variations the isotopes of boron. They are identified by

isotope number = # of protons + # of neutrons

For boron it is also possible to make atoms of 8B, 9B, 12B, and 13B. However, they can exist only for a very short time before their nuclei decay. An atom of boron-11 is just as much an atom of boron as an atom of boron-10. It is the number of protons that determine the elemental chemical nature of the atom. The number of neutrons changes the mass but does not change the chemical nature of the atom in a significant way.

When we go out into the field and dig up a macroscopic sample of boron, we have a mixture of isotopes of boron. If we examine that sample, we will find that 19.78% of the sample is composed of boron-10 and the remaining 80.22% of the sample is composed of boron-11. The molar mass of a sample of boron with a natural abundance of isotopes is then

0.1978 x 10.082 g/mol + 0.8022 x 11.091 g/mol = 10.89 g/mol > mB

Surely you're joking! The mass is still not the correct molar mass of boron. What can be missing from our calculation?

Missing mass

The mass of a mole of atoms of boron in the correct natural abundance is 10.81 g/mol but the mass of a collection of protons, neutrons, and electrons in the same abundance has a mass of 10.89 g/mol. What gives?

We can find the same problem in our method for computing the mass of a single atom of a single isotope. If we consider a single atom of 4He we find that it has a mass of 6.64648 x 10-27 kg. However, when we add the masses of the subatomic particles we find

2 mp + 2 me + 2 mn = 6.69693 x 10-27 kg

As in the case of boron, the sum of the masses of the subatomic particles is greater than the mass of the atom that they compose.

The answer to this problem is found in a remarkable concept that was first understood by Albert Einstein. His understanding and the subsequent applications of the idea have changed the face of war and life on Earth. It is that

2 p + 2 n + 2 e → 4He + energy

How much energy? Einstein proposed the equivalence

E = m c2

where m is the mass of a resting particle and c = 2.99 x 108 m/s is the speed of light.

Let's take the difference in mass between the atom of 4He and the six particles that it is composed of. That difference is m = 0.05044 x 10-27 kg and the corresponding energy is E = m c2 = 4.5 x 10-12 J = 2.7 x 1012 J/mol.

Yow! Chemical reactions that yield a significant amount of energy, such as combustion reactions, produce on the order of 500,000 Joules of energy per mole. But that is a mass equivalent of m = 500,000 J/mol / c2 = 5.5 x 10-9 g/mol which is very, very small.

This is what Einstein realized. Small changes in the arrangements of nucleons can lead to the release of vast quantities of energy that far exceed what can be produced by chemical means. Nuclear bombs can far exceed the power of conventional chemical explosives. That knowledge led to the production of the atomic bombs that were exploded over the cities of Hiroshima, Japan on August 6, 1945 and over Nagasaki, Japan on August 9, 1945.